Question

An analytical chemist is titrating 200.3mLof a 0.7600M solution of acetic acid HCH3CO2 with a 0.7800M solution of KOH. The pKa of acetic acid is 4.70 Calculate the pH of the acid solution after the chemist has added 53.22mL of the KOH solution to it

Answer 1

Given:

M(HCH3CO2) = 0.76 M

V(HCH3CO2) = 200.3 mL

M(KOH) = 0.78 M

V(KOH) = 53.22 mL

mol(HCH3CO2) = M(HCH3CO2) * V(HCH3CO2)

mol(HCH3CO2) = 0.76 M * 200.3 mL = 152.228 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.78 M * 53.22 mL = 41.5116 mmol

We have:

mol(HCH3CO2) = 152.228 mmol

mol(KOH) = 41.5116 mmol

41.5116 mmol of both will react

excess HCH3CO2 remaining = 110.7164 mmol

Volume of Solution = 200.3 + 53.22 = 253.52 mL

[HCH3CO2] = 110.7164 mmol/253.52 mL = 0.4367M

[CH3CO2-] = 41.5116/253.52 = 0.1637M

They form acidic buffer

acid is HCH3CO2

conjugate base is CH3CO2-

use:

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

= 4.7+ log {0.1637/0.4367}

= 4.274

Answer: 4.27