Chapter 4 Lecture- Solution Stoich
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Chapter 4 Lecture- Solution Stoich

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Chapter 4 Lecture- Solution Stoich for AP chem

Chapter 4 Lecture- Solution Stoich for AP chem

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  • 1. Solutions:
    • Homogeneous mixtures of two or more pure substances.
    • The solvent is present in greatest abundance.
    • All other substances are solutes .
  • 2. Dissociation
    • When an ionic substance dissolves in water, the solvent pulls the individual ions from the crystal and solvates them.
    • This process is called dissociation .
  • 3. Electrolytes
    • Substances that dissociate into ions when dissolved in water.
    • A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.
  • 4. Electrolytes and Nonelectrolytes
    • Soluble ionic compounds tend to be electrolytes.
  • 5. Electrolytes and Nonelectrolytes
    • Molecular compounds tend to be nonelectrolytes, except for acids and bases.
  • 6. Electrolytes
    • A strong electrolyte dissociates completely when dissolved in water.
    • A weak electrolyte only dissociates partially when dissolved in water.
  • 7. Strong Electrolytes Are…
    • Strong acids
  • 8. Strong Electrolytes Are…
    • Strong acids
    • Strong bases
  • 9. Strong Electrolytes Are…
    • Strong acids
    • Strong bases
    • Soluble ionic salts
  • 10. Answers: C 6 H 12 O 6 (nonelectrolyte) < HC 2 H 3 O 2 (weak electrolyte, existing mainly in the form of molecules with few ions) < NaC 2 H 3 O 2 (strong electrolyte that provides two ions, and (strong electrolyte that provides three ions, Ca 2+ and 2 NO 3 – PRACTICE EXERCISE Consider solutions in which 0.1 mol of each of the following compounds is dissolved in 1 L of water: Ca(NO 3 ) 2 (calcium nitrate), C 6 H 12 O 6 (glucose), NaC 2 H 3 O 2 (sodium acetate), and HC 2 H 3 O 2 (acetic acid). Rank the solutions in order of increasing electrical conductivity, based on the fact that the greater the number of ions in solution, the greater the conductivity.
  • 11. Precipitation Reactions
    • When one mixes ions that form compounds that are insoluble (as could be predicted by the solubility guidelines), a precipitate is formed.
  • 12. PRACTICE EXERCISE Classify the following compounds as soluble or insoluble in water: (a) cobalt(II) hydroxide, (b) barium nitrate, (c) ammonium phosphate. Answers:   (a) insoluble, (b) soluble, (c) soluble
  • 13.
    • PRACTICE EXERCISE
    • What compound precipitates when solutions of Fe 2 (SO 4 ) 3 and LiOH are mixed?
    • (b) Write a balanced equation for the reaction.
    • (c) Will a precipitate form when solutions of Ba(NO 3 ) 2 and KOH are mixed?
    (c) no (both possible products are water soluble)
  • 14. Metathesis (Exchange) Reactions
    • Metathesis comes from a Greek word that means “to transpose”
    • AgNO 3 ( aq ) + KCl ( aq )  AgCl ( s ) + KNO 3 ( aq )
  • 15. Metathesis (Exchange) Reactions
    • Metathesis comes from a Greek word that means “to transpose”
    • It appears the ions in the reactant compounds exchange, or transpose, ions
    • Ag NO 3 ( aq ) + K Cl ( aq )  AgCl ( s ) + KNO 3 ( aq )
  • 16. Metathesis (Exchange) Reactions
    • Metathesis comes from a Greek word that means “to transpose”
    • It appears the ions in the reactant compounds exchange, or transpose, ions
    • Ag NO 3 ( aq ) + K Cl ( aq )  AgCl ( s ) + KNO 3 ( aq )
  • 17. Solution Chemistry
    • It is helpful to pay attention to exactly what species are present in a reaction mixture (i.e., solid, liquid, gas, aqueous solution).
    • If we are to understand reactivity, we must be aware of just what is changing during the course of a reaction.
  • 18. Molecular Equation
    • The molecular equation lists the reactants and products in their molecular form.
    • AgNO 3 ( aq ) + KCl ( aq )  AgCl ( s ) + KNO 3 ( aq )
  • 19. Ionic Equation
    • In the ionic equation all strong electrolytes (strong acids, strong bases, and soluble ionic salts) are dissociated into their ions.
    • This more accurately reflects the species that are found in the reaction mixture.
    • Ag + ( aq ) + NO 3 - ( aq ) + K + ( aq ) + Cl - ( aq ) 
    • AgCl ( s ) + K + ( aq ) + NO 3 - ( aq )
  • 20. Net Ionic Equation
    • To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right.
