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Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
Ch4 z5e reactions
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Ch4 z5e reactions

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  • Z5e 132 Section 4.1Water, the Common Solvent Alternative link if can’t get through to YouTube: http://dailycackle.com/2007/01/20/science-experiment-gone-wrong/
  • Z5e 132 Fig. 4.1
  • Z5e 133 Figure 4.2
  • Z5e 134 Figure 4.3
  • Z5e 134 Section 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes
  • Z5e 135 Figure 4.5
  • Z5e 136 Figure 4.6
  • Z5e 136 Figure 4.7
  • Z5e 137 Figure 4.8
  • Z5e 137 Figure 4.9
  • Z5e 139 Section 4.3 The Composition of Solutions (34.6g)(1 mole/58 g)/0.125L = 4.77 M
  • Z5e 139 Section 4.3 The Composition of Solutions (34.6g)(1 mole/58 g)/0.125L = 4.77 M
  • (0.050 L)(2.7 moles/L)(36.5 g/mole) = 4.9 g HCl (27 g)(1 mole/110.986)(1/0.500L) = 0.49 M (about 0.5 mole) Ca = 0.50 M ; Cl = 1.00 M
  • (0.050 L)(2.7 moles/L)(36.5 g/mole) = 4.9 g HCl (27 g)(1 mole/110.986)(1/0.500L) = 0.49 M (about 0.5 mole) Ca = 0.50 M ; Cl = 1.00 M
  • (45.6 g)(1 mole/400. g)/.475 L = 0.240 M Fe = 0.48 M ; S = 0.72 M ; O = 2.88 M
  • Z5e 143 Figure 4.10 Weighed amount of solute is put into volumetric flask and small quantity of H2O added Solid dissolved by gently swirling with stopper in place More water added until level just reaches mark on neck (d)
  • Z5e 142 a. (1.0 mole/L)(246 g/mole)(0.100 L) = dissolve 24.6 g in 90 ml distilled water; qsad to 100 ml b. (0.250L)(2.0 mole/L )(196 g/mole) = dissolve 98 g in 200 ml distilled water; qsad to 250 ml
  • Z5e 142 a. (1.0 mole/L)(246 g/mole)(0.100 L) = dissolve 24.6 g in 90 ml distilled water; qsad to 100 ml b. (0.250L)(2.0 mole/L )(196 g/mole) = dissolve 98 g in 200 ml distilled water; qsad to 250 ml
  • Z5e 144 Dilution
  • (1.7)(x) = (250)(0.50) so x = 73.5 ml add to distilled H20 to 250 ml 18.5 * 2.3 = 250 * x; x = 0.17 M Tricky: find moles of HCl 0.0185 L * 2.3 moles/L = 0.0426 moles; then M = 0.0426 moles/(0.250 + 0.0185) L = 0.16 M
  • (1.7)(x) = (250)(0.50) so x = 73.5 ml add to distilled H20 to 250 ml 18.5 * 2.3 = 250 * x; x = 0.17 M Tricky: find moles of HCl 0.0185 L * 2.3 moles/L = 0.0426 moles; then M = 0.0426 moles/(0.250 + 0.0185) L = 0.16 M
  • Z5e 145 Figure 4.11 Measuring pipet is graduated & can measure various volumes accurately. Volumetric (transfer) pipet measures one volume accurately.
  • 1(.75) = 4(x); x = 190 ml diluted with dist. H20 to 1.0 L Convert to moles or mmoles (since Stoich) then use spider. So 25 * .67 = 16.75 H2SO4; 35 * 0.4 = 14 moles CaCl2; CaCl2 is limiting; 0.014 moles * 136 g/mole = 1.9 g (note: IF this were NOT a ppt, then conc = 0.014 moles/(0.025 + 0.035); = 0.23 M
  • 1(.75) = 4(x); x = 190 ml diluted with dist. H20 to 1.0 L Convert to moles or mmoles (since Stoich) then use spider. So 25 * .67 = 16.75 H2SO4; 35 * 0.4 = 14 moles CaCl2; CaCl2 is limiting; 0.014 moles * 136 g/mole = 1.9 g (note: IF this were NOT a ppt, then conc = 0.014 moles/(0.025 + 0.035); = 0.23 M
  • 1(.75) = 4(x); x = 190 ml diluted with dist. H20 to 1.0 L Convert to moles or mmoles (since Stoich) then use spider. So 25 * .67 = 16.75 H2SO4; 35 * 0.4 = 14 moles CaCl2; CaCl2 is limiting; 0.014 moles * 136 g/mole = 1.9 g (note: IF this were NOT a ppt, then conc = 0.014 moles/(0.025 + 0.035); = 0.23 M
  • Section 4.4 Types of reactions
  • Z5e Section 4.5 Precipitation reactions
  • Rules for Solubility (Taken from Cornell University Chemistry Department) Sing to Rhythm of “99 Bottles” or to “Gilligan’s Isle”
  • Rules for Solubility (Taken from Cornell University Chemistry Department) Sing to Rhythm of “99 Bottles” or to “Gilligan’s Isle”
  • Rules for Solubility (Taken from Cornell University Chemistry Department) Sing to Rhythm of “99 Bottles” or to “Gilligan’s Isle”
  • Rules for Solubility (Taken from Cornell University Chemistry Department) Sing to Rhythm of “99 Bottles” or to “Gilligan’s Isle”
  • Z5e 150 Table 4.1
  • Z5e 153
  • Z5e Section 4.7 Stoichiometry of Precipitation Reactions BaCl 2 MW 208; NaOH 40; Ba(OH) 2 171; HCl 37; Ag 108 NaOH is limiting; 0.85 g of Ba(OH) 2 12.25 ml (derived from 1st getting 0.0025 mol), then using mole ratio, etc.
  • Z5e Section 4.7 Stoichiometry of Precipitation Reactions BaCl 2 MW 208; NaOH 40; Ba(OH) 2 171; HCl 37; Ag 108 NaOH is limiting; 0.85 g of Ba(OH) 2 12.25 ml (derived from 1st getting 0.0025 mol), then using mole ratio, etc.
  • Z5e Section 4.7 Stoichiometry of Precipitation Reactions BaCl 2 MW 208; NaOH 40; Ba(OH) 2 171; HCl 37; Ag 108 NaOH is limiting; 0.85 g of Ba(OH) 2 12.25 ml (derived from 1st getting 0.0025 mol), then using mole ratio, etc.
