Question

An analytical chemist is titrating 142.0 mL of a 0.3500 M solution of acetic acid (HCH3CO2) with a 1.100 M solution of KOH. the pKa of acetic acid is 4.70. Calculate the pH of the acid solution after the chemist has added 49.90 mL of the KOH solution to it. Please round your answer to 2 decimal places

Answer 1

Given molarity of acetic acid=0.35 M=0.35 mol/L and volume=142 mL=0.142 L. (Since 1 L=1000 mL)

Moles of acetic acid=molarity x volume=0.35 mol/L x 0.142 L=0.0497 mol.

And molarity of KOH=1.1 M=1.1 mol/L and volume=49.90 L=0.0499 L.

Moles of KOH=1.1 mol/ L x 0.0499 L=0.05489 mol.

The neutralization reaction is

CH3COOH+KOH -----> CH3COOK + H2O.

Therefore moles of acid =moles of base or moles of H^+ =moles of OH^-.

But here moles of KOH (OH^-) are more.

Therefore moles of OH^- remaining=0.05489 mol - 0.0497 mol=0.00519 mol OH^-.

Total volume=142 mL+49.90 mL=191.9 mL=0.1019 L

Therefore [OH^-]=moles /volume=0.00519mol/0.1919 L

[OH^-]=0.027 M

Since KOH is a strong acid then pOH=-log[OH^-]=- log(0.027)

pOH=1.567

Then pH=14-pOH=14-1.567

pH=12.43.

Please let me know if you have any doubt. Thanks.