Question

An analytical chemist is titrating 60.1 ml of a 0.2600 M solution of diethylamine ((CH), NH with a 0.4000 M solution of HNO,. The pK, of diethylamine is 2.89. Calculate the pH of the base solution after the chemist has added 26.7 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 HNO, solution added. Round your answer to 2 decimal places. pH = 0 X 5 ?

Answer 1

use:

pKb = -log Kb

2.89= -log Kb

Kb = 1.288*10^-3

Given:

M(HNO3) = 0.4 M

V(HNO3) = 26.7 mL

M((C2H5)2NH) = 0.26 M

V((C2H5)2NH) = 60.1 mL

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

mol(HNO3) = 0.4 M * 26.7 mL = 10.68 mmol

mol((C2H5)2NH) = M((C2H5)2NH) * V((C2H5)2NH)

mol((C2H5)2NH) = 0.26 M * 60.1 mL = 15.626 mmol

We have:

mol(HNO3) = 10.68 mmol

mol((C2H5)2NH) = 15.626 mmol

10.68 mmol of both will react

excess (C2H5)2NH remaining = 4.946 mmol

Volume of Solution = 26.7 + 60.1 = 86.8 mL

[(C2H5)2NH] = 4.946 mmol/86.8 mL = 0.057 M

[(C2H5)2NH2+] = 10.68 mmol/86.8 mL = 0.123 M

They form basic buffer

base is (C2H5)2NH

conjugate acid is (C2H5)2NH2+

Kb = 1.288*10^-3

pKb = - log (Kb)

= - log(1.288*10^-3)

= 2.89

use:

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

= 2.89+ log {0.123/5.698*10^-2}

= 3.224

use:

PH = 14 - pOH

= 14 - 3.2243

= 10.7757

Answer: 10.78