Question

Given a titration between 30.00 mL of 0.400 M acid, HA (Ka = 8.0 x 10–6), and the titrant sodium hydroxide, NaOH, whose concentration is 0.300 M. Calculate the pH of the resulting solution at the following points of the titration curve: 40.00 mL of the titrant, NaOH, have been added.

Answer 1

30.00 mL of 0.400 M acid, HA (Ka = 8.0 x 10–6),

Base NaOH, whose concentration is 0.300 M and volume = 40.00 mL

This is a equivalent point ,
all HA moles will have been converted
to NaA, so find pH using Kb:
Find resulting concentration of NaA, conj. base:
0.030 L x 0.4 M = 0.012 moles HA = moles NaA
0.012 moles/ (0.040L + 0.030 L) = 0.171

Ka = 8.0 x 10–6

pKa = -log(Ka)

Pka = 5.1
14 - pKa = pKb
14 – 5.1 = 8.9= pKb
10^-pKb = Kb
10^-8.9 = 1.26 x 10^-9

Kb = x^2/0.171
1.26*10^-9 = x^2/0.171-x
Ignoring -x,

x^2 = 2.15 x 10^-10
x = [OH-] = 1.46 x 10^-5
pOH = -log[OH-] = 4.8
pH = 14 – 4.8= 9.2