Question

Calculating the pit of a weak base titrated with a strong acid An analytical chemist is troting 79.6 ml. of a 0.08800 M solution of ammonia (NH,) with a 0.5400 M solution of HNO,. The pK, of ammonia is 4.74. Calculate the pH of the base solution after the chemist has added 9.0 mL of the HNO, solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO, solution added. Round your answer to 2 decimal places. D#- D ?

Answer 1

use:

pKb = -log Kb

4.74= -log Kb

Kb = 1.82*10^-5

Given:

M(HNO3) = 0.54 M

V(HNO3) = 9 mL

M(NH3) = 0.088 M

V(NH3) = 79.6 mL

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

mol(HNO3) = 0.54 M * 9 mL = 4.86 mmol

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

mol(NH3) = 0.088 M * 79.6 mL = 7.0048 mmol

We have:

mol(HNO3) = 4.86 mmol

mol(NH3) = 7.0048 mmol

4.86 mmol of both will react

excess NH3 remaining = 2.1448 mmol

Volume of Solution = 9 + 79.6 = 88.6 mL

[NH3] = 2.1448 mmol/88.6 mL = 0.0242 M

[NH4+] = 4.86 mmol/88.6 mL = 0.0549 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.82*10^-5

pKb = - log (Kb)

= - log(1.82*10^-5)

= 4.74

use:

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

= 4.74+ log {5.485*10^-2/2.421*10^-2}

= 5.095

use:

PH = 14 - pOH

= 14 - 5.0952

= 8.9048

Answer: 8.90