Question

What pH of the solution will be if exactly 12.50mL of 0.01 M NaOH solution is added to 25.0 mL of 0.01M of acetic acid solution?

Answer 1

Given:

M(CH3COOH) = 0.01 M

V(CH3COOH) = 25 mL

M(NaOH) = 0.01 M

V(NaOH) = 12.5 mL

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

mol(CH3COOH) = 0.01 M * 25 mL = 0.25 mmol

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

mol(NaOH) = 0.01 M * 12.5 mL = 0.125 mmol

We have:

mol(CH3COOH) = 0.25 mmol

mol(NaOH) = 0.125 mmol

0.125 mmol of both will react

excess CH3COOH remaining = 0.125 mmol

Volume of Solution = 25 + 12.5 = 37.5 mL

[CH3COOH] = 0.125 mmol/37.5 mL = 0.0033M

[CH3COO-] = 0.125/37.5 = 0.0033M

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 {3.333*10^-3/3.333*10^-3}

= 4.745

Answer: 4.745