Question

Sodium hydroxide is used to titrate 50.0 mL of 0.100 M benzoic acid.(Ka6.3 x10-5) What is the pH after addition of 10.0mL of 0.200M NaOH?

Answer 1

Given:

M(C6H6COOH) = 0.1 M

V(C6H6COOH) = 50 mL

M(NaOH) = 0.2 M

V(NaOH) = 10 mL

mol(C6H6COOH) = M(C6H6COOH) * V(C6H6COOH)

mol(C6H6COOH) = 0.1 M * 50 mL = 5 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.2 M * 10 mL = 2 mmol

We have:

mol(C6H6COOH) = 5 mmol

mol(NaOH) = 2 mmol

2 mmol of both will react

excess C6H6COOH remaining = 3 mmol

Volume of Solution = 50 + 10 = 60 mL

[C6H6COOH] = 3 mmol/60 mL = 0.05M

[C6H5COO-] = 2/60 = 0.0333M

They form acidic buffer

acid is C6H6COOH

conjugate base is C6H5COO-

Ka = 6.3*10^-5

pKa = - log (Ka)

= - log(6.3*10^-5)

= 4.201

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.201+ log {3.333*10^-2/5*10^-2}

= 4.025

Answer: 4.03