Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.Solubility&Preci...
Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.                ...
Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.                ...
Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.                ...
Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.                ...
Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.                ...
Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.                ...
Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.                ...
Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.                ...
Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.                ...
Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.                ...
Outcome 1-01 Explain observed examples of solubility and precipitation at the molecular and symbolic levels. Precipitation...
Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.Precipitation • ...
Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.                ...
Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.Precipitation   ...
Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.Precipitation
Outcome 1-02Perform a lab to develop a set of solubility rules.Outcome 1-03Use a table of solubility rules to predict the ...
Outcome 1-02Perform a lab to develop a set of solubility rules.Outcome 1-03Use a table of solubility rules to predict the ...
Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstr...
Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstr...
Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstr...
Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstr...
Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstr...
Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstr...
Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstr...
Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstr...
Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstr...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S Outcome 1-04                                                         Unit 3 – Acid/BaseEquilibria neutraliza...
Chemistry 40S                                                            Unit 3 – Acid/BaseEquilibria1-08 Outcome Define o...
Chemistry 40S                                                            Unit 3 – Acid/BaseEquilibria1-08 Outcome Define o...
Chemistry 40S                                                            Unit 3 – Acid/BaseEquilibria1-08 Outcome Define o...
Chemistry 40S                                                            Unit 3 – Acid/BaseEquilibria1-08 Outcome Define o...
Chemistry 40S                                                            Unit 3 – Acid/BaseEquilibria1-08 Outcome Define o...
Chemistry 40S                                                            Unit 3 – Acid/BaseEquilibria1-08 Outcome Define o...
Chemistry 40S                                                            Unit 3 – Acid/BaseEquilibria1-08 Outcome Define o...
Chemistry 40S                                                            Unit 3 – Acid/BaseEquilibria1-08 Outcome Define o...
Chemistry 40S                                                            Unit 3 – Acid/BaseEquilibria1-08 Outcome Define o...
Chemistry 40S                                                            Unit 3 – Acid/BaseEquilibria1-08 Outcome Define o...
Chemistry 40S                                                            Unit 3 – Acid/BaseEquilibria1-08 Outcome Define o...
Chemistry 40S                                                            Unit 3 – Acid/BaseEquilibria1-08 Outcome Define o...
Chemistry 40S                                                            Unit 3 – Acid/BaseEquilibria1-08 Outcome Define o...
Chemistry 40S                                                   Unit 3 – Acid/BaseEquilibria1-09 Outcome Determine the oxi...
Chemistry 40S                                                   Unit 3 – Acid/BaseEquilibria1-09 Outcome Determine the oxi...
Chemistry 40S                                                   Unit 3 – Acid/BaseEquilibria1-09 Outcome Determine the oxi...
Chemistry 40S                                                     Unit 3 – Acid/BaseEquilibria1-10 Outcome Identify and de...
Chemistry 40S                                                     Unit 3 – Acid/BaseEquilibria1-10 Outcome Identify and de...
Chemistry 40S                                                     Unit 3 – Acid/BaseEquilibria1-10 Outcome Identify and de...
Chemistry 40S                                                     Unit 3 – Acid/BaseEquilibria1-10 Outcome Identify and de...
Chemistry 40S                                                 Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-r...
Chemistry 40S                                                 Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-r...
Chemistry 40S                                                 Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-r...
Chemistry 40S                                                 Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-r...
Chemistry 40S                                                 Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-r...
Chemistry 40S                                                  Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-...
Chemistry 40S                                                 Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-r...
Chemistry 40S                                                 Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-r...
Chemistry 40S                                                 Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-r...
Chemistry 40S                                                 Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-r...
Chem 40S Uunit 1 Notes
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Chem 40S Uunit 1 Notes

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  • In this example, 25.0 mL of an HCl solution is titrated with a 0.100 mol/L solution of NaOH. The equation for this reaction is as follows: HCl( aq ) + NaOH( aq )  NaCl( aq ) + H 2 O( l ) At equivalence, the hydronium ion and hydroxide ion concentrations are equal; as are the sodium and chloride ion concentrations. Since sodium ions are a weaker acid than water and chloride ions are a weaker base than water, the solution is neutral. That is, at 25°C the pH will be 7.0. The expected titration curve is shown above. The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L NaOH, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [HCl] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L HCl The pink on the graph indicates the colour change of phenolphthalein with pH change. Despite the colour change at a pH higher than equivalence, the difference is not significant, because the curve is so steep at the equivalence point.