    • Ag + (aq) + NO 3 - ( aq ) + K + (aq) + Cl - ( aq ) 
    • AgCl ( s ) + K + (aq) + NO 3 - ( aq )
  • 21. Net Ionic Equation
    • To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right.
    • The only things left in the equation are those things that change (i.e., react) during the course of the reaction.
    • Ag + (aq) + Cl - ( aq )  AgCl ( s )
  • 22. Net Ionic Equation
    • To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right.
    • The only things left in the equation are those things that change (i.e., react) during the course of the reaction.
    • Those things that didn’t change (and were deleted from the net ionic equation) are called spectator ions .
    • Ag + (aq) + NO 3 - ( aq ) + K + (aq) + Cl - ( aq ) 
    • AgCl ( s ) + K + (aq) + NO 3 - ( aq )
  • 23. Writing Net Ionic Equations
    • Write a balanced molecular equation.
    • Dissociate all strong electrolytes.
    • Cross out anything that remains unchanged from the left side to the right side of the equation.
    • Write the net ionic equation with the species that remain.
  • 24. PRACTICE EXERCISE Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of silver nitrate and potassium phosphate are mixed.
  • 25. Acids:
    • Substances that increase the concentration of H + when dissolved in water (Arrhenius).
    • Proton donors (Brønsted – Lowry).
  • 26. Acids
    • There are only seven strong acids:
      • Hydrochloric (HCl)
      • Hydrobromic (HBr)
      • Hydroiodic (HI)
      • Nitric (HNO 3 )
      • Sulfuric (H 2 SO 4 )
      • Chloric (HClO 3 )
      • Perchloric (HClO 4 )
  • 27. Bases:
    • Substances that increase the concentration of OH − when dissolved in water (Arrhenius).
    • Proton acceptors (Brønsted – Lowry).
  • 28. Bases
    • The strong bases are the soluble salts of hydroxide ion:
      • Alkali metals
      • Calcium
      • Strontium
      • Barium
  • 29. Acid-Base Reactions
    • In an acid-base reaction, the acid donates a proton (H + ) to the base.
  • 30. Neutralization Reactions
    • Generally, when solutions of an acid and a base are combined, the products are a salt and water.
    • HCl ( aq ) + NaOH ( aq )  NaCl ( aq ) + H 2 O ( l )
  • 31. Neutralization Reactions
    • When a strong acid reacts with a strong base, the net ionic equation is…
    • HCl ( aq ) + NaOH ( aq )  NaCl ( aq ) + H 2 O ( l )
    • H + ( aq ) + Cl - ( aq ) + Na + ( aq ) + OH - ( aq ) 
    • Na + ( aq ) + Cl - ( aq ) + H 2 O ( l )
  • 32. Neutralization Reactions
    • When a strong acid reacts with a strong base, the net ionic equation is…
    • HCl ( aq ) + NaOH ( aq )  NaCl ( aq ) + H 2 O ( l )
    • H + ( aq ) + Cl - ( aq ) + Na + ( aq ) + OH - ( aq ) 
    • Na + ( aq ) + Cl - ( aq ) + H 2 O ( l )
    • H + ( aq ) + Cl - ( aq) + Na + ( aq ) + OH - ( aq ) 
    • Na + ( aq ) + Cl - ( aq ) + H 2 O ( l )
  • 33. PRACTICE EXERCISE (a) Write a balanced molecular equation for the reaction of carbonic acid (H 2 CO 3 ) and potassium hydroxide (KOH). (b) Write the net ionic equation for this reaction. (H 2 SO 3 is a weak acid and therefore a weak electrolyte, whereas KOH, a strong base, and K 2 CO 3 , an ionic compound, are strong electrolytes.)
  • 34. Gas-Forming Reactions
    • These metathesis reactions do not give the product expected.
    • The expected product decomposes to give a gaseous product (CO 2 or SO 2 ).
    • CaCO 3 ( s ) + HCl ( aq )  CaCl 2 ( aq ) + CO 2 ( g ) + H 2 O ( l )
    • NaHCO 3 ( aq ) + HBr ( aq )  NaBr ( aq ) + CO 2 ( g ) + H 2 O ( l )
    • SrSO 3 ( s ) + 2 HI ( aq )  SrI 2 ( aq ) + SO 2 ( g ) + H 2 O ( l )
  • 35. Gas-Forming Reactions
    • This reaction gives the predicted product, but you had better carry it out in the hood, or you will be very unpopular!
