Question

Calculate the pH in the titration of 45.0 mL of 0.350 M accetic acid by sodium hydroxide after the addition to the acid solution of 10.5 mL of 0.250 M NaOH. The ka of acetic acid is 1.8 x 10^-5

Answer 1

Given:

M(CH3COOH) = 0.35 M

V(CH3COOH) = 45 mL

M(NaOH) = 0.25 M

V(NaOH) = 10.5 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.35 M * 45 mL = 15.75 mmol

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

mol(NaOH) = 0.25 M * 10.5 mL = 2.625 mmol

We have:

mol(CH3COOH) = 15.75 mmol

mol(NaOH) = 2.625 mmol

2.625 mmol of both will react

excess CH3COOH remaining = 13.125 mmol

Volume of Solution = 45 + 10.5 = 55.5 mL

[CH3COOH] = 13.125 mmol/55.5 mL = 0.2365M

[CH3COO-] = 2.625/55.5 = 0.0473M

They form acidic buffer

acid is CH3COOH

conjugate base is CH3COO-

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {4.73*10^-2/0.2365}

= 4.046

Answer: 4.05