  • This is very similar to the previous example except, a 25.0 mL sample of strong base, NaOH, is added to the flask and 0.100 mol/L HCl is slowly added from a burette. The curve for this reaction is shown above. Once again, the equivalence volume is at 25.0 mL so the concentration of the NaOH and HCl must be equal at 0.100 mol/L. For both the strong acid with strong base and strong base with strong acid titrations, phenolphtalein (pH range = 8.0 - 9.8), phenol red (pH range = 6.8 - 8.4) or bromothymol blue (pH range = 6.0 - 7.6) would be suitable indicators.
  • In this example, 25.0 mL of an acetic acid solution is titrated with a 0.100 mol/L solution of NaOH. The equation for this reaction is as follows: HC 2 H 3 O 2 ( aq ) + NaOH( aq ) NaC 2 H 3 O 2 ( aq ) + H 2 O( l ) At equivalence, the [HC 2 H 3 O 2 ] = [OH¯]; [Na + ] = [C 2 H 3 O 2 ¯]; and since sodium ions are a weaker acid than water while C 2 H 3 O 2 ¯ ions are a stronger base than water, the solution will be basic. That is, at 25°C the pH will be greater than 7.0. The expected titration curve is shown above. The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L NaOH, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [HC 2 H 3 O 2 ] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L HC 2 H 3 O 2 The pink on the graph indicates the colour change of phenolphthalein with pH change. Phenolphthalein would be a suitable indicator for this titration. At the half-equivalence volume ( 0.5 V eq ) , half the acid is neutralized, so [HC 2 H 3 O 2 ] = [C 2 H 3 O 2 ¯]. If this is the case, according to the equilibrium law for acetic acid K a = [H 3 O + ]; therefore, pH = pK a (pK a = -logK a ) at half-equivalence volume. A titration curve can be used to find the kA of a weak acid.
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • This is very similar to the weak acid/strong base titration, except that the acid is added to the base. In this example, 25.0 mL of an aqueous ammonia solution is titrated with a 0.100 mol/L HCl solution. The equation for the reaction is: HC l ( aq ) + NH 3 ( aq ) NH 4 Cl( aq ) The titration curve appears below: The equivalence point is found in the centre of the steepest portion of the curve, or where the curve appears to change directions. The equivalence volume is 25.0 mL of 0.100 mol/L HCl, since the acid:base stoichiometry is 1:1, moles H + = moles OH¯ = (0.100 mol/L)(0.025 L) = 0.0025 moles [NH 3 (aq)] = 0.0025 moles ÷ 0.025 L = 0.100 mol/L NH 3 (aq) At the half-equivalence volume, [NH 4 + ]=[NH 3 ], and [H 3 O + ]=K a for the conjugate acid NH 4 + . If K a = K W /K b , then at half equivalence volume, pH = pK W - pK b Phenolphthalein would not be a suitable indicator for this reaction, since the equivalence point is at a pH of about 5.3. Suitable indicators would be methyl red (pH range = 4.8 - 6.0) or bromocresol green (pH range = 3.6 - 5.2).
  • Chem 40S Uunit 1 Notes

    1. 1. Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.Solubility&Precipitation Compare and Contrast Solubility vs Precipitation
    2. 2. Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels. • When a solute dissolves in a solvent,The the individual particles of the soluteSolution separate from the other particles ofProcess the solute and move between the spaces of the solvent particles. The solvent particles collide with theA Review from solute particles and forces ofChemistry 30S attraction between solute and solvent particles "hold" the solute particles in the spaces. • For a solute to be dissolved in a solvent, the attractive forces between the solute and solvent molecules must be greater than the forces of attraction between the solute molecules.
    3. 3. Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels. 1. The solvent particles must move apart toThe make room for solute particles. This process requires energy to overcomeSolution forces of attraction between solventProcess particles. The first step in the dissolving process is endothermic.A 3 Step 2. The solute particles must separate form the other solute particles. This processProcess also requires energy to overcome the forces of attraction between the solute particles. The second step in the dissolving process is endothermic. 3. When the solute particles move between the solvent particles the forces of attraction between solute and solvent take hold and the particles "snap" back and move closer. This process releases energy. The final step in the dissolving process is exothermic.
    4. 4. Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels. • The total heat change in theEnergy dissolving process is the sum of theChanges three heat changes. If the sum ofDuring the heat absorbed in the first twoDissolving steps of the dissolving process is greater than the heat released in the last step, the dissolving of that substance will be endothermic. If the dissolving process for a substance is endothermic, the container will feel cooler as the substance dissolves. • In some cases, the exothermic process is larger than the sum of the two endothermic processes. In these cases, as the solid dissolves large amounts of heat are evolved.