  • Z5e 157 Section 4.8 Acid-Base Reactions
  • Z5e 161 Acid-Base Titrations
  • Z15e 162 Neutralization Titrations # of H + x M A x V A = # of OH - x M B x V B (1)(0.0980)(34.65) = (2)(50)(x) 0.0340 M ; but look at Zumdahl method page 158!!!
  • Mmole HCl = 75 * 0.25 = 18.75; since 1:1 H:Cl; H = 18.75 mmole Mmole Ba(OH)2 = 225 * 0.055; since 2:1 OH:Ba; OH = 24.75 mmole H:OH = 1:1; so 6 mmoles xs OH in (75 + 225)ml = 0.02 M OH
  • Z5e 164 Section 4.9 Oxidation-Reduction Reactions
  • Hrw 591 Z5e 167 Table 4.2 Rules for Assigning Oxidation States
  • Hrw 591 Z5e 167 Table 4.2 Rules for Assigning Oxidation States
  • Hrw 591.
  • Hrw 591
  • Hrw 592
  • See, Z5e 167 SE 4.16 for partial solutions on these.
  • Hrw 592
  • Z5e 169 figure 4.19
  • Hrw 602
  • Hrw 602
  • Hrw 603 Z5e 170
  • Hrw 603 Z5e 170
  • Hrw 593 (Z5e 171 Section 4.10 Balancing Oxidation-Reduction Equations)
  • Hrw 593 (Z5e 171 Section 4.10 Balancing Oxidation-Reduction Equations)
  • Hrw 593 (Z5e 171 Section 4.10 Balancing Oxidation-Reduction Equations)
  • Hrw 593 (Z5e 171 Section 4.10 Balancing Oxidation-Reduction Equations)
  • Hrw 594
  • Hrw 597 Z5e 172 acid redox steps
  • Hrw 598
  • Hrw 601 Add SP 19-1
  • Hrw 601 Add SP 19-1
  • Hrw 601 Add SP 19-1
  • Hrw 597 Z5e 172 acid redox steps
  • Hrw 597 Z5e 172 acid redox steps
  • Hrw 597 Z5e 172 acid redox steps
  • Hrw 597 Z5e 172 acid redox steps
  • Hrw 597 Z5e 172 acid redox steps
  • Hrw 597 Z5e 172 acid redox steps
  • 1, 5 --> 1, 5 2,3,4 --> 2,3,5 (basic solution!) b. 3,2,8 --> 3,2,4
  • 188 H + + 5 Fe(CN) 6 4- + 61 MnO 4 1- --> 94 H2O + 61 Mn 2+ + 5 Fe 3+ + 30 CO 2 + 30 NO 3 1-
  • Hrw 598 Z5e 176 Basic solution redox balancing (see, SE 4.20 p. 177, if needed) 2,3,4 --> 2,3,5
  • Z5e 176 Basic solution redox balancing (see, SE 4.20 p. 177, if needed) ,2 --> 1, 2 1, 2, 1, 2 --> 1, 2,
  • Z5e 176 Basic solution redox balancing (see, SE 4.20 p. 177, if needed) 2,3,4 --> 2,3,5
  • a. write NR Fe2+ w/MnO4 1- (8H + 5Fe2+ + MnO4 1- -->5Fe3+ + Mn2+ + 4H20 b. Do stoich (calc moles MnO4 1st) --> .0086 moles MnO4 & .043 moles Fe2+ c. Find grams Fe2+ = 0.043 x 55.847 = 2.40 g d. (2.40 g/6.128 g ore) x 100% = 39.2%
  • a. write NR Fe2+ w/MnO4 1- (8H + 5Fe2+ + MnO4 1- -->5Fe3+ + Mn2+ + 4H20 b. Do stoich (calc moles MnO4 1st) --> .0086 moles MnO4 & .043 moles Fe2+ c. Find grams Fe2+ = 0.043 x 55.847 = 2.40 g d. (2.40 g/6.128 g ore) x 100% = 39.2%
  • Transcript

    1. Chapter 4 Aqueous solutions Types of reactions pp http://www.youtube.com/watch?v=h2UczzWu-Qs&mode=related&search =
    2. 4.1 Water, the Common Solvent - Parts of Solutions <ul><li>Solution - homogeneous mixture. </li></ul><ul><li>Solute - what gets dissolved. </li></ul><ul><li>Solvent - what does the dissolving. </li></ul><ul><li>Soluble - Can be dissolved. </li></ul><ul><li>Miscible - liquids dissolve in each other . </li></ul>
    3. Aqueous solutions <ul><li>Dissolved in water. </li></ul><ul><li>Water is a good solvent because the molecules are polar. </li></ul><ul><li>The oxygen atoms have a partial negative charge. </li></ul><ul><li>The hydrogen atoms have a partial positive charge. </li></ul><ul><li>The angle is 105º. </li></ul>
    4. Figure 4.1 The Water Molecule is Polar
    5. Hydration <ul><li>The process of breaking the ions of salts apart. </li></ul><ul><li>Ions have charges and attract the opposite charges on the water molecules. </li></ul>
    6. Hydration H H O H H O H H O H H O H H O H H O H H O H H O H H O
    7. Figure 4.2 Polar Water Molecules Interact with the Positive and Negative Ions of a Salt
    8. Solubility <ul><li>How much of a substance will dissolve in a given amount of water. </li></ul><ul><li>Usually in g/100 mL </li></ul><ul><li>Solubility varies greatly, but if they do dissolve the ions are separated , </li></ul><ul><li>and they can move around . </li></ul><ul><li>Water can also dissolve non -ionic compounds if they have polar bonds. </li></ul>
    9. Figure 4.3 Polar Bond
    10. 4.2 Nature of Aqueous Solutions Strong and Weak Electrolytes <ul><li>Electricity is moving charges. </li></ul><ul><li>The ions that are dissolved can move. </li></ul><ul><li>Solutions of ionic compounds can conduct electricity. </li></ul><ul><li>Called “electrolytes”. </li></ul><ul><li>Solutions are classified three ways. </li></ul>
    11. Types of solutions <ul><li>Strong electrolytes - completely dissociate (fall apart into ions). </li></ul><ul><li>Many ions- Conduct well. </li></ul><ul><li>Weak electrolytes - Partially fall apart into ions. </li></ul><ul><li>Few ions -Conduct electricity slightly. </li></ul><ul><li>Non-electrolytes - Don’t fall apart. </li></ul><ul><li>No ions- Don’t conduct. </li></ul>
    12. Figure 4.5 BaCI 2 Dissolving
