Question

A chemist titrates 200.0mL of a 0.7669M ammonia NH3 solution with 0.4750M HBr solution at 25°C . Calculate the pH at equivalence. The pKb of ammonia is 4.75 . Round your answer to 2 decimal places.

Answer 1

use:

pKb = -log Kb

4.75= -log Kb

Kb = 1.778*10^-5

find the volume of HBr used to reach equivalence point

M(NH3)*V(NH3) =M(HBr)*V(HBr)

0.7669 M *200.0 mL = 0.475M *V(HBr)

V(HBr) = 322.9053 mL

At half equivalence point, volume of HBr required will be half of the above value

So, volume of HBr = 322.90526315789475/2 mL

= 161.45263157894738 mL

Given:

M(HBr) = 0.475 M

V(HBr) = 161.4526 mL

M(NH3) = 0.7669 M

V(NH3) = 200 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.475 M * 161.4526 mL = 76.69 mmol

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

mol(NH3) = 0.7669 M * 200 mL = 153.38 mmol

We have:

mol(HBr) = 76.69 mmol

mol(NH3) = 153.38 mmol

76.69 mmol of both will react

excess NH3 remaining = 76.69 mmol

Volume of Solution = 161.4526 + 200 = 361.4526 mL

[NH3] = 76.69 mmol/361.4526 mL = 0.2122 M

[NH4+] = 76.69 mmol/361.4526 mL = 0.2122 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.778*10^-5

pKb = - log (Kb)

= - log(1.778*10^-5)

= 4.75

use:

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

= 4.75+ log {0.2122/0.2122}

= 4.75

use:

PH = 14 - pOH

= 14 - 4.75

= 9.25

Answer: 9.25