Question

A 102.0 mL sample of 0.100 M methylamine (CH3NH2;Kb=3.7×10−4) is titrated with 0.270 M HNO3. Calculate the pH after the addition of each of the following volumes of acid:

Part A: 0.0 mL

Part B: 18.9 mL

Part C: 37.8 mL

Part D: 56.7 mL

Answer 1

0.102 L x 0.100 mol/L = 0.0102 mol methylamine (a monobase)

0.0102 mol / 0.270 mol/L = 0.0378 L = 37.8 mL HNO3 used for a complete neutralization (EP).

A. pKb = 3.43

[HO-] = (Kb.Cbase)^1/2 =(3.7×10−4 x 0.1 )^1/2 = 6.08x10^-3

pOH = 2.22

pH = 14-pOH = 11.78

B.   18.9 mL/ 37.8 mL = 0.5      18.9 mL is half EP

    pH = pKa = 14 – pKb = 10.57

C. end point, the solution contains the conjugate acid, pKa=10.57

Cacid = 0.100 M x102mL/139.8mL = 0.0730

pH = 0.5pKa – 0.5logCacid = 5.28 + 0.57 = 5.85

d. The excess of HNO3 is

56.7-37.8 = 18.9 mL acid containing

0.0189 L x 0.270 mol/L = 0.00510 mol exces acid (or H+) in a total volume 102+56.7=158.7 mL

[H+] = 0.00510 mol / 0.1587 L = 0.0321 mol/L

pH = -log0.0321 = 1.49