    5. 5. Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels. • When we talk about the mixing of twoSolubility or more substances together in solution we must consider solubility. Simply defined, it is a measure of how much solute will dissolve into the solvent. Not all substances will dissolve in all solvents. • Example 1: NaCl(s) dissolved in water Molecular level: NaCl (solid) + H2O (liquid) NaCl(aq)
    6. 6. Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels. Student TASKSolubility • Create a series of drawings that outline what happens in the solution process involving NaCl dissolving in water.
    7. 7. Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels. • Example 1: NaCl(s) dissolved in waterSolubility Symbolic level: NaCl(s) → NaCl(aq)
    8. 8. Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels. • A precipitation reaction is a commonPrecipitation type of chemical reaction in solution chemistry. Basically, two or more solutions are combined resulting in a reaction that produces an insoluble product, a precipitate. Typically, these types of reactions involve ionic compounds in aqueous solution. The precipitation reaction is said to occur because of the strong attractive forces certain ions have for each other. These ions combine and fall out of solution in the formation of a solid. • The ability to predict precipitate formation is based upon solubility. When the product of a reaction is insoluble, it precipitates.
    9. 9. Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels. • Example 1: NaCl(s) and AgNO3 combined togetherPrecipitation Molecular level: •
    10. 10. Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels. • Example 1: NaCl(s) and AgNO3 combined togetherPrecipitation Molecular level: •
    11. 11. Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels. Student TASKPrecipitation • Create a series of 3 diagrams that outline what happens when NaCl and AgNO3 are combined in a precipitation reaction. Diagram 1 – draw Na+ and Cl- ions circulating amidst the water molecules Diagram 2 – AgNO3 is drawn with the Ag+ and NO3- ions floating around the water molecules Diagram 3 – Shows the mixing of the two solutions. The Ag+ and Cl- combine to form a precipitate.
    12. 12. Outcome 1-01 Explain observed examples of solubility and precipitation at the molecular and symbolic levels. Precipitation• Symbolic level: Molecular equation: NaCl (aq) + AgNO3 (aq) → AgCl (s) + NaNO3 (aq) Ionic equation: Na+ (aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq) → AgCl (s) + Na+ (aq) + NO3- (aq) Net ionic equation: Ag+ (aq) + Cl- (aq) → AgCl (s)
    13. 13. Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.Precipitation • When certain solutions of electrolytes are mixed, the cation of one solute may combine with the anion of the otherHow to write solute to form an insoluble compound.chemical This leads to the formation of aequations for precipitate.precipitation • A precipitate is an insoluble solid thatreactions settles from a solution. For instance, when a solution of Pb(NO3)2 is mixed with a solution of HCl, a precipitate of insoluble PbCl2(s) forms. This occurs because combining the solutions results in a new solution containing more PbCl2(aq) than is soluble. A condition referred to as supersaturation. In this case, solid PbCl2 must settle out of the solution.
    14. 14. Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels. • In the molecular equation thePrecipitation formulas of the compounds are written as their usual chemical formulas. Pb(NO3)2 (aq) + 2HCl (aq) PbCl2 (s) + 2HNO3 (aq) • The precipitation reaction is more accurately represented by the ionic equation. The ionic equation shows the compounds as being dissociated in solution. Pb2+ (aq) + 2NO3- (aq) + 2H+ (aq) + 2Cl– (aq) PbCl2 (s) + 2H+ (aq) + 2NO3- (aq)
    15. 15. Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.Precipitation Pb2+ (aq) + 2NO3- (aq) + 2H+ (aq) + 2Cl– (aq) PbCl2 (s) + 2H+ (aq) + 2NO3- (aq) • If you examine the ionic equation, you will notice that the H+ ions and NO3-ions are not involved in the formation of the precipitate. We refer to the ions which are not involved in the reaction as spectator ions. Identical species on both sides of the equation can be omitted from the equation. The net ionic equation shows only the species that actually undergo a chemical change. Pb2+ (aq) + 2Cl– (aq) PbCl2 (s) • In order to predict whether a precipitate will form when two solutions are mixed, you need to know the solubility rules for ionic compounds.
