A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of

Question

zalisa [80]

3 years ago

11

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of

NaOH.

Chemistry

1 answer:

Rus_ich [418] · Answer 1 · 2022-06-13T16:01:17+03:00

52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.

31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.

After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:


pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) 
= -log(1.8x10^-5*5.8/12.4) = 5.07