Question

Calculate the pH of a buffer consisting of 50.00 mL of 0.12 M NH3 and 3.50 mL of 1.0 M HCl. (Given: Kb for NH3 = 1.76 x 10^(-5).

I'm mostly confused because it doesn't mention a conjugate acid for NH3, and HCl isn't its conjugate acid.

Answer 1

Given:

M(HCl) = 1 M

V(HCl) = 3.5 mL

M(NH3) = 0.12 M

V(NH3) = 50 mL

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

mol(HCl) = 1 M * 3.5 mL = 3.5 mmol

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

mol(NH3) = 0.12 M * 50 mL = 6 mmol

We have:

mol(HCl) = 3.5 mmol

mol(NH3) = 6 mmol

3.5 mmol of both will react

excess NH3 remaining = 2.5 mmol

Volume of Solution = 3.5 + 50 = 53.5 mL

[NH3] = 2.5 mmol/53.5 mL = 0.0467 M

[NH4+] = 3.5 mmol/53.5 mL = 0.0654 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.76*10^-5

pKb = - log (Kb)

= - log(1.76*10^-5)

= 4.754

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.754+ log {6.542*10^-2/4.673*10^-2}

= 4.901

use:

PH = 14 - pOH

= 14 - 4.9006

= 9.0994

Answer: 9.10