Question 14.8: 50.00 mL of 1.000 M acetic acid, HC2H3O2, is titrated with 0...
50.00 mL of 1.000 M acetic acid, HC_{2}H_{3}O_{2}, is titrated with 0.8000 M NaOH.
a Find the pH of the solution before any base is added.
ANALYSIS
Information given: HC_{2}H_{3}O_{2} (HAc): V (0.05000 L), M (1.000) NaOH: M (0.8000)
Information implied: K_{a} for HAc
Asked for: pH before titration starts (no base added)
STRATEGY
1. This is simply determining the pH of a weak acid. Recall Example 13.7.
2. Let x = [H^+] = [Ac^-] at equilibrium. HAc at equilibrium = [HAc]_{o} – x. Make the assumption that x = [H^+] << [HAc]_{o}.
3. Substitute into Equation 13.5, solve for x, and check the assumption.
K_a = \frac{[H^+][B^-]}{[HB]} (13.5)
4. Find pH.
b Find the pH of the solution when half the acetic acid has been neutralized.
ANALYSIS
Information given: HC_{2}H_{3}O_{2}(HAc): V (0.05000 L), M (1.000) NaOH: M (0.8000)
Information implied: K_{a} for HAc
Asked for: pH at half-neutralization
c Find the pH of the solution at the equivalence point.
ANALYSIS
Information given: HC_{2}H_{3}O_{2} (HAc): V (0.05000 L), M (1.000); from part (a): mol HAc (0.05000 mol) NaOH: M (0.8000)
Information implied: K_{a} for HAc and K_{b}\> for\> Ac^{-}
Asked for: pH at the equivalence point
STRATEGY
1. Write the reaction for the titration.
2. Find the volume of NaOH required to reach the equivalence point.
For the titration: mol HAc = mol OH^-
3. At the equivalence point, all the acetic acid has been converted to acetate ions and mol Ac^- = mol HAc at the start. Find [Ac^-] .
[Ac^-] = \frac { mol\> Ac^-}{V_{HAc} + V_{OH^{-}}}
4. Find [OH^-] by substituting into Equation 13.8.
K_b = \frac{[HB] \times [OH^-]}{[B^-]} (13.8)
5. Find [H^+] and pH.
