Question

An analytical chemist is titrating 142.4 mL of a 0.6900 M solution of propylamine (CzH NH) with a 0.1100 M solution of HIO 3. The pk of propylamine is 3.46. Calculate the pH of the base solution after the chemist has added 507.7 mL of the HIOZ 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. PH = 0 | x ?

Answer 1

use:

pKb = -log Kb

3.46= -log Kb

Kb = 3.467*10^-4

Given:

M(HIO3) = 0.11 M

V(HIO3) = 507.7 mL

M(C3H7NH2) = 0.69 M

V(C3H7NH2) = 142.4 mL

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

mol(HIO3) = 0.11 M * 507.7 mL = 55.847 mmol

mol(C3H7NH2) = M(C3H7NH2) * V(C3H7NH2)

mol(C3H7NH2) = 0.69 M * 142.4 mL = 98.256 mmol

We have:

mol(HIO3) = 55.847 mmol

mol(C3H7NH2) = 98.256 mmol

55.847 mmol of both will react

excess C3H7NH2 remaining = 42.409 mmol

Volume of Solution = 507.7 + 142.4 = 650.1 mL

[C3H7NH2] = 42.409 mmol/650.1 mL = 0.0652 M

[C3H7NH3+] = 55.847 mmol/650.1 mL = 0.0859 M

They form basic buffer

base is C3H7NH2

conjugate acid is C3H7NH3+

Kb = 3.467*10^-4

pKb = - log (Kb)

= - log(3.467*10^-4)

= 3.46

use:

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

= 3.46+ log {8.591*10^-2/6.523*10^-2}

= 3.58

use:

PH = 14 - pOH

= 14 - 3.5795

= 10.4205

Answer: 10.42