Question

A 115.2 mL sample of 0.100 M methylamine (CH3NH2;Kb=3.7×10−4) is titrated with 0.265 M HNO3. Calculate the pH after the addition of each of the following volumes of acid.

43.5ml-

65.2 ml-

Answer 1

1)when 43.5 mL of HNO3 is added

Given:

M(HNO3) = 0.265 M

V(HNO3) = 43.5 mL

M(CH3NH2) = 0.1 M

V(CH3NH2) = 115.2 mL

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

mol(HNO3) = 0.265 M * 43.5 mL = 11.5275 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)

mol(CH3NH2) = 0.1 M * 115.2 mL = 11.52 mmol

We have:

mol(HNO3) = 11.5275 mmol

mol(CH3NH2) = 11.52 mmol

11.52 mmol of both will react

excess HNO3 remaining = 0.0075 mmol

Volume of Solution = 43.5 + 115.2 = 158.7 mL

[H+] = 0.0075 mmol/158.7 mL = 0 M

use:

pH = -log [H+]

= -log (4.726*10^-5)

= 4.3255

Answer: 4.33

2)when 65.2 mL of HNO3 is added

Given:

M(HNO3) = 0.265 M

V(HNO3) = 65.2 mL

M(CH3NH2) = 0.1 M

V(CH3NH2) = 115.2 mL

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

mol(HNO3) = 0.265 M * 65.2 mL = 17.278 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)

mol(CH3NH2) = 0.1 M * 115.2 mL = 11.52 mmol

We have:

mol(HNO3) = 17.278 mmol

mol(CH3NH2) = 11.52 mmol

11.52 mmol of both will react

excess HNO3 remaining = 5.758 mmol

Volume of Solution = 65.2 + 115.2 = 180.4 mL

[H+] = 5.758 mmol/180.4 mL = 0.0319 M

use:

pH = -log [H+]

= -log (3.192*10^-2)

= 1.496

Answer: 1.50