Question

An analytical Chemist is titrating 61.0 ml of a 0.1400 M solution of piperidine (C H, NH with a 0.1800 M solution of HIO, The pk, of piperidine is 2.89. Calculate the ph of the base solution after the chemist has added 55.4 ml of the HIO, solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HIO, solution added. Round your answer to 2 decimal places.

Answer 1

use:

pKb = -log Kb

2.89= -log Kb

Kb = 1.288*10^-3

Given:

M(HIO3) = 0.18 M

V(HIO3) = 55.4 mL

M(C5H1ONH) = 0.14 M

V(C5H1ONH) = 61 mL

mol(HIO3) = M(HIO3) * V(HIO3)

mol(HIO3) = 0.18 M * 55.4 mL = 9.972 mmol

mol(C5H1ONH) = M(C5H1ONH) * V(C5H1ONH)

mol(C5H1ONH) = 0.14 M * 61 mL = 8.54 mmol

We have:

mol(HIO3) = 9.972 mmol

mol(C5H1ONH) = 8.54 mmol

8.54 mmol of both will react

excess HIO3 remaining = 1.432 mmol

Volume of Solution = 55.4 + 61 = 116.4 mL

[H+] = 1.432 mmol/116.4 mL = 0.0123 M

use:

pH = -log [H+]

= -log (1.23*10^-2)

= 1.91

Answer: 1.91