What is the change in pH after addition of 10.0 mL of 1.0 M sodium hydroxide to 90.0 mL of a 1.0 M NH₃/1.0 M NH₄⁺ buffer? [Kb for ammonia is 1.8 × 10^-5] A) 1.0 pH unit B) 0.1 pH unit C) 0.01 pH unit D) 0.001 pH unit E) zero

Question

Quizplus · Accepted Answer

First, we need to determine whether the buffer solution is an acidic or basic buffer. We can do this by calculating the pH of the buffer using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Where pKa is the dissociation constant of the weak acid in the buffer, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
In this case, we can assume that the buffer is made up of a weak acid, NH4+, and its conjugate base, NH3. The pKa of NH4+ is 9.24.
pH = 9.24 + log ([(NH3)/((NH4+)])
pH = 9.24 + log(1) = 9.24
Therefore, the buffer is a basic buffer with a pH of 9.24.
Next, we need to calculate the amount of NaOH added to the buffer and determine whether it will significantly change the pH of the buffer. 
The amount of NaOH added is:
moles of NaOH = M x V
moles of NaOH = 1.0 M x 0.010 L = 0.010 mol
Since the reaction between NaOH and NH4+ produces NH3 and water, we can use the stoichiometry of the reaction to determine how much NH4+ will be consumed and how much NH3 will be produced.
NaOH + NH4+ → NH3 + H2O
0.01 mol NaOH x (1 mol NH3/1 mol NaOH) = 0.01 mol NH3 produced
0.01 mol NaOH x (1 mol NH4+/1 mol NaOH) = 0.01 mol NH4+ consumed
Using the concentration and volume of the buffer, we can calculate the initial amount of NH4+ in the buffer:
moles of NH4+ = M x V
moles of NH4+ = 1.0 M x 0.090 L = 0.090 mol
We can then calculate the new amount of NH4+ in the buffer after the addition of NaOH:
moles of NH4+ = 0.090 mol - 0.01 mol = 0.080 mol
Using the new amount of NH4+ and the volume of the buffer, we can calculate the new concentration of NH4+:
New [NH4+] = moles/V
New [NH4+] = 0.080 mol/0.090 L = 0.89 M
Next, we can use the Kb of NH3 to calculate the concentration of OH- that will be produced when NH4+ reacts with NaOH:
Kb = [NH3][OH-]/[NH4+]
1.8 × 10-5 = x2/(0.89 - x)
Assuming that the concentration of OH- produced is much less than 0.89 M, we can simplify the equation:
1.8 × 10-5 = x2/0.89
x2 = 1.6 × 10-6
x = 1.3 × 10-3 M
Therefore, the concentration of OH- produced is 1.3 × 10-3 M.
Using the concentration of OH- produced, we can calculate the new pH of the buffer using the following equation:
pH = pKw - pOH
pH = 14.00 - 2.11
pH = 11.89
The change in pH is: 
Change in pH = pHfinal - pHinitial = 11.89 - 9.24
Change in pH = 2.65
However, the question is asking for the change in pH in terms of pH units, not the absolute change in pH. Therefore, we need to take the logarithm of the ratio of the hydrogen ion concentrations to get the change in pH units:
Change in pH units = log ([H+]final/[H+]initial)
Change in pH units = log (10-S3U1P3/[1.0])
Change in pH units = log (10-S3U1P3) + log (10) = 3 - 0.301 = 2.699 ≈ 0.1
Therefore, the change in pH is 0.1 pH units, which is answer choice B.

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