Question

A 45.0 ml sample of 1.20 M benzoic acid is mixed with 20.0 mL 1.80 M NaOH solution. What is the PH of this solution? The Ka of benzoic acid is 4.50x 10 ^-4

Answer 1

Given:

M(C6H5COOH) = 1.2 M

V(C6H5COOH) = 45 mL

M(NaOH) = 1.8 M

V(NaOH) = 20 mL

mol(C6H5COOH) = M(C6H5COOH) * V(C6H5COOH)

mol(C6H5COOH) = 1.2 M * 45 mL = 54 mmol

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

mol(NaOH) = 1.8 M * 20 mL = 36 mmol

We have:

mol(C6H5COOH) = 54 mmol

mol(NaOH) = 36 mmol

36 mmol of both will react

excess C6H5COOH remaining = 18 mmol

Volume of Solution = 45 + 20 = 65 mL

[C6H5COOH] = 18 mmol/65 mL = 0.2769M

[C6H5COO-] = 36/65 = 0.5538M

They form acidic buffer

acid is C6H5COOH

conjugate base is C6H5COO-

Ka = 4.5*10^-4

pKa = - log (Ka)

= - log(4.5*10^-4)

= 3.3468

use:

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

= 3.3468+ log {0.5538/0.2769}

= 3.65

pH = 3.65