Question

Calculate the pH when 15.0 mL of 0.20 M benzoic acid (HC7H5O2, Ka = 6.5 ×10–5) is titrated with 20.0 mL of 0.15 M sodium hydroxide (NaOH). The answer is 8.56 but I do not know how to get to that point.

Answer 1

moles of C6H5COOH added= 0.20 M * 15.0 mL = 3 mmol
mol of NaOH added = 20.0 mL * 0.15 M = 3 mmol

whole of C6H5COOH will be converted to C6H5COONa
mol of C6H5COONa formed = 3 mmol

total volume = 15 mL + 20 mL = 35 mL

[C6H5COO-] = number of mol / volume
= 3 mmol / 35 mL
= 0.0857 M

Kb of C6H5COO- = 10^-14 / (6.5*10^-5)
= 1.54*10^-10

C6H5COO- + H2O <---------> C6H5COOH + OH-
0.0857 0 0 (initial)
0.0857-x x x (at equilibrium)

Kb = x*x / (0.0857-x)
since Kb is small, x will be small and it be ignored as compared to 0.0857
1.54*10^-10 = x^2 / (0.0857)
x = 3.63*10^-6 M

[OH-] = x = 3.63*10^-6 M
pOH = -log [OH-]
= -log (3.63*10^-6)
= 5.44

pH = 14 - pOH
= 14 - 5.44
= 8.56

Answer: 8.56