    • Just as in the previous examples, a gas is formed as a product of this reaction:
    • Na 2 S ( aq ) + H 2 SO 4 ( aq )  Na 2 SO 4 ( aq ) + H 2 S ( g )
  • 36. BELLWORK
    • Write the molecular equation, the complete ionic equation, and the net ionic equation for the reaction that occurs when the following are mixed. Identify the spectator ions.
    • HF(aq) + NaOH(aq) 
  • 37. Oxidation-Reduction Reactions
    • An oxidation occurs when an atom or ion loses electrons.
    • A reduction occurs when an atom or ion gains electrons.
  • 38. Oxidation-Reduction Reactions
    • One cannot occur without the other.
  • 39. Oxidation Numbers
    • To determine if an oxidation-reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity.
  • 40. Oxidation Numbers
    • Elements in their elemental form have an oxidation number of 0.
    • The oxidation number of a monatomic ion is the same as its charge.
  • 41. Oxidation Numbers
    • Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.
      • Oxygen has an oxidation number of − 2, except in the peroxide ion in which it has an oxidation number of − 1.
      • Hydrogen is − 1 when bonded to a metal, +1 when bonded to a nonmetal.
  • 42. Oxidation Numbers
    • Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.
      • Fluorine always has an oxidation number of − 1.
      • The other halogens have an oxidation number of − 1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions.
  • 43. Oxidation Numbers
    • The sum of the oxidation numbers in a neutral compound is 0.
    • The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.
  • 44. Oxidation Numbers
  • 45. Displacement Reactions
    • In displacement reactions, ions oxidize an element.
    • The ions, then, are reduced.
  • 46. Displacement Reactions
    • In this reaction,
    • silver ions oxidize
    • copper metal.
    • Cu ( s ) + 2 Ag + ( aq )  Cu 2+ ( aq ) + 2 Ag ( s )
  • 47. Displacement Reactions
    • The reverse reaction,
    • however, does not
    • occur.
    • Cu 2+ ( aq ) + 2 Ag ( s )  Cu ( s ) + 2 Ag + ( aq )
    x
  • 48. Activity Series
  • 49. Bellwork Question4-style #1
    • a)A solution of ethanoic (acetic) acid is added to a solution of barium hydroxide.
    • b)Explain why a mixture of equal volumes of equimolar solutions of acetic acid and barium hydroxide is basic.
  • 50. Molarity
    • Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different.
    • Molarity is one way to measure the concentration of a solution.
    moles of solute volume of solution in liters Molarity ( M ) =
  • 51. Mixing a Solution
  • 52. Dilution
  • 53. Answer:  0.278 M SAMPLE EXERCISE 4.11 PRACTICE EXERCISE Calculate the molarity of a solution made by dissolving 5.00 g of glucose (C 6 H 12 O 6 ) in sufficient water to form exactly 100 mL of solution.
  • 54. PRACTICE EXERCISE What is the molar concentration of K + ions in a 0.015 M solution of potassium carbonate? Answer:  0.030 M K +
  • 55. PRACTICE EXERCISE (a) How many grams of Na 2 SO 4 are there in 15 mL of 0.50 M Na 2 SO 4 ? (b) How many milliliters of 0.50 M Na 2 SO 4 solution are needed to provide 0.038 mol of this salt? Answers:  (a) 1.1 g, (b) 76 mL
  • 56. PRACTICE EXERCISE (a) What volume of 2.50 M lead(II)nitrate solution contains 0.0500 mol of Pb 2+ ? (b) How many milliliters of 5.0 M K 2 Cr 2 O 7 solution must be diluted to prepare 250 mL of 0.10 M solution? (c) If 10.0 mL of a 10.0 M stock solution of NaOH is diluted to 250 mL, what is the concentration of the resulting stock solution? Answers:  (a) 0.0200 L = 20.0 mL, (b) 5.0 mL, (c) 0.40 M
  • 57. Using Molarities in Stoichiometric Calculations
  • 58. PRACTICE EXERCISE (a) How many grams of NaOH are needed to neutralize 20.0 mL of 0.150 M H 2 SO 4 solution? (b) How many liters of 0.500 M HCl( aq ) are needed to react completely with 0.100 mol of Pb(NO 3 ) 2 ( aq ), forming a precipitate of PbCl 2 ( s )? Answers:  (a) 0.240 g, (b) 0.400 L
  • 59. Titration
    • The analytical technique in which one can calculate the concentration of a solute in a solution.
  • 60. Titration
  • 61. PRACTICE EXERCISE What is the molarity of an NaOH solution if 48.0 mL is needed to neutralize 35.0 mL of 0.144 M H 2 SO 4 ? Answers:  0.210 M