Question

A 50.00 mL aliquot of a 0.1000 M propenoic acid solution, H₂CCHCO₂H, is titrated with 0.1250 M NaOH. Calculate the pH when 25.00 mL of NaOH is added. K_a = 5.52 × 10⁻⁵

pKa= -log(Ka)

So pKa = 4.2581

Propenoic Acid= 0.1*0.05= 0.005 mols

NaOH= 0.125*0.025= 0.003125 mols

pH= 4.2581 + log(0.003125/0.005)

pH=4.05

Is this answer correct?

Answer 1

Given:

M(C2H5COOH) = 0.1 M

V(C2H5COOH) = 50 mL

M(NaOH) = 0.125 M

V(NaOH) = 25 mL

mol(C2H5COOH) = M(C2H5COOH) * V(C2H5COOH)

mol(C2H5COOH) = 0.1 M * 50 mL = 5 mmol

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

mol(NaOH) = 0.125 M * 25 mL = 3.125 mmol

We have:

mol(C2H5COOH) = 5 mmol

mol(NaOH) = 3.125 mmol

3.125 mmol of both will react

excess C2H5COOH remaining = 1.875 mmol

Volume of Solution = 50 + 25 = 75 mL

[C2H5COOH] = 1.875 mmol/75 mL = 0.025M

[C2H5COO-] = 3.125/75 = 0.0417M

They form acidic buffer

acid is C2H5COOH

conjugate base is C2H5COO-

Ka = 5.52*10^-5

pKa = - log (Ka)

= - log(5.52*10^-5)

= 4.258

use:

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

= 4.258+ log {4.167*10^-2/2.5*10^-2}

= 4.48

Answer: 4.48