    13. Types of solutions <ul><li>Acids - form H + ions when dissolved. </li></ul><ul><li>Strong acids fall apart completely. </li></ul><ul><li>Many ions </li></ul><ul><li>H 2 SO 4 HNO 3 HCl HBr HI HClO 4 </li></ul><ul><li>Weak acids- don’t dissociate completely. </li></ul><ul><li>Bases - form OH - ions when dissolved. </li></ul><ul><li>Strong bases - many ions. </li></ul><ul><li>KOH, NaOH, Ca(OH) 2 </li></ul>
    14. Figure 4.6 HCI ( aq ) is Completely Ionized
    15. Figure 4.7 An Aqueous Solution of Sodium Hydroxide
    16. Figure 4.8 Acetic Acid in Water
    17. Figure 4.9 The Reaction of NH 3 in Water
    18. 4.3 Composition of Solutions <ul><li>Concentration - how much is dissolved. </li></ul><ul><li>Molarity = Moles of solute Liters of solution </li></ul><ul><li>abbreviated M </li></ul><ul><li>1 M = 1 mol solute / 1 liter solution </li></ul><ul><li>Calculate the molarity of a solution with 34.6 g of NaCl dissolved in 125 mL of solution. Steps . . . </li></ul>
    19. 4.3 Composition of Solutions pp <ul><li>Calculate the molarity of a solution with 34.6 g of NaCl dissolved in 125 mL of solution. </li></ul><ul><li>Convert grams to moles. How? Convert mL to L. How? Divide moles by liters </li></ul><ul><li>(34.6 g)(1 mol) (58.5 g)(0.125 L) </li></ul><ul><li>= 4.73 M </li></ul>
    20. Molarity <ul><li>How many grams of HCl (M = 36.5) would be required to make 50.0 mL of a 2.7 M solution? Ans. . . </li></ul><ul><li>(0.050 L)(2.7 moles/L)(36.5 g/mole) = 4.9 g HCl </li></ul>
    21. Molarity <ul><li>What would the concentration of CaCl 2 be if you used 27g of CaCl 2 (M = 111) to make 500. mL of solution? Ans. . . </li></ul><ul><li>(27 g)(1 mole/110.986)(1/0.500L) = 0.49 M (about 0.50 molarity) </li></ul><ul><li>What is the approximate concentration of each ion? </li></ul><ul><li>Ca 2+ = 0.50 M ; Cl 1- = 1.00 M </li></ul>
    22. Molarity <ul><li>Calculate the concentration of a solution made by dissolving 45.6 g of Fe 2 (SO 4 ) 3 to a total volume of 475 mL. (M of Fe 2 (SO 4 ) 3 = 400. g/mol) Ans. . . </li></ul><ul><li>(45.6 g)(1 mole/400. g)/.475 L = 0.240 M </li></ul><ul><li>What is the concentration of each ion ? </li></ul><ul><li>Fe = 0.48 M ; SO 4 = 0.72 M </li></ul>
    23. Figure 4.10 Preparation of a Standard Solution <ul><li>Weighed amount of solute is put into volumetric flask and small quantity of H2O added </li></ul><ul><li>Solid dissolved by gently swirling with stopper in place </li></ul><ul><li>More water added until level just reaches mark on neck (d) </li></ul>
    24. Making solutions <ul><li>Describe how to make 100.0 mL of a 1.0 M K 2 CrO 4 solution. </li></ul><ul><li>(1.0 mole/L)(246 g/mole)(0.100 L) = dissolve 24.6 g in 85 ml distilled water; qsad to 100 ml </li></ul>
    25. Making solutions <ul><li>Describe how to make 250. mL of an 2.0 M copper(II) sulfate dihydrate solution. </li></ul><ul><li>What is formula for copper(II) sulfate dihydrate? </li></ul><ul><li>CuSO 4 •2H 2 O (M = 196) </li></ul><ul><li>What is the answer? . . . </li></ul><ul><li>(0.250L)(2.0 mole/L )(196 g/mole) = dissolve 98 g in 200 ml distilled water; qsad to 250 ml </li></ul>
    26. Dilution <ul><li>Adding more solvent to a known solution. </li></ul><ul><li>The moles of solute stay the same. </li></ul><ul><li>moles = M x L </li></ul><ul><li>M 1 V 1 = M 2 V 2 </li></ul><ul><li>moles = moles </li></ul><ul><li>Stock solution is a solution of known concentration used to make more dilute solutions </li></ul>
    27. Dilution <ul><li>What volume of a 1.7 M solution is needed to make 250. mL of a 0.50 M solution? Steps. </li></ul><ul><li>V 1 M 1 = V 2 M 2 </li></ul><ul><li>( x )( 1.7 ) = ( 250. )( 0.50 ) </li></ul><ul><li>So, x = 74 mL, (rounded from 73.5 mL) of the 1.7 M solution combined with enough distilled H 2 0 to make a total volume of 250. mL </li></ul><ul><li>Be careful! If you just add it to 250. mL then you get a volume of 323 mL and this throws off your molarity. </li></ul>
    28. Dilution <ul><li>18.5 mL of 2.3 M HCl is diluted to 250 mL of water. What’s solution concentration? Ans. </li></ul><ul><li>(18.5)(2.3) = 250(x) ; x = 0.17 M </li></ul><ul><li>18.5 mL of 2.3 M HCl is added to 250 mL of water. Solution concentration? (Tricky) </li></ul><ul><li>Find moles of HCl (0.0185 L)(2.3 moles/L) = 0.0426 moles; then M = 0.0426 moles/(0.250 + 0.0185 L ) = 0.16 M (compare with above) </li></ul>
    29. Fig. 4.11 p. 145 <ul><li>A Measuring Pipet is graduated & can measure various volumes accurately. </li></ul><ul><li>A Volumetric (transfer) Pipet measures one volume accurately. </li></ul>
    30. More dilution (if time) <ul><li>You have a 4.0 M stock solution. Describe how to make 1.0L of a .75 M solution. </li></ul><ul><li>(1)(.75) = 4(x); x = 190 ml diluted with distilled H 2 0 to 1.0 L </li></ul>
