Question

An analytical chemist is titrating 221.6 ml of a 0.2800 M solution of dimethylamine (CH), NH) with a 0.2700 M solution of HNO, The pK, of dimethylamine is 3.27. Calculate the pH of the base solution after the chemist has added 190.6 ml of the HINO, solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HINO, solution added. Round your answer to 2 decimal places. pH = 0 x 5 ?

Answer 1

use:

pKb = -log Kb

3.27= -log Kb

Kb = 5.37*10^-4

Given:

M(HNO3) = 0.27 M

V(HNO3) = 190.6 mL

M((CH3)2NH) = 0.28 M

V((CH3)2NH) = 221.6 mL

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

mol(HNO3) = 0.27 M * 190.6 mL = 51.462 mmol

mol((CH3)2NH) = M((CH3)2NH) * V((CH3)2NH)

mol((CH3)2NH) = 0.28 M * 221.6 mL = 62.048 mmol

We have:

mol(HNO3) = 51.462 mmol

mol((CH3)2NH) = 62.048 mmol

51.462 mmol of both will react

excess (CH3)2NH remaining = 10.586 mmol

Volume of Solution = 190.6 + 221.6 = 412.2 mL

[(CH3)2NH] = 10.586 mmol/412.2 mL = 0.0257 M

[(CH3)2NH2+] = 51.462 mmol/412.2 mL = 0.1248 M

They form basic buffer

base is (CH3)2NH

conjugate acid is (CH3)2NH2+

Kb = 5.37*10^-4

pKb = - log (Kb)

= - log(5.37*10^-4)

= 3.27

use:

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

= 3.27+ log {0.1248/2.568*10^-2}

= 3.957

use:

PH = 14 - pOH

= 14 - 3.9568

= 10.0432

Answer: 10.04