Question

What is the pH of a solution that results from mixing together equal volumes of a 0.040 M solution of ammonia and a 0.020 M solution of hydrochloric acid? (pKa of NH4+ = 9.25)

Answer 1

Let assume that 1 mL of acid and base are mixed.

Given:

M(HCl) = 0.02 M

V(HCl) = 1 mL

M(NH3) = 0.04 M

V(NH3) = 1 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.02 M * 1 mL = 0.02 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.04 M * 1 mL = 0.04 mmol

We have:

mol(HCl) = 0.02 mmol

mol(NH3) = 0.04 mmol

0.02 mmol of both will react

excess NH3 remaining = 0.02 mmol

Volume of Solution = 1 + 1 = 2 mL

[NH3] = 0.02 mmol/2 mL = 0.01 M

[NH4+] = 0.02 mmol/2 mL = 0.01 M

They form buffer

use:

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

pH = pKa + log {[NH3]/[NH4+]}

= 9.25 + log {1*10^-2/1*10^-2}

= 9.25 + log (1)

= 9.25 + 0

= 9.25

Answer: 9.25