Question

Titrations

Part A

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 19.0 mL of HNO3.

Express your answer numerically.

Part B

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of NaOH.

Express your answer numerically.

Answer 1

1) The initial solution contains 0.075 L X 0.200 M = 0.015 mol NH3
You've added 0.019 L X 0.500 M = 9.5 X 10^-3 mol HNO3.
The HNO3 has reacted completely with NH3 forming 9.5X10^-3 mol of NH4+. This has left 0.015 - 9.5 X 10^-3 = 5.5X10^-3 mol of NH3.
The molar concentrations of NH4+ and NH3 are:
[NH4+ ] 9.5 X 10^-3 / 0.094 = 0.101 M
[NH3] = 5.5X10^-3 / 0.094 = 0.0585

So, using the expression for Kb:

Kb = [NH4+][OH-] / [NH3] = 1.8 X 10^-5

1.8 X 10^-5 = (0.101) [OH-] / 0.0585
[OH-] = 1.4 X 10^-5 M
pOH = 4.98
pH = 14.00 - pOH = 9.02

2)

52.0ml of 0.35M CH3COOH is: 52.0ml*0.35M = 18.2mMol of CH3COOH.
23.0ml of 0.40M NaOH is: 23.0ml*0.40M = 9.2 mMol of NaOH.
After reaction, 9.2 mMol CH3COO- is generated and 8 mMol CH3COOH is left un-reacted. The concentration would be 9.2/V and 9/V, respectively. Hence:
pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-])
= -log(1.8x10^-5*9/9.2) = 4.75