CH302 Spring 2009 Practice Quiz 3 Answer Key—The TA Version
1. Which of the following pairs of solutions would result in a buffer upon mixing?
1. 25 mL of 4 M HCl & 15 mL of 4 M HNO2
2. 200 mL of 0.5 M LiOH & 100 mL of 0.5 M H2SO4
3. 100 mL of 1 M NH3 & 10 mL of 10 M HNO3
4. 150 mL of 3 M Ba(OH)2 & 200 mL of 2 M HClO
5. 100 mL of 1 M CH3COOH & 50 mL of 1 M NaOH Correct
Explanation: 100 mL of 1 M CH3COOH & 50 mL of NaOH, would equate to 0.1 mol of a weak acid and
0.05 mol of a strong base. Upon neutralization, the resulting solution would contain 0.05 mol each of
acetic acid and its conjugate base, acetate - resulting in a buffered system.
2. What would be the pH of a solution prepared from 2 L of H2O, 85 g of NH3 and 98 g of NH4Br?
Assume the Kb of ammonia is 2x10-5.
1. 4
2. 5.4
3. 10 Correct
4. 8.6
5. 7
Explanation: 85 g of NH3 x 1 mol/17 g = 5 mol NH3
5 mol NH3 / 2 L H2O = 2.5 M NH3
98 g of NH4Br x 1 mol/98 g = 1 mol NH4Br
1 mol NH4Br / 2 L H2O = 0.5 M NH4Br
[OH-] = Kb(Cb\Ca) = 2x10-5(2.5 M/0.5 M) = 10-4
pH = 10
3. Two liters of a buffer containing 0.6 M CH3NH2 and 0.8 M CH3NH3Cl has 102.4 g of HI added to it.
What is the new pH? Assume the Kb of CH3NH3 is 6x10-4.
1. 6
2. 3
3. 11
4. 4
5. 10 Correct
6. 8
Explanation: 102.4 g of HI x 1 mol/128 g = 0.8 mol HI
[OH-] = Kb(Cb\Ca) = 6x10-4(0.2 M/1.2 M) = 10-4
pH = 10
4. A 0.08 M CH3NH2 solution is titrated against a 0.08 M HCl solution. Assuming the Kb of CH3NH2 is
4x10-10, what is the pH at the equivalence point?
1. 3 Correct
2. 7
3. 9
4. 5
5. not enough information
Explanation: Because the titrant and analyte are equimolar, the volume of the system at the
equivalance point will be double its initial value and the concentration of the conjugate acid will be half
the initial concentration of the base.
Vtotal = 2 x Vinitial
Ca = 0.04 M CH3NH3+
Ka = Kw / Kb = 10-14 / 4x10-10 = 2.5x10-5
[H+] = (Ka·Ca)1/2 = (2.5x10-5·0.04)1/2 = (10-6)1/2 = 10-3
pH = 3