    16. 16. Outcome 1-01Explain observed examples of solubility and precipitation at the molecular and symboliclevels.Precipitation
    17. 17. Outcome 1-02Perform a lab to develop a set of solubility rules.Outcome 1-03Use a table of solubility rules to predict the formation of a precipitate.SolubilityRules
    18. 18. Outcome 1-02Perform a lab to develop a set of solubility rules.Outcome 1-03Use a table of solubility rules to predict the formation of a precipitate.ProcessNotes forWriting NetIonicEquationsRules
    19. 19. Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong base and astrong acid.Acid/Base Definitions of acids and basesReview • Arrhenius acid: generates [H+] in solution base: generates [OH-] in solution • normal Arrhenius equation: acid + base salt + water • example: HCl + NaOH NaCl + H2O
    20. 20. Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong base and astrong acid.Acid/Base • A strong acid is defined as anReview acid that completely dissociates into ions. This means that if there are 100 molecules of HCl dissolved in water, 100 ions of H+ and 100 ions of Cl- are produced. • It should be stressed that there are only 6 strong acids. These are: – hydrochloric acid (HCl), – hydrobromic acid (HBr), – hydroiodic acid (HI), – sulfuric acid (H2SO4), – nitric acid (HNO3), – perchloric acid (HClO4).
    21. 21. Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand astrong acid.Acid/Base • A strong base is defined as aReview base that completely dissociates into ions. This means that if there are 100 formula units of NaOH dissolved in water, 100 ions of Na+ and 100 ions of OH- are produced. • Strong bases include any ionic compound that contains the hydroxide (OH-) ion. • When combined with the hydroxide ion, elements found in groups 1 and 2 form strong bases.
    22. 22. Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand astrong acid.Naming Acids • To name a binary acid follow the steps given: 1. the prefix "hydro" is used 2. the root of the anion is used 3. the suffix "ic" is used 4. the word "acid" is used as the second word in the name • Example of naming a binary acid: HCl Step 1: hydro- Step 2: -chloride Step 3: -chloric Step 4: hydrochloric acid
    23. 23. Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand astrong acid.Naming Acids • Naming polyatomic acids follow a different set of rules. Many of the oxygen-rich polyatomic negative ions form acids that are named by replacing the suffix -ate with -ic and the suffix -ite with -ous. • To name oxyacids (acids containing the element oxygen) you recognize them by the general formula HaXbOc where X represents an element other than hydrogen or oxygen. If enough H+ ions are added to a (root)ate polyatomic ion to completely neutralize its charge, the (root)ic acid is formed • Examples of polyatomic acids: – If one H+ ion is added to nitrate, NO3- – HNO3 is formed. This is named nitric acid. – If two H+ ions are added to sulfate, SO42- – H2SO4 is formed. This is named sulfuric acid.
    24. 24. Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand astrong acid.Neutralization • Most common acid-baseReactions reactions take place in water solutions (commonly referred to as aqueous solutions). • In the reaction of an acid with a base in aqueous solution, the hydrogen ions of the acid react with the hydroxide ions of the base to give water. The second product is a salt, which is composed of the positive metal ion from the base and the negative ion from the acid. • In general, the reaction for a neutralization reaction is given by acid + base → salt + water
    25. 25. Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand astrong acid.Neutralization • For example,Reactions • HCl(aq) + KOH(aq) → H2O(l) + KCl(aq) • Since HCl(aq) and KOH(aq) are fully ionized in solution, the preceding equation can be written as • H+ (aq) + Cl− (aq) K+(aq) + OH−(aq) → H2O (l) K+(aq) + Cl−(aq) • Ions common to both sides (spectator ions) can be canceled to yield • H+(aq) + OH−(aq) → H2O(l) • This is referred to as the net ionic equation for the neutralization reaction. If H3O+ is substituted for H+ (aq) the neutralization equation becomes • H3O+(aq) + OH−(aq) → 2 H2O (l)
    26. 26. Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand astrong acid.ProcessNotes forWriting NetIonicEquationsRules
    27. 27. Outcome 1-04Write balanced neutralization reactions involving strong acids and bases.Outcome 1-05Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand astrong acid.ProcessNotes forWriting NetIonicEquationsRules
    28. 28. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand a strong acid. Titrations • A titration is a carefully controlled neutralization reaction. • To perform a titration, a standard solution is required. A standard solution is a solution of a strong acid or base of known concentration. To determine the concentration of an acid, a basic standard solution is required, and vice versa for an unknown base. The standard solution is usually added to a sample of unknown concentration, until neutralization has occurred.
    29. 29. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand a strong acid. Titrations • A buret is used to control the additions of acids and bases in a titration. The buret determines the volumes of acids and bases added in a titration. • If you recall, when an acid or base is just neutralized, the moles of hydronium ions from the acid and moles of hydroxide ions from the base are equal. The point at which the amount of standard acid or base solution added just neutralizes the unknown sample is called the equivalence point.
