Question

1. (1 point) A buffer solution is prepared by adding 20.0 mL of 0.100 M sodium hydroxide solution to 50.0 mL of 0.100 M acetic acid solution. What is the pH of the buffer?

Answer 1

Given:
M(CH3COOH) = 0.1 M
V(CH3COOH) = 50 mL
M(NaOH) = 0.1 M
V(NaOH) = 20 mL


mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.1 M * 50 mL = 5 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 20 mL = 2 mmol


We have:
mol(CH3COOH) = 5 mmol
mol(NaOH) = 2 mmol

2 mmol of both will react

excess CH3COOH remaining = 3 mmol
Volume of Solution = 50 + 20 = 70 mL
[CH3COOH] = 3 mmol/70 mL = 0.0429M

[CH3COO-] = 2/70 = 0.0286M

They form acidic buffer
acid is CH3COOH
conjugate base is CH3COO-


Ka = 1.8*10^-5

pKa = - log (Ka)
= - log(1.8*10^-5)
= 4.745

use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.745+ log {2.857*10^-2/4.286*10^-2}
= 4.569


Answer: 4.57