Question

A 300.0 mL buffer solution is 0.240 M in acetic acid and 0.240 M in sodium acetate.

Part A

What is the initial pH of this solution?

Part B

What is the pH after addition of 0.0150 mol of HCl?

Part C

What is the pH after addition of 0.0150 mol of NaOH?

Answer 1

A) pH = pKa + log [sodium acetate] / [acetic acid]

pKa of acetic acid = 4.74

pH = 4.74 + log [0.240] / [0.240]

pH = 4.74

B) Molarity of HCl added = 0.0150 / 0.3 = 0.05M

after HCl added

[acetic acid] = 0.240 + 0.05 =0.29 M

[sodium acetate] = 0.240 - 0.05 = 0.19 M

pH = 4.74 + log [0.19] / [0.29]

pH = 4.56

C) molarity of NaOH added = 0.015 / 0.3 = 0.05 M

after NaOH added   

[acetic acid] = 0.240 - 0.05 = 0.19 M

[sodium acetate] = 0.240 + 0.05 = 0.29 M

pH = 4.74 + log [0.29] / [0.19]

pH = 4.92