Question

Calculate the pH at the equivalence point when 10.0 mL of 0.100 M HC9H7O2 (Ka = 3.6 × 10–5) is titrated against 0.200 M KOH

Answer 1

for equivalence point,
moles of HC9H7O2 = mol of KOH
0.10 M * 10 ml = 0.2 M * V(KOH)
V(KOH) = 5 mL

at equivalence point, whole of HC9H7O2 will be converted to HC9H6O2 -
mol of HC9H6O2 - formed = 0.10 M * 10 mL = 1 mmol

total volume = 10 mL + 5 mL = 15 mL

[HC9H6O2 -] = number of mol / volume
= 1 mmol / 15 mL
= 0.067 M

Kb of HC9H6O2 - = 10^-14 / (3.6*10^-5)
= 2.78*10^-10

HC9H6O2 - + H2O <---------> HC9H7O2 + OH-
0.067 0 0 (initial)
0.067-x x x (at equilibrium)

Kb = x*x / (0.067-x)
since Kb is small, x will be small and it be ignored as compared to 0.067
2.78*10^-10 = x^2 / (0.067)
x = 4.31*10^-6 M

[OH-] = x = 4.31*10^-6 M
pOH = -log [OH-] = 5.4

pH = 14 - pOH
= 14 - 5.4
= 8.6

Answer: 8.6