    30. 30. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand a strong acid. Titrations • The equivalence point can be determined by adding a pH indicator to the unknown sample or by using a pH meter to measure the pH change as standard solution is added to the sample. If an indicator is used, the point in the titration at which the desired colour forms is called the endpoint of the titration. The indicator is chosen so that the endpoint and the equivalence point are very close. • The endpoint and the equivalence point are NOT the same thing. The equivalence point is a single point defined by the reaction stoichiometry. The endpoint is determined by the choice of indicators.
    31. 31. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand a strong acid. Titrations Titration Procedure • When performing an acid-base titration on an unknown solution, both the standard and unknown solutions are placed into separate burets. For this example, we will determine the concentration of a hydrochloric acid solution using a 0.100 mol/L solution of NaOH. • Each buret is filled and the initial volume readings are taken from the bottom of the meniscus (the smile-shaped surface of the solution in the container). • A sample or aliquot of the acid solution is released into an Erlenmeyer flask. The acid volume in the buret is read and called the "final acid volume". The volume of the aliquot is equal to: final volume reading - initial volume reading
    32. 32. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand a strong acid. Titrations Titration Procedure - continued • We will use phenolphthalein as our indicator. The endpoint will be reached when the sample turns a light pink. Even though phenolphthalein turns pink at a pH of about 8.2, this is not a significant problem, as we will see later. • We slowly add the standard base solution to the flask, with stirring or swirling, until the solution in the flask turns (and stays) a light pink. We take the reading on the base buret and call this the "final base volume". • This procedure is repeated until there are several pieces of consistent data.
    33. 33. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand a strong acid. Titration Procedure
    34. 34. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand a strong acid. Titration Video Demo
    35. 35. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand a strong acid. Determining Example 1 Concentratio n • A titration was performed using a standard solution of 0.100 mol/L NaOH into and unknown HCl solution. The following data was collected: Base Acid Final Volume Reading 14.45 mL 12.57 mL Initial Volume Reading 0.62 mL 1.13 mL Determine the concentration of the acid.
    36. 36. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand a strong acid. Determini Example 2 ng the Mass of • A student receives a solid an sample of sulfamic acid (molar unknown mass is 97.09 g/mol), a Sample monoprotic acid, and dissolves the sample in enough water to make 100.0 mL of solution. The students takes a 12.00 mL aliquot and titrates with 0.0985 mol/L sodium hydroxide. If 13.38 mL of the base is needed to reach the endpoint, what is the mass of the sample of acid?
    37. 37. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand a strong acid. Titration • If pH readings are taken during a Curves titration and those pH values are plotted, a titration curve is generated. The general shape of the curves generated may be grouped into families, according to the solution titrated and the titrating solutions. We will discuss four classes of titrations: –1. Strong acid titrated with strong base. –2. Strong base titrated with strong acid. –3. Weak acid titrated with strong base. –4. Weak base titrated with strong acid.
    38. 38. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong base and a strong acid. Titration 1. Strong Acid Titrated with a Strong Base Curves
    39. 39. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand a strong acid. Titration 2. Strong Base Titrated with a Strong Acid Curves
    40. 40. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand a strong acid. Titration 3. Titration of a Weak Acid with a Strong Base Curves
    41. 41. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand a strong acid. Titration 4. Titration of a Weak Base with a Strong Acid Curves
    42. 42. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong baseand a strong acid. Titration Curves
    43. 43. Chemistry 40S Outcome 1-04 Unit 3 – Acid/BaseEquilibria neutralization reactions involving strong acids and bases. Write balanced Outcome 1-05 Perform a lab to demonstrate the stoichiometry of a neutralization reaction between a strong base and a strong acid. Titration Curves
    44. 44. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-08 Outcome Define oxidation and reduction. Include: gain and loss of electrons, oxidizing agent, reducing agent. Oxidation What is oxidation and reduction? Reduction • The term “oxidation” was first applied to the combining of oxygen with other elements (ex. rusting iron or burning carbon or methane). Burning is another name for rapid oxidation. • The term “reduction” originally meant the removal of oxygen from a compound. The term reduction comes from the fact that the free metal has a lower mass than its oxide compound. There is a decrease or reduction in the mass of the material as the oxygen is removed.