    31. More dilution (if time) <ul><li>25 mL 0.67 M of H 2 SO 4 added to 35 mL of 0.40 M CaCl 2 . What mass CaSO 4 is formed? Steps. . . </li></ul><ul><li>1st get moles of both (moles = V • M ) </li></ul><ul><li>0.025 L • 0.67 M = 0.017 mol H 2 SO 4 ; 0.035 L • 0.40 M = 0.014 moles CaCl 2 </li></ul><ul><li>Then, need to find limiting reactant (2 stoichs) </li></ul>
    32. More dilution (if time) <ul><li>25 mL 0.67 M of H 2 SO 4 added to 35 mL of 0.40 M CaCl 2 . What mass CaSO 4 formed? </li></ul><ul><li>0.01675 mol H 2 SO 4 and 0.014 moles CaCl 2 </li></ul><ul><li>Then, need to find limiting reactant (2 stoichs) </li></ul><ul><li>H 2 SO 4 + CaCl 2  CaSO 4 + 2HCl </li></ul><ul><li>CaCl 2 is LR and the mole ratio to CaSO 4 = 1:1 </li></ul><ul><li>(0.014 mol CaCl 2 )(1 mol CaSO 4 )(136 g CaSO 4 ) = 1.9 g CaSO 4 (1 mol CaCl 2 )(1 mol CaSO 4 ) </li></ul><ul><li>IF this were NOT a ppt, then conc = 0.014 moles/(0.025 + 0.035) = 0.23 M </li></ul>
    33. 4.4 Types of Reactions <ul><li>Precipitation reactions </li></ul><ul><li>When aqueous solutions of ionic compounds are poured together a solid forms. </li></ul><ul><li>A solid that forms from mixed solutions is a precipitate </li></ul><ul><li>If you’re not a part of the solution, your part of the precipitate </li></ul>
    34. 4.5 Precipitation reactions <ul><li>NaOH(aq) + FeCl 3 (aq)  NaCl(aq) + Fe(OH) 3 (s) is really . . . </li></ul><ul><li>Na + (aq) + OH - (aq) + Fe +3 (aq) + Cl - (aq)  Na + (aq) + Cl - (aq) + Fe(OH) 3 (s) </li></ul><ul><li>So all that really happens is . . . </li></ul><ul><li>Na + (aq) + OH - (aq) + Fe +3 (aq) + Cl - (aq)  Na + (aq) + Cl - (aq) + Fe(OH) 3 (s) </li></ul><ul><li>OH - (aq) + Fe +3 (aq)  Fe(OH) 3 (s) </li></ul><ul><li>Then, balance: 3 OH - (aq) + Fe +3 (aq)  Fe(OH) 3 (s) </li></ul><ul><li>Double replacement reaction </li></ul>
    35. Precipitation reaction <ul><li>We can predict the products </li></ul><ul><li>Can only be certain by experimenting </li></ul><ul><li>The anion and cation switch partners </li></ul><ul><li>AgNO 3 ( aq ) + KCl( aq )  (molecularly) .  </li></ul><ul><li> AgCl + KNO 3 </li></ul><ul><li>Zn(NO 3 ) 2 ( aq ) + BaCr 2 O 7 ( aq )  </li></ul><ul><li>ZnCr 2 O 7 + Ba(NO 3 ) 2 </li></ul><ul><li>CdCl 2 ( aq ) + Na 2 S( aq )  </li></ul><ul><li>CdS + 2NaCl </li></ul>
    36. Precipitations Reactions <ul><li>Only happen if one of the products is insoluble </li></ul><ul><li>Otherwise all the ions stay in solution- nothing has happened. </li></ul><ul><li>Need to memorize the rules for solubility (see, p. 150 or use solubility poem) </li></ul>
    37. Let’s Sing! <ul><li>Potassium, sodium and ammonium salts, Whatever they may be, Can always be depended on for solubility. </li></ul><ul><li>When asked about the nitrates The answer is always clear, They each and all are soluble, Is all we want to hear. </li></ul><ul><li>Most every chloride’s soluble At least we’ve always read Save silver, mercurous mercury And (slightly) chloride of lead </li></ul>
    38. Let’s Sing! <ul><li>Every single sulfate Is soluble, ‘Tis said ‘Cept barium and strontium And calcium and lead. </li></ul><ul><li>Hydroxides of metals won’t dissolve That is, all but three Potassium, sodium and ammonium Dissolve quite readily </li></ul>
    39. Let’s Sing! <ul><li>And then you must remember That you must not “forgit” Calcium, barium, strontium Dissolve a little bit. </li></ul><ul><li>The carbonates are insoluble, It’s lucky that it’s so, Or else, our marble buildings Would melt away like snow. </li></ul>
    40. Repeat with feeling!! <ul><li>Potassium, sodium, and ammonium salts Whatever they may be Can always be depended on For solubility </li></ul>
    41. Solubility Rules <ul><li>All nitrates are soluble </li></ul><ul><li>Alkali metals ions and NH 4 + ions are soluble </li></ul><ul><li>Halides are soluble except Ag + , Pb +2 , and Hg 2 +2 </li></ul><ul><li>Most sulfates are soluble, except Pb +2 , Ba +2 , Hg +2 ,and Ca +2 </li></ul>
    42. Solubility Rules <ul><li>Most hydroxides are slightly soluble (insoluble) except NaOH and KOH </li></ul><ul><li>Sulfides, carbonates, chromates, and phosphates are insoluble </li></ul><ul><li>Lower number rules supersede so Na 2 S is soluble </li></ul>
    43. 4.6 Describing Reactions in Solution <ul><li>Molecular Equation - written as whole formulas, not the ions. </li></ul><ul><li>K 2 CrO 4 ( aq ) + Ba(NO 3 ) 2 ( aq )  NO 3 ( aq ) + BaCrO 4 (s) </li></ul><ul><li>Complete Ionic equation - show dissolved electrolytes as the ions. </li></ul><ul><li>2K + + CrO 4 -2 + Ba +2 + 2 NO 3 -  BaCrO 4 (s) + 2K + + 2 NO 3 - </li></ul><ul><li>Net Ionic equation - includes only solution components undergoing a change. Spectator ions are those that don’t react. </li></ul><ul><li>CrO 4 -2 + Ba +2  BaCrO 4 (s) </li></ul>
    44. Three Type of Equations <ul><li>Write the three types of equations for the reactions when these solutions are mixed. (Whiteboard) </li></ul><ul><li>iron (III) sulfate and potassium sulfide </li></ul><ul><li>lead (II) nitrate and sulfuric acid. </li></ul>
