CHEM 1P81 Assignment 9 Hints

Hints for Assignment 9, CHEM 1P81

Acid-Base Reactions and Titrations

1 and 2. Hints. See Chapter 17.1 and the examples in it. Exercise 17.1 has similar questions and the answers are in the back of the book.

3. Example. A solution of 0.300 M HCl (25.0 mL) is mixed with 35.0 mL of 0.250 M NaOH. Assume that the final volume is the sum of the initial volumes, calculate: (a) the molarity of the Na⁺ cation; (b) the molarity of the Cl^-anion; (c) the pH of the final solution; (d) the pOH of the final solution.

Solution. The first thing to recognize is that a strong acid (HCl) is reacting with a strong base (NaOH) to produce a salt and water. The equation for the reaction is
HCl (aq) + NaOH (aq) ---> NaCl (aq) + H₂O

Strong acids, strong bases, and salts dissociate completely into their ions in aqueous solution. Thus the net ionic equation is

H⁺ (aq) + OH^- (aq) ---> H₂O

Next, you need to calculate the moles of the two reactants. Then determine which reactant (if either) is the limiting reactant, and calculate the concentrations of the products and reactants after the reactant has occurred.
moles H⁺ = moles Cl^- = 0.00750 moles
moles Na⁺ = moles OH^- = 0.00875 moles
Final volume = 25.0 mL + 35.0 mL = 60.0 mL = 0.0600 L
Thus H⁺ is limiting. All of it will react and leave excess hydroxide in solution, so the solution will be basic.
Final Cl^- concentration = 0.00750 mol/0.0600 L = 0.125 M
Final Na⁺ concentration = 0.00875 mol/0.0600 L = 0.146 M
Final OH^- moles = (0.00875 mol OH^- initially) - (0.00750 mol to react with 0.00750 mol H⁺) = 0.00125 mol
Final OH^- concentration = 0.0125 mol/0.0600 L = 0.208 M
pOH = 0.68; pH = 13.32

4 (a). Example. Calculate the pH of a buffer solution made by dissolving 5.00 g of sodium carbonate and 5.00 g of sodium hydrogen carbonate in 100 mL of water. K_a2 for carbonic acid is 4.8 x 10^-11.

Partial solution. The key is to recognize what constant you are being given. The constant K_a2 refers to the second dissociation of carbonic acid,
H₂O + HCO₃^- ---> H₃O⁺ + CO₃^2-

K_a2 = 4.8 x 10^-11 = [H₃O⁺] [CO₃^2-] / [HCO₃^-]

So all you need to do is to calculate the molarity of the carbonate and hydrogen carbonate ions, and solve for the hydronium concentration.

4(b). Hint 1. Write the chemical equation for which the dissociation constant is given. K_a's refer to weak acids reacting with water to give their conjugate base and the hydronium ion; K_b's refer to weak bases reacting with water to give their conjugate acid and the hydroxide ion.

Hint 2. Calculate the moles of each buffer component from their respective molarities and volumes before mixing. Then recalculate the molarity of the each buffer component from its moles and the final volume (which is assumed to be the sum of the volumes of the two solutions).

5. Example. The weak monoprotic acid lactic acid, CH₃CHOHCO₂H (K_a = 8.40 x 10^-4) is titrated with the strong base NaOH. Suppose that 32.74 mL of 0.1328 M NaOH is required to reach the equivalence point when 40.00 mL of lactic acid is titrated.

(a) What is the original molarity of the lactic acid?
(b) What is the concentration of the lactate ion when the equivalence point is reached?
(c) What is the concentration of the hydroxide ion when the equivalence point is reached?
(d) What is the pH at the equivalence point of the reaction?

Overview. There are several calculations involved here. In part (a) you are doing a standard kind of analysis in which you find (by titration) the concentration of a solution whose molarity is unknown. Then you make use of this data to find out the pH at the equivalence point of the titration (part d). Parts (b) and (c) are essential steps on the way to getting the answer to part (d).