    45. 45. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-08 Outcome Define oxidation and reduction. Include: gain and loss of electrons, oxidizing agent, reducing agent. Oxidation What is oxidation and reduction? Reduction • Chemists recognized that other nonmetallic elements unite with substances in a manner similar to that of oxygen (ex. hydrogen, antimony, and sodium will burn in chlorine; iron will burn in fluorine). • Therefore, the term oxidation was redefined as the process by which electrons are removed from an atom or ion. • Reduction was then defined as the process by which any atom or ion gains electrons.
    46. 46. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-08 Outcome Define oxidation and reduction. Include: gain and loss of electrons, oxidizing agent, reducing agent. Oxidation Reduction
    47. 47. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-08 Outcome Define oxidation and reduction. Include: gain and loss of electrons, oxidizing agent, reducing agent. Oxidation What is oxidation and reduction? Reduction • Oxidation occurs when an atom loses one or more electrons and reduction is when an atom gains one or more electrons. There are two mnemonics to help us remember these terms: • "OIL RIG": Oxidation Is Losing electrons, Reduction Is Gaining electrons Or • "LEO says GER": Losing Electrons is Oxidation, Gaining Electrons is Reduction
    48. 48. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-08 Outcome Define oxidation and reduction. Include: gain and loss of electrons, oxidizing agent, reducing agent. Oxidation- What are REDOX Reactions? Reduction • Oxidation-Reduction reactions, or Reactions Redox reactions, are defined as chemical changes that occur when electrons are transferred from one reactant to another. • For example, 2 Mg(s) + O2(g) → 2 MgO(s) • If this reaction is written in ionic form it becomes: 2 Mg0 + O20 → Mg2+ O2- • Magnesium and oxygen gases are both elements and have no charge • Non-scientists usually refer to this reaction as burning or combustion but as scientists, we refer to this reaction as oxidation.
    49. 49. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-08 Outcome Define oxidation and reduction. Include: gain and loss of electrons, oxidizing agent, reducing agent. Oxidation- What are REDOX Reactions? Reduction • For example, Reactions 2 Mg(s) + O2(g) → 2 MgO(s) • We say that the magnesium has been oxidized to MgO by the reaction with oxygen gas • Considering the charges, the metal has gone from a 0 charge to 2+ and the non-metal from 0 charge to 2- • In this reaction magnesium begins as a neutral atom and loses two electrons to become a Mg2+ ion in MgO. Magnesium is oxidized. Oxygen begins as a neutral atom and gains the two electrons from magnesium to become an O2-¯ion in MgO. Oxygen is reduced.
    50. 50. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-08 Outcome Define oxidation and reduction. Include: gain and loss of electrons, oxidizing agent, reducing agent. Oxidation- What are REDOX Reactions? Reduction • For example, Reactions 2 Mg(s) + O2(g) → 2 MgO(s) • If we look at the change in ion charge as a function of electrons, the following relationships can be written: • 2 Mg → 2Mg 2+ + 4 e- in this equation, charge is conserved, 0 charge on both sides • O2 + 4 e- → 2O 2- in this equation, charge is conserved, 4- charge on both sides
    51. 51. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-08 Outcome Define oxidation and reduction. Include: gain and loss of electrons, oxidizing agent, reducing agent. Oxidation- What are REDOX Reactions? Reduction Reactions • Similarly, the reaction of magnesium combining with chlorine to produce magnesium chloride is also a redox reaction. Mg(s) + Cl2(g) → MgCl2(s) • Magnesium loses two electrons to become a Mg2+ ion and is oxidized. Each chlorine atom gains one electron from magnesium to become Cl- ions. Each chlorine atom is reduced.
    52. 52. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-08 Outcome Define oxidation and reduction. Include: gain and loss of electrons, oxidizing agent, reducing agent. Oxidation- What are REDOX Reactions? Reduction Reactions
    53. 53. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-08 Outcome Define oxidation and reduction. Include: gain and loss of electrons, oxidizing agent, reducing agent. Oxidation- What are REDOX Reactions? Reduction Reactions
    54. 54. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-08 Outcome Define oxidation and reduction. Include: gain and loss of electrons, oxidizing agent, reducing agent. Reducing What are Reducing Agents? Agents
    55. 55. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-08 Outcome Define oxidation and reduction. Include: gain and loss of electrons, oxidizing agent, reducing agent. Oxidizing What are Oxidizing Agents? Agents
    56. 56. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-08 Outcome Define oxidation and reduction. Include: gain and loss of electrons, oxidizing agent, reducing agent. Reducing What are Reducing & Oxidizing & Agents? Oxidizing Agents • Any substance which causes the reduction of another substance is called the reducing agent. • Any substance which causes the oxidation of another substance is called an oxidizing agent. 2 Mg(s) + O2(g) → 2 MgO(s)
    57. 57. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-09 Outcome Determine the oxidation numbers for atoms in compounds and ions. Oxidation Assigning Oxidation Numbers Numbers • The oxidation number represents the charge the atom would have if every bond were ionic. Not every bond is ionic, but chemists assume they are for this system. The oxidation number is not always the actual charge, but it is very helpful to follow electrons in redox reactions.
