Question

A solution of 0.1 M acetic acid (CH3COOH, pKa 4.76) was titrated with NaOH.

(a) What is the Ka of acetic acid?

(b) What is the starting pH of this solution? (Assume that at equilibrium, [HA] = 0.1 M.)

(c) What is the pH at the titration midpoint?

(d) At this midpoint, [CH3COO- ] = [CH3COOH]. Did [H+ ] increase or decrease when OH- was added (relative to the starting point) and how did this change occur?

(e) What is the Kw at the midpoint?

(f) what is [OH-] at the midpoint?

Answer 1

a) Ka = 10^-pKa  = 10^-4.76

Ka = 1.73 x 10^-5

b) initially only acid present so

pH = 1/2 [pKa - logC]

pH = 1/2 [4.76 - log 0.1]

pH = 2.88

c) at titration mid point

50 % acid becomes salt so

[acid] = [salt]

pH = pKa + log [salt] /[acid]

as [salt] = [acid]

pH = pKa

pH = 4.76

d) relative to starting point [H⁺] will decrease,

the added NaOH reacts with H⁺  so that decreasing in H⁺

e) at any point

Kw = 1.0 x 10^-14

at mid point

pH = 4.76

pOH = 14 - 4.76

pOH = 9.24

[OH^-] = 10^-pOH  = 10^-9.24

[OH^-] = 5.75 x 10^-10 M