Solution hint (part a). The acid is monoprotic; that means it has one ionizable (titratable) hydrogen ion. Thus the equation for the reaction is
CH₃CHOHCO₂H + OH^- ---> CH₃CHOHCO₂^- + H₂O

The equation shows that each mole of acid requires a mole of base for titration. Thus moles acid present = moles base for titration. You know the molarity and volume of the base, and the volume of the acid solution initially. Thus the initial molarity of the acid is 0.1087 M.
Solution hint (part b). Now you know the molarity of the acid initially from part (a), so you can calculate the number of moles of acid initially present.
When titration is complete, you know that all lactic acid has been converted to lactate so you know the number of moles of lactate at the end of titration. You also know the final volume at the end of the titration (assuming the volumes are additive). Calculate the molarity of lactate from these two quantities. (= 0.05977 M).
Solution hint (parts c and d). To get the equivalence point, you want to calculate the pH when lactic acid has been exactly titrated by base and all you have in solution is sodium lactate at a concentration of 0.1087 M. The lactate ion, being the conjugate base of a weak acid, will hydrolyze according to the equation

CH₃CHOHCO₂^- + H₂O ---> CH₃CHOHCO₂H + OH^-

The K_b for lactate ion is K_w/K_a = 1.19 x 10^-11 (the dissociation constant for water is the product of the acid dissociation constant and the dissociation constant for the conjugate base of the acid). See Example 17.9 in Kotz and Treichel for calculating the pH for an aqueous solution of a salt of a weak acid. Answer 7.93 for pH at the equivalence point.

6. Hints. This problem also has many different parts, just like the previous one. All calculations are based on knowing how to make use of 3 equations: the reaction of a base with an acid (stoichiometry), the base dissociation constant (equilibrium), and the conjugate acid dissociation constant (equilibrium).

Sample question. You have 15.00 mL of a 0.1000 M aqueous solution of the weak monoprotic base B (K_b = 7.00 x 10^-7). This solution will be titrated with 0.1000 M HCl.

(a) How many mL of acid must be added to reach the equivalence point?

(b) What is the pH of the solution before any acid is added?

(d) What is the pH of the solution at the equivalence point of the titration?

(e) What is the pH of the solution when 20.00 mL of acid has been added?

(a) Hint. This is a stoichiometry question. A monoprotic base can accept only one proton. Thus it reacts with acid in a 1:1 mole ratio:
B + HCl ---> BH⁺ + Cl^- (Reaction of base with acid; stoichiometry)

So do (a) in the following order: Calculate the moles of base present, then calculate the number of moles of acid needed to react with it, then calculate the volume of acid needed given that you know the molarity. ANSWER 15.00 mL 0.1000 M HCl to titrate to the equivalence point.
(b) Hint. This is an equilibrium calculation involving the pH of a solution of a weak base. Remember the equilibrium expression for a weak base reacting with water:

B + H₂O ---> BH⁺ + OH^- (Base dissociation; equilibrium)

ANSWER: pH = 10.42
(c) Hint. You know from your answer to part (a) that you haven't reached the equivalence point. So you have a solution that contains some unreacted base and the salt (conjugate acid) of the base. This is a buffer calculation. Calculate how many moles of the base and its conjugate acid are present, then determine molarity.
ANSWER: pH = 7.22
(d) Hint. At equivalence, the conjugate acid of the base is the dominant species. What is its concentration? Let it hydrolyze and calculate the pH.

BH⁺ + H₂O ---> B + H₃O⁺ (Hydrolysis of conjugate acid; equilibrium)

ANSWER: 4.57
(e) Hint. Now you are beyond the equivalence point and there is only excess HCl. How much is in excess? Calculate its molarity, then determine the pH.
ANSWER: 1.85

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Last revised: March 28 2001 by M. F. Richardson
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