    58. 58. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-09 Outcome Determine the oxidation numbers for atoms in compounds and ions. Oxidation Assigning Oxidation Numbers Numbers – Oxidation numbers are always assigned PER ATOM. – The oxidation numbers of all uncombined elements is zero. (ex. O2, K(s), H2, – The oxidation number of monatomic ions equals the charge of that ion. – In compounds, the oxidation number for alkali metals (ex. Li, Na, K, etc.) is always +1. – In compounds, the oxidation number for the alkaline earth metals (ex. Be, Mg, Ca, etc.) is always +2. – In compounds, the oxidation number of aluminum is always +3.
    59. 59. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-09 Outcome Determine the oxidation numbers for atoms in compounds and ions. Oxidation Assigning Oxidation Numbers Numbers – In compounds, the oxidation number of fluorine is -1. – In compounds, the oxidation number of hydrogen is +1. An exception is in metal hydrides, such as NaH or MgH2, when hydrogen is -1. – In compounds, the oxidation number of oxygen is -2. An exception is in peroxides, such as H2O2 or Na2O2, when its oxidation number is -1. – For any neutral compound, the sum of the oxidation numbers for each atom must be zero. – For a poly atomic ion, the sum of the oxidation numbers for each atom must be the charge of that ion. Note the difference between ion charge and oxidation numbers: for example, the magnesium ion, Mg 2+ ion charge = 2+ oxidation number = +2
    60. 60. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-10 Outcome Identify and describe reactions as redox or non-redox. Include: oxidizing agent, reducing agent, oxidzed substance, and reduced substance Redox Recognizing Redox Reactions Reactions • In order to recognize a redox reaction, we must follow the electrons. We can follow the electrons by determining if the oxidation numbers change. If no oxidation numbers change, the reaction is NOT a redox reaction.
    61. 61. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-10 Outcome Identify and describe reactions as redox or non-redox. Include: oxidizing agent, reducing agent, oxidzed substance, and reduced substance Redox Recognizing Redox Reactions Reactions Example – Is the reaction below a redox reaction? SO2 + H2O → H2SO3 Solution – If we assign oxidation numbers to each atom, we find – the oxidation number of S remains at +4 in the reactants and in the products, – O remains unchanged in the products at -2 and H remains unchanged at +1. – The oxidation numbers do not change, so electrons were not transferred. Therefore, this is not a redox reaction.
    62. 62. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-10 Outcome Identify and describe reactions as redox or non-redox. Include: oxidizing agent, reducing agent, oxidzed substance, and reduced substance Redox Recognizing Redox Reactions Reactions Example – Is the reaction below a redox reaction? Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s) Solution – Assign oxidation numbers to each atom, – Cu is zero in the reactants and +2 in the products. – Ag is +1 in the reactants and zero in the products. – N = +5 in both products and reactants. – O = -2 in both products and reactants.
    63. 63. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-10 Outcome Identify and describe reactions as redox or non-redox. Include: oxidizing agent, reducing agent, oxidzed substance, and reduced substance Redox Recognizing Redox Reactions Reactions Example – Is the reaction below a redox reaction? Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s) Solution – This is redox reaction, since the oxidation numbers of Cu and Ag change. – Cu is oxidized because its oxidation number becomes more positive, indicating it has lost electrons. . – The oxidation number of Ag becomes more negative, indicating a gain of electrons. Ag is reduced.
    64. 64. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-reduction reactions using redox methods. Include acidic and basic solutions Balancing Balancing Redox Reactions Redox Reactions • Redox reactions transfer electrons from one substance to another. Balancing a redox reaction balances the number of electrons lost and gained, then ensures the reaction obeys the Law of Conservation of Mass. • There are two methods we will use to balance redox reactions: – Oxidation Number Method – Half-Reactions Method
    65. 65. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-reduction reactions using redox methods. Include acidic and basic solutions Balancing Oxidation Number Method Redox Reactions • We will learn this method by balancing the following reaction: K2Cr2O7 + H2O + S SO2 + KOH + Cr2O3 • Step 1: Assign oxidation numbers. • Eventually, you will be able to recognize those just substances that are oxidized and reduced. You can just label those substances.