    45. 4.7 Stoichiometry of Precipitation pp <ul><li>Exactly the same, except you may have to figure out what the pieces are. </li></ul><ul><li>What mass of solid is formed when 100.00 mL of 0.100 M Barium chloride is mixed with 100.00 mL of 0.100 M sodium hydroxide? . . . </li></ul><ul><li>Strategy: need to convert both reactants to moles, then determine LR then use stoich. </li></ul><ul><li>Have 0.01 moles of both </li></ul><ul><li>BaCl 2 + 2NaOH  Ba(OH) 2 + 2NaCl </li></ul>
    46. BaCl 2 continued pp <ul><li>What mass of solid is formed when 100.00 mL of 0.100 M Barium chloride is mixed with 100.00 mL of 0.100 M sodium hydroxide? Have 0.010 moles of both BaCl 2 + 2NaOH  Ba(OH) 2 + 2NaCl </li></ul><ul><li>NaOH is limiting & makes 0.005 mol Ba(OH) 2 </li></ul><ul><li>0.005 mol BaCl 2 • 208 g/mol = 0.85 g formed </li></ul>
    47. Another Stoich of Precipitation problem <ul><li>What volume of 0.204 M HCl is needed to precipitate the silver from 50.ml of 0.0500 M silver nitrate solution? </li></ul><ul><li>HCl + AgNO 3  AgCl + HNO 3 </li></ul><ul><li>50.ml of 0.0500 M AgNO 3 = 0.0025 moles </li></ul><ul><li>Mole ratio = 1:1; need 0.0025 mol HCl also </li></ul><ul><li>Since M = mol/vol; vol = mol/ M </li></ul><ul><li>Vol = 0.0025 ÷ 0.204 = 0.0122 L = 12.2 mL </li></ul>
    48. Types of Reactions <ul><li>4.8 Acid-Base </li></ul><ul><li>For our purposes an acid is a proton donor . </li></ul><ul><li>A base is a proton acceptor (usually OH) </li></ul><ul><li>What is the net ionic equation for the reaction of HCl(aq) and KOH(aq)? </li></ul><ul><li>Acid + Base  salt + water </li></ul><ul><li>H + + OH -  H 2 O </li></ul>
    49. 4.8 Acid - Base Reactions <ul><li>Often called a neutralization reaction because the acid neutralizes the base. </li></ul><ul><li>Often titrate to determine concentrations. </li></ul><ul><li>Solution of known concentration (titrant) . . . </li></ul><ul><li>is added to the unknown (analyte) . . . </li></ul><ul><li>until the equivalence point is reached where enough titrant has been added to neutralize it. </li></ul>
    50. Titration pp <ul><li>Where indicator changes color is endpoint . </li></ul><ul><li>Not always at the equivalence point. </li></ul><ul><li>A 50.00 mL sample of aqueous Ca(OH) 2 requires 34.66 mL of 0.0980 M Nitric acid for neutralization. What is [Ca(OH) 2 ]? </li></ul><ul><li># of H + x M A x V A = # of OH - x M B x V B </li></ul><ul><li>(look at formulas, not at the balanced reaction) </li></ul><ul><li>(1)(0.0980)(34.66) = (2)(50)(x) = 0.0340 M </li></ul><ul><li>but look at Zumdahl method page 150! </li></ul>
    51. Acid-Base Reaction pp <ul><li>75 mL of 0.25 M HCl is mixed with 225 mL of 0.055 M Ba(OH) 2 . What is the concentration of the excess H + or OH - ? We can do this in mmoles since the units will cancel. Use mL and mmoles is same as L and moles. Steps. . . </li></ul><ul><li>Mmole HCl = 75 mL • 0.25 M = 18.75 mmole; </li></ul><ul><li>Since 1:1 mole ratio of H + : HCl; H 1+ = 18.75 mmole </li></ul><ul><li>Mmole Ba(OH) 2 = 225 mL • 0.055 M = 12.37 mmole </li></ul><ul><li>2:1 mole ratio OH - : Ba(OH) 2 ; OH - = 24.75 mmole </li></ul><ul><li>H : OH mole ratio = 1:1; so 6.00 mmoles xs OH in (75 + 225) ml = 0.0200 M OH 1- </li></ul>
    52. Types of Reaction <ul><li>4.9 Oxidation-Reduction called Redox </li></ul><ul><li>Ionic compounds are formed through the transfer of electrons. </li></ul><ul><li>An Oxidation-reduction reaction involves the transfer of electrons. </li></ul><ul><li>We need a way of keeping track. </li></ul>
    53. Oxidation States - Memorize! <ul><li>A way of keeping track of the electrons. </li></ul><ul><li>Not necessarily true of what is in nature, but it works (i.e ., a model only). </li></ul><ul><li>Need the rules for assigning (memorize). </li></ul>
    54. Oxidation States - Memorize! <ul><li>Elements & Ions </li></ul><ul><li>The oxidation state of elements in their standard states is zero . Na (s) = 0; Cl 2(g) = 0, but O 2( s ) ≠ 0! </li></ul><ul><li>Oxidation states for mono atomic ions are the same as their charge. Na 1+ = 1; Cl 1- = 1 - ; N 3- = 3 - </li></ul>
    55. Oxidation states - memorize! <ul><li>Oxygen & Hydrogen </li></ul><ul><li>Oxygen is assigned an oxidation state of -2 in its covalent compounds (H 2 O) except as peroxide (H 2 O 2 ) where = -1 or with fluorine (OF 2 ) where = +2 </li></ul><ul><li>Hydrogen in compounds with: non metals is assigned +1 (H 2 O) with metals it is -1 (CaH 2 ) </li></ul>
    56. Oxidation states - memorize! <ul><li>fluorine is always -1 in compounds (HF). </li></ul><ul><li>The sum of the oxidation states must be: zero in compounds (Na 2 SO 4 ) or equal the charge of the ion (SO 4 2- ) </li></ul>
