A pH 9.56 buffer was prepared by mixing 2.00 moles of ammonia (K_b for ammonia is 1.8 × 10^-5)and 1.00 mol of ammonium chloride in water to form a solution with a volume of 1.00 L.To a 200.0 mL aliquot of this solution was added 10.0 mL of 10.0 M aqueous sodium hydroxide.What was the resulting pH? A)9.28 B)9.56 C)9.95 D)10.50 E)13.7

Question

Quizplus · Accepted Answer

First, we need to determine the initial concentrations of ammonia and ammonium chloride in the buffer solution.

- Ammonia: 
2.00 moles / 1.00 L = 2.00 M

- Ammonium chloride: 
1.00 mole / 1.00 L = 1.00 M

Next, we need to determine the reaction that occurs when sodium hydroxide is added:

NH3 + NaOH → NH2- + H2O

This reaction consumes ammonia and produces ammonium ions, which will shift the equilibrium of the buffer solution.

We can use the Henderson-Hasselbalch equation to calculate the resulting pH:

pH = pKa + log([base]/[acid])

The pKa of ammonia is 9.25.

Before the addition of NaOH, the buffer solution has:

[base] = [NH3] = 2.00 M 
[acid] = [NH4+] = 1.00 M

Substituting into the Henderson-Hasselbalch equation:

pH = 9.25 + log(2.00/1.00) 
pH = 9.56

After the addition of 10.0 mL of 10.0 M NaOH, the amount of ammonia will decrease and the amount of ammonium ions will increase.

The initial amount of NH3 in the 200.0 mL aliquot is:

2.00 M x 0.200 L = 0.400 moles

The amount of NaOH added is:

10.0 mL x (1 L / 1000 mL) x 10.0 M = 0.100 moles

Since NH3 and NaOH react in a 1:1 ratio, the amount of NH3 remaining is:

0.400 moles - 0.100 moles = 0.300 moles

The amount of NH4+ produced is also 0.100 moles.

Therefore, the new concentrations of base and acid in the buffer solution are:

[base] = [NH3] = 0.300 moles / 0.200 L = 1.50 M 
[acid] = [NH4+] = 1.00 M + 0.100 moles / 0.200 L = 1.50 M

Substituting into the Henderson-Hasselbalch equation:

pH = 9.25 + log(1.50/1.50) 
pH = 9.25 + log(1) 
pH = 9.25

However, we need to consider the effect of adding a strong base (NaOH) to the buffer solution. The strong base will react with the weak acid (NH4+) in the buffer solution and increase the amount of NH3.

NH4+ + OH- → NH3 + H2O

The amount of OH- added is:

10.0 mL x (1 L / 1000 mL) x 10.0 M = 0.100 moles

Since NH4+ and OH- react in a 1:1 ratio, the amount of NH4+ remaining is:

1.00 moles - 0.100 moles = 0.900 moles

The amount of NH3 produced is also 0.100 moles.

Therefore, the new concentrations of base and acid in the buffer solution are:

[base] = [NH3] = 1.50 M + 0.100 moles / 0.200 L = 2.00 M 
[acid] = [NH4+] = 0.900 moles / 0.200 L = 4.50 M

Substituting into the Henderson-Hasselbalch equation:

pH = 9.25 + log(2.00/4.50) 
pH = 9.25 - 0.22 
pH = 9.03

Therefore, the resulting pH after the addition of NaOH is 9.03, which is closest to choice C.

A PH 956 Buffer Was Prepared by Mixing 2