    66. 66. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-reduction reactions using redox methods. Include acidic and basic solutions Balancing Oxidation Number Method Redox K2Cr2O7 + H2O + S SO2 + KOH + Cr2O3 Reactions • Step 2: Identify substances oxidized and reduced, then balance electrons lost and gained. • S is oxidized and Cr is reduced. • Notice that there are two Cr atoms and each gains three electrons, for a total of six.
    67. 67. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-reduction reactions using redox methods. Include acidic and basic solutions Balancing Oxidation Number Method Redox K2Cr2O7 + H2O + S SO2 + KOH + Cr2O3 Reactions • Step 2 (continued): We then determine the lowest common multiple of 6 and 4, which is 12. We multiply the S atoms by 3 and the Cr atoms by 2, so the total number of electrons lost and gained is 12.
    68. 68. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-reduction reactions using redox methods. Include acidic and basic solutions Balancing Oxidation Number Method Redox K2Cr2O7 + H2O + S SO2 + KOH + Cr2O3 Reactions • Step 3: Balance all other atoms by inspection. • We need 4 K atoms on the left, and 4 H atoms on the right. • The final balanced equation is 2K2Cr2O7 + 2H2O + 3S 3SO2 + 4KOH + 2Cr2O3
    69. 69. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-reduction reactions using redox methods. Include acidic and basic solutions Balancing Half-Reactions Method Redox • Half-reactions, as their name suggests, show the Reactions oxidation or reduction reaction. • For example: Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s) • The oxidation half reaction shows the oxidation of copper to copper (II) ions and the loss of electrons: Cu(s) Cu2+(aq) + 2e¯ • The loss of electrons is indicated in the equation by putting electrons on the product side of the equation. • The reduction half reaction shows the reduction of silver ions to metallic silver and the gaining of electrons: Ag+ (aq) + 1e¯ Ag (s) • Gaining electrons is indicated in the equation by placing electrons on the reactant side of the equation.
    70. 70. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-reduction reactions using redox methods. Include acidic and basic solutions Balancing Half-Reactions Method Redox • Redox reactions often occur in the Reactions presence of acids or bases. The half- reaction method is the preferred method for balancing these types of reactions. • In general, the method requires the entire equation to be separated into the oxidation and reduction half- reactions. The electrons are balanced and the overall balanced equation is produced by adding the two balanced half-reactions.
    71. 71. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-reduction reactions using redox methods. Include acidic and basic solutions Balancing Half-Reactions Method Redox • Steps for Balancing Redox Reactions in Reactions Acidic Solutions • This method will be demonstrated by balancing the following reaction in an acidic solution. Cr2O72¯(aq) + SO3 2¯(aq) Cr3+(aq) + SO42-(aq) • Step 1: Assign oxidation numbers. Identify then write the oxidation and reduction half- reactions. • Step 2: Balance all elements except hydrogen and oxygen. Add electrons lost and gained. • Step 3: Balance oxygen atoms by using H2O. • Step 4: Balance hydrogen atoms using H+ ions. • Step 5: Balance the number of electrons lost and gained. • Step 6: Add the two half-reactions.
    72. 72. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-reduction reactions using redox methods. Include acidic and basic solutions Half-Reactions Method Balancing Redox • The first part of balancing redox Reactions reactions in basic solutions follows the same steps as that for acidic solutions. Before adding the two half- reactions, you add the same number of hydroxide ions as hydrogen ions to BOTH sides of the equation. We eliminate the hydrogen ions by forming water (H+ + OH- H2O)
    73. 73. Chemistry 40S Unit 3 – Acid/BaseEquilibria1-11 Outcome Balance oxidation-reduction reactions using redox methods. Include acidic and basic solutions Half-Reactions Method Balancing Redox • For example, balance the following Reactions reaction in a basic solution. MnO4¯ + C2O42¯ CO2 + MnO2 • Step 1: Assign oxidation numbers. Identify then write the oxidation and reduction half-reactions. • Step 2: Add water and hydrogen ions to balance oxygen and hydrogen. • Step 3: Add hydroxide ions to each side for each hydrogen ion. • Step 4: Change the hydroxides and hydrogens to waters . • Step 5: Balance electrons lost and gained . • Step 6: Add the two half-reactions.

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