    57. Tr103 Assigning Oxidation Numbers What is the oxidation # of Mn in KMnO 4 ? 1•1 + 1•Mn + 4•-2 = 0; Mn = +7 What is the oxidation # of Cr in BaCr 2 O 7 ? 1•2 + 2•Cr + 7•-2 = 0; Cr = +6 What is the oxidation # of Cl in Sr(ClO 4 ) 2 ? 1•2 + 2*Cl + 8•-2 = 0; Cl = +7 ! Even though as an ion is would be Cl 1-
    58. Assign the oxidation states to each element <ul><li>CO 2 </li></ul><ul><li>NO 3 -1 </li></ul><ul><li>H 2 SO 4 </li></ul><ul><li>Fe 2 O 3 </li></ul><ul><li>Fe 3 O 4 </li></ul><ul><li>C = +4 ; O = -2 </li></ul><ul><li>N = +5 ; O = -2 </li></ul><ul><li>H = +1; S = +6 ; O = -2 </li></ul><ul><li>Fe = +3 ; O = -2 </li></ul><ul><li>Fe = +2.67 ; O = -2 </li></ul><ul><li>Remember: This isn’t real. It’s only a model to help us. </li></ul>
    59. Oxidation-Reduction <ul><li>Electrons transfer, so oxidation states change. </li></ul><ul><li>2Na + Cl 2  2NaCl </li></ul><ul><li>CH 4 + 2O 2  CO 2 + 2H 2 O </li></ul><ul><li>Oxidation is the loss of electrons (an increase in the oxidation state). </li></ul><ul><li>Reduction is the gain of electrons (a decrease in the oxidation state). </li></ul><ul><li>OIL RIG </li></ul><ul><li>LEO GER </li></ul><ul><li>Coefficients or subscripts do NOT change the oxidation values. </li></ul>
    60. Oxidation-Reduction <ul><li>Oxidation means an increase in oxidation state - lose electrons. </li></ul><ul><li>Na  Na 1+ + 1 e - </li></ul><ul><li>Reduction means a decrease in oxidation state - gain electrons. </li></ul><ul><li>2 e - + Cl 2  2Cl 1- </li></ul><ul><li>The substance that is oxidized is called the reducing agent . </li></ul><ul><li>The substance that is reduced is called the oxidizing agent . </li></ul>
    61. A Summary of an Oxidation-Reduction Process
    62. Agents <ul><li>Oxidizing agent - has the potential to cause another substance to get oxidized. </li></ul><ul><li>Other substance loses electrons , but . . . </li></ul><ul><li>In causing that other substance to get reduced the oxidizing agent gets reduced . </li></ul><ul><li>The oxidizing agent gains electrons (gets a more negative oxidation state). </li></ul><ul><li>MnO 4 1- will oxidize Fe 2+ to Fe 3+ so here it is an oxidizing agent. </li></ul><ul><li>But the Mn in MnO 4 1- gets reduced from +7 to +2 (MnO 4 1- + 5e -  Mn 2+ ) +7 +2 </li></ul>
    63. Agents <ul><li>Reducing agent - has the potential to cause another substance to get reduced. </li></ul><ul><li>Other substance gains electrons , but . . . </li></ul><ul><li>In causing that other substance to get reduced the reducing agent gets oxidized . </li></ul><ul><li>The reducing agent loses electrons (gets a more positive oxidation state). </li></ul><ul><li>Na will reduce Zn 2+ to Zn so here it is a reducing agent. </li></ul><ul><li>But the Na gets oxidized from 0 to +1 (Na  Na 1+ + 1e - ) </li></ul>
    64. Identify the <ul><li>Substance oxidized Substance reduced Oxidizing agent Reducing agent in the following reactions: </li></ul><ul><li>Fe( s ) + O 2 ( g )  Fe 2 O 3 ( s ) oxidation #s?… </li></ul><ul><li>0 0 +3 -2 O, R, OA, RA? </li></ul><ul><li>O R RA OA </li></ul>
    65. Identify the <ul><li>O, R, OA, RA in the following reactions: </li></ul><ul><li>Fe 2 O 3 ( s )+ 3 CO( g )  2 Fe( s ) + 3 CO 2 ( g ) </li></ul><ul><li>Fe 3+ reduced to Fe o so Fe 2 O 3 is the OA C 2+ oxidized to C 4+ so CO is the RA. </li></ul><ul><li>SO 3 2- + H + + MnO 4 1-  SO 4 2- + H 2 O + Mn +2 </li></ul><ul><li>Sulfur goes from +4 to +6 (from SO 3 2- to SO 4 2- ) so it is oxidized and its compound, SO 3 2- is the reducing agent which reduces Mn in MnO 4 1- from +7 to +2. </li></ul><ul><li>Mn is reduced & MnO 4 1- is the Oxidizing agent (it oxidizes S from +4 to +6). </li></ul>
    66. Half-Reactions <ul><li>All redox reactions can be thought of as happening in two halves. </li></ul><ul><li>One produces electrons - Oxidation half. </li></ul><ul><li>Other requires electrons - Reduction half. </li></ul><ul><li>We write the half-reactions to help us balance the equation. </li></ul><ul><li>Then we add the half-reactions to cancel out the electrons. </li></ul><ul><li>Every redox reaction has an oxidation half-reaction & a reduction half-reaction. </li></ul>
    67. Half-Reactions <ul><li>Write the half reactions for the following. </li></ul><ul><li>Na + Cl 2  Na + + Cl - Steps. . . </li></ul><ul><li>Na  Na 1+ + 1e- </li></ul><ul><li> 2e- + Cl 2  2Cl 1- </li></ul>
    68. Half-Reactions (more steps) <ul><li>Write the half reactions for: SO 3 -2 + H + + MnO 4 -  SO 4 -2 + H 2 O + Mn +2 (steps …) </li></ul><ul><li>1st, identify the elements that are changing their oxidation numbers, they are . . . </li></ul><ul><li>S from +4 to +6 so it is losing 2 e- (it is being oxidized so it is the reducing agent) </li></ul><ul><li>Mn from +7 to +2 so it is gaining 5 e- (being reduced so it is the oxidizing agent). </li></ul><ul><li>S O 3 -2 + H + + Mn O 4 -  S O 4 -2 + H 2 O + Mn +2 +4 +7 +6 +2 </li></ul><ul><li>So, the half reactions are the stuff with sulfur and manganese. </li></ul>
