Question

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 25.0 mL of HNO3.

Answer 1

Given:

M(HNO3) = 0.5 M

V(HNO3) = 25 mL

M(NH3) = 0.2 M

V(NH3) = 75 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.5 M * 25 mL = 12.5 mmol

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

mol(NH3) = 0.2 M * 75 mL = 15 mmol

We have:

mol(HNO3) = 12.5 mmol

mol(NH3) = 15 mmol

12.5 mmol of both will react

excess NH3 remaining = 2.5 mmol

Volume of Solution = 25 + 75 = 100 mL

[NH3] = 2.5 mmol/100 mL = 0.025 M

[NH4+] = 12.5 mmol/100 mL = 0.125 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {0.125/2.5*10^-2}

= 5.444

use:

PH = 14 - pOH

= 14 - 5.4437

= 8.56

Answer: 8.56