Question 14.4: Consider the buffer described in Example 14.1, where nHLac =...
Consider the buffer described in Example 14.1, where n_{HLac} = n_{Lac^-} = 1.00 \>mol (K_{a} HLac = 1.4 \times 10^{-4}). You will recall that in this buffer the pH is 3.85.
a Calculate the pH after addition of 0.08 mol of HCl.
Information given: mol HLac (1.00); mol Lac^- (1.00); mol HCl = mol H^+ (0.08) pH of the buffer (3.85) K_{a} for HLac (1.4 \times 10^{-4})
Asked for: pH of the buffer after the addition of acid
STRATEGY
1. Write the reaction between the strong acid H^+ and the conjugate base, Lac^-.
2. Adding H^+ uses up the conjugate base in a 1:1 stoichiometric ratio.
mol Lac^- after addition = mol Lac^- – mol H^+
3. Adding H^+ produces more weak acid in a 1:1 stoichiometric ratio.
mol HLac after addition = mol HLac + mol H^+
4. Substitute into Equation 14.2 to find [H^+] and pH.
[H^+]= K_{a} \times \frac{n_{HB}}{n_{B^-}}\blacktriangleleft ⓘ (14.2)
ⓘIf you start with 0.20 mol HB and 0.10 mol B^-, [H^+] = 2K_{a}.
b Calculate the pH after addition of 0.08 mol of NaOH.
Information given: mol HLac (1.00); mol Lac^- (1.00); mol NaOH = mol OH^- (0.08) pH of the buffer (3.85) K_{a}for HLac (1.4 \times 10^{-4})
Asked for: pH of the buffer after the addition of strong base
STRATEGY
1. Write the reaction between the strong base OH^- and the weak acid, HLac.
2. Adding OH^- uses up the weak acid in a 1:1 stoichiometric ratio.
mol HLac after addition = mol HLac – mol OH^-
3. Adding OH^- produces more conjugate base in a 1:1 stoichiometric ratio.
mol Lac^- after addition = mol Lac^- + mol OH^-
4. Substitute into Equation 14.2 to find [H^+] and pH.