    69. Half-Reactions (more steps) <ul><li>S O 3 -2 + H + + Mn O 4 -  S O 4 -2 + H 2 O + Mn +2 +4 +7 +6 +7 S from +4 to +6 (losing) and . . . Mn from +7 to +2 (gaining) </li></ul><ul><li>The half-reactions are . . . </li></ul><ul><li>S O 3 -2  S O 4 -2 + 2 e- +4 +6 5 e- + Mn O 4 -  Mn +2 +7 +2 </li></ul>
    70. Not every reaction is redox <ul><li>Only reactions that change oxidation state (gain or lose electrons) are redox. </li></ul><ul><li>Is SO 2 + H 2 O  H 2 SO 3 a redox? </li></ul><ul><li>+4 -2 +1 -2 +1+4-2 No! </li></ul><ul><li>Is H 2 + CuO  Cu + H 2 O a redox? </li></ul><ul><li>0 +2 -2 0 +1 -2 Yes! </li></ul><ul><li>(Discuss Rainbow Matrix ) </li></ul>
    71. 4.10 Balancing Redox Equations <ul><li>In aqueous solutions the key is the number of electrons produced must be the same as those required . </li></ul><ul><li>For reactions in acidic solution: 8-step procedure. </li></ul><ul><li>Write separate half-reactions (assign oxidation numbers, temporarily delete substances that don’t change) </li></ul><ul><li>For each half reaction balance all reactants except H and O </li></ul><ul><li>Balance O using H 2 O </li></ul>
    72. Acidic Solution <ul><li>Balance H using H + </li></ul><ul><li>Balance charge using e - </li></ul><ul><li>Multiply equations to make electrons equal </li></ul><ul><li>Add equations and cancel identical species (cancel coefficients as needed) </li></ul><ul><li>Check that charges and elements are balanced. </li></ul>
    73. Example pp <ul><li>A breathalyzer uses potassium dichromate to test for ethanol because the orange potassium dichromate changes to the green chromium 3 + ion in the presence of alcohol. </li></ul><ul><li>Write and balance the Breathalyzer reaction given the following: </li></ul><ul><li>The reactants are K 2 Cr 2 O 7 , HCl and C 2 H 5 OH. </li></ul><ul><li>The products are CrCl 3 , CO 2 , KCl & H 2 O. </li></ul>
    74. Example pp <ul><li>Write and balance the Breathalyzer equation given the following: Reactants : K 2 Cr 2 O 7 , HCl and C 2 H 5 OH. Products : CrCl 3 , CO 2 , KCl & H 2 O. </li></ul><ul><li>Write the formula equation: </li></ul><ul><li>K 2 Cr 2 O 7 + HCl + C 2 H 5 OH  Cr Cl 3 + CO 2 + KCl + H 2 O </li></ul><ul><li>Write the ionic equation (solubility poem): </li></ul><ul><li>2K 1+ Cr 2 O 7 2- + H 1+ + Cl 1- + C 2 H 5 OH  Cr 3+ 3Cl 1- + CO 2 + K 1+ + Cl 1- + H 2 O </li></ul><ul><li>Assign oxidation numbers </li></ul><ul><li>2K 1+ Cr 2 O 7 2- + H 1+ + Cl 1- + C 2 H 5 OH  Cr 3+ 3Cl 1- + CO 2 + K 1+ + Cl 1- + H 2 O +1 +6 -2 +1 -1 -2+1 -2+1 +3 -1 +4 -2 +1 -1 +1 -2 </li></ul>
    75. Example pp <ul><li>Write the formula equation: K 2 Cr 2 O 7 + HCl + C 2 H 5 OH  Cr Cl 3 + CO 2 + KCl + H 2 O Write the ionic equation (solubility poem): 2K 1+ Cr 2 O 7 2- + H 1+ + Cl 1- + C 2 H 5 OH  Cr 3+ 3Cl 1- + CO 2 + K 1+ + Cl 1- + H 2 O Assign oxidation numbers 2K 1+ Cr 2 O 7 2- + H 1+ + Cl 1- + C 2 H 5 OH  Cr 3+ + 3Cl 1- + CO 2 + K 1+ + Cl 1- + H 2 O +1 +6 -2 +1 -1 -2 +1 -2+1 +3 -1 +4 -2 +1 -1 +1 -2 </li></ul><ul><li>Delete substances where no element changes its oxidation number: </li></ul><ul><li>2K 1+ Cr 2 O 7 2- + H 1+ Cl 1- + C 2 H 5 OH  Cr 3+ + 3Cl 1- + CO 2 + K 1+ + Cl 1- + H 2 O +1 +6 -2 +1 -1 -2 +1 -2+1 +3 -1 +4 -2 +1 -1 +1 -2 </li></ul><ul><li>Only those where oxidation number changes : Cr 2 O 7 2- + C 2 H 5 OH  Cr 3+ + CO 2 +6 -2 +3 +4 </li></ul>
    76. Balancing Redox Equations pp <ul><li>K 2 Cr 2 O 7 + HCl + C 2 H 5 OH  Cr Cl 3 + CO 2 + KCl + H 2 O </li></ul><ul><li>Only those where oxidation number changes : Cr 2 O 7 2- + C 2 H 5 OH  Cr 3+ + CO 2 +6 -2 +3 +4 </li></ul><ul><li>For reactions in acidic solution: 8-step procedure. </li></ul><ul><li>Write separate half-reactions </li></ul><ul><li>Cr 2 O 7 2-  Cr 3+ C 2 H 5 OH  CO 2 </li></ul>
    77. Balancing Redox Equations pp <ul><li>Cr 2 O 7 2-  Cr 3+ - C 2 H 5 OH  CO 2 </li></ul><ul><li>For each half-reaction balance all reactants except H and O </li></ul><ul><li>Cr 2 O 7 2-   Cr 3+ C 2 H 5 OH   CO 2 </li></ul>
    78. Balancing Redox Equations pp <ul><li>Cr 2 O 7 2-   Cr 3+ - C 2 H 5 OH   CO 2 </li></ul><ul><li>Balance O using H 2 O </li></ul><ul><li>Cr 2 O 7 2-  Cr 3+ + 7H 2 O 3 H 2 O + C 2 H 5 O H   C O 2 </li></ul><ul><li>Balance H using H 1+ </li></ul><ul><li>14H 1+ + Cr 2 O 7 2-  Cr 3+ + 7H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ </li></ul>
    79. Example pp <ul><li>14H 1+ + Cr 2 O 7 2-  Cr 3+ + 7H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ </li></ul><ul><li>Balance the charge using e - </li></ul><ul><li>6e - + 14H 1+ + Cr 2 O 7 2-  Cr 3+ + 7H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ + 12e - </li></ul><ul><li>Multiply equations to equalize the charges </li></ul><ul><li>2 (6e - + 14H 1+ + Cr 2 O 7 2-  Cr 3+ + 7H 2 O) 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ + 12e - </li></ul><ul><li>Get: 12 e - + 28 H 1+ + 2 Cr 2 O 7 2-   Cr 3+ + 14 H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ + 12e - </li></ul>
    80. Example pp <ul><li>12 e - + 28 H 1+ + 2 Cr 2 O 7 2-   Cr 3+ + 14 H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ + 12e - </li></ul><ul><li>Add equations and cancel identical species (cancel coefficients if needed) </li></ul><ul><li>12e - + 16 28H 1+ + 2Cr 2 O 7 2-  Cr 3+ + 11 14H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ + 12e - </li></ul><ul><li>16 H 1+ + 2Cr 2 O 7 2- + C 2 H 5 OH  CO 2 +  Cr 3+ + 11 H 2 O </li></ul>
    81. Acidic Solution pp <ul><li>16 H 1+ + 2 Cr 2 O 7 2- + C 2 H 5 OH   CO 2 +  Cr 3+ + 11 H 2 O </li></ul><ul><li>Combine the net ions to form the original compounds and check that everything is balanced. </li></ul><ul><li>The original compounds were: K 2 Cr 2 O 7 + H Cl + C 2 H 5 OH  Cr Cl 3 + CO 2 + KCl + H 2 O </li></ul><ul><li>So the balanced equation is . . .(write it out) </li></ul><ul><li>2 K 2 Cr 2 O 7 + 16 HCl + C 2 H 5 OH   CrCl 3 + 2 CO 2 + 4 KCl + 11 H 2 O </li></ul>
    82. Balance the following in acidic solution <ul><li>MnO 4 1 - + Fe +2  Mn +2 + Fe +3 Ans. . . </li></ul><ul><li>8H 1+ + MnO 4 - + 5Fe +2  Mn +2 + 5Fe +3 + 4H 2 O </li></ul><ul><li>Cu + NO 3 1-  Cu +2 + NO (g) Ans. . . </li></ul><ul><li>3Cu + 2NO 3 1- + 8H 1+  3Cu +2 + 2NO (g) + 4H 2 O </li></ul>
    83. More Practice if time <ul><li>The following reactions occur in aqueous solution. Balance them </li></ul><ul><li>Pb + PbO 2 + SO 4 -2  PbSO 4 (not done) </li></ul><ul><li>Mn +2 + NaBiO 3  Bi +3 + MnO 4 - (not done) </li></ul>
    84. Now for a tough one <ul><li>Fe(CN) 6 -4 + MnO 4 -  Mn +2 + Fe +3 + CO 2 + NO 3 - . . . </li></ul><ul><li>188 H + + 5 Fe(CN) 6 4- + 61 MnO 4 1-  94 H 2 O + 61 Mn 2+ + 5 Fe 3+ + 30 CO 2 + 30 NO 3 1- </li></ul>
    85. Basic Solution pp <ul><li>Do everything you would with acid, but we have to add a step because . . . </li></ul><ul><li>Because there is no H 1+ in basic solution. </li></ul><ul><li>So, add OH 1- sufficient to convert the H 1+ to water (OH 1- + H 1+  H 2 O) </li></ul><ul><li>Be sure to add the OH 1- to both sides of the reaction (conservation of mass law) </li></ul><ul><li>Cr(OH) 3 + OCl - + OH -  CrO 4 -2 + Cl - + H 2 O we’ll do this on the board to get . . . </li></ul><ul><li>2Cr(OH) 3 + 3OCl - + 4OH -  CrO 4 -2 + 3Cl - + 5H 2 O </li></ul>
    86. Basic Solution <ul><li>After balancing like it was in acidic solution, add OH 1- (to both sides) to cancel all the H 1+ in basic solution. </li></ul><ul><li>You try CrO 2 1- + ClO 1-  CrO 4 1- + Cl 1- in basic solution to get . . . </li></ul><ul><li>CrO 2 1- + 2ClO 1-  CrO 4 1- + 2Cl 1- </li></ul><ul><li>Try NO 2 1- + Al  NH 3 + AlO 2 1- in basic solution to get . . . </li></ul><ul><li>OH 1- + H 2 O + NO 2 1- + 2Al  NH 3 + 2AlO 2 1- </li></ul>
    87. Basic Solution - more problems if time <ul><li>Do everything you would with acid, but add one more step. </li></ul><ul><li>Add enough OH - to both sides to neutralize the H + (by forming water) </li></ul><ul><li>CrI 3 + Cl 2  CrO 4 - + IO 4 - + Cl - </li></ul><ul><li>Fe(OH) 2 + H 2 O 2  Fe(OH) - </li></ul>
    88. Redox Titrations <ul><li>Same as any other titration. </li></ul><ul><li>The permanganate ion is used often because it is its own indicator. MnO 4 - is purple, Mn +2 is colorless. When reaction solution remains clear, MnO 4 - is gone. (Know for AP lab portion) </li></ul><ul><li>Chromate ion (CrO 4 1- ) is also useful, but color change, orangish yellow to green, is harder to detect. </li></ul>
    89. Redox Titration Example pp <ul><li>The iron content of iron ore can be determined by titration with standard KMnO 4 solution. The iron ore is dissolved in excess HCl, and the iron reduced to Fe +2 ions. This solution is then titrated with KMnO 4 solution, producing Fe +3 and Mn +2 ions in acidic solution. </li></ul><ul><li>If it requires 41.95 mL of 0.205 M KMnO 4 to titrate a solution made with 6.128 g of iron ore, what percent of the ore was iron? Steps. . . </li></ul>
    90. Example pp <ul><li>Iron ore + xs HCl  iron reduced to Fe +2 . Then titrated w/ KMnO 4 solution  Fe +3 + Mn +2 ions in acidic solution. </li></ul><ul><li>41.95 mL of 0.205 M KMnO 4 titrates 6.128 g of iron ore , what percent is iron? </li></ul><ul><li>Write NR Fe 2+ reacting with MnO 4 1- 8H 1+ + 5Fe 2+ + MnO 4 1-  5Fe 3+ + Mn 2+ + 4H 2 0 </li></ul><ul><li>Do stoich (calc moles MnO 4 1- 1st) to get 0.0086 moles MnO 4 1- and 0.043 moles Fe 2+ </li></ul><ul><li>Find g of Fe 2+ = 0.043 mol x 55.847 g/mol = 2.40 g </li></ul><ul><li>(2.40 g Fe/6.128 g Fe ore ) x 100% = 39.2% </li></ul>

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