The dissociation equilibrium constants for the protonated form of alanine (a diprotic amino acid,H₂X⁺)are K_a₁ = 4.6 × 10^-3 and K_a2 = 2.0 × 10^-10.What is the pH of 50.00 mL of a 0.0500 M solution of alanine after 25.00 mL of 0.100 M NaOH has been added? A)2)34 B)4)85 C)5)59 D)6)72

Question

Quizplus · Accepted Answer

First, let's write out the dissociation reactions for alanine:
HS1U1B12S1U1B0XS1U1P1+S1S1P0 ⇆ H+ + HS1U1B11S1U1B0XS1U1P1+S1S1P0
HS1U1B11S1U1B0XS1U1P1+S1S1P0 ⇆ H+ + S1U1B10XS1U1P1+S1S1P0
We can set up an ICE table for each reaction.
For the first reaction:
HS1U1B12S1U1B0XS1U1P1+S1S1P0 ⇆ H+ + HS1U1B11S1U1B0XS1U1P1+S1S1P0
I:    0.050 M         0              0  
C: -x              +x           +x
E: 0.050-x        x              x
Ks1 = 4.6 × 10S1U1P1-3S1S1P0
Ks1 = [H+][HS1U1B11S1U1B0XS1U1P1+S1S1P0]/[HS1U1B12S1U1B0XS1U1P1+S1S1P0]
4.6 × 10S1U1P1-3S1S1P0 = xS1U1P1S1S1P0 / (0.050 - x)S1U1P1S1S1P0
Assumption: x is negligible compared to 0.050, so 0.050 - x ≈ 0.050
4.6 × 10S1U1P1-3S1S1P0 = xS1U1P1S1S1P0 / 0.050
x = 1.15 × 10S1U1P1-S1S1P0 M = [H+]
Now we can calculate the pH:
pH = -log[H+] = -log(1.15 × 10S1U1P1-S1S1P0) = 5.59
Next, we need to consider the second dissociation. However, since the equilibrium constant for KS1U1B1a2S1U1B0 is much smaller than that for KS1U1B1aS1U1B0, we can assume that the concentration of the fully deprotonated form is negligible compared to the concentration of the singly deprotonated form. Therefore, we can ignore the second dissociation in this problem.
Next, we need to consider the addition of NaOH. NaOH reacts with H+ to form water, which means that we are effectively removing H+ ions from the solution. Since we initially had a solution with excess H+ ions, the addition of NaOH will cause the pH to rise, but not above the pKa1 of alanine.
The balanced equation for the reaction between NaOH and H+ is:
NaOH + H+ → Na+ + H2O
We can use stoichiometry to determine how many moles of H+ ions are neutralized by 25.00 mL of 0.100 M NaOH:
0.100 M NaOH × 0.02500 L = 2.5 × 10S1U1P1-S1S1P0 mol NaOH
Since NaOH and H+ react in a 1:1 ratio, we neutralize 2.5 × 10S1U1P1-S1S1P0 mol H+ ions.
What is the concentration of H+ ions remaining after the addition of NaOH?
We started with x = 1.15 × 10S1U1P1-S1S1P0 M H+ ions, but we just neutralized 2.5 × 10S1U1P1-S1S1P0 mol H+ ions. Therefore, the number of moles of H+ ions remaining is:
x - 2.5 × 10S1U1P1-S1S1P0 mol H+ ions
The total volume of the solution is now 50.00 mL + 25.00 mL = 75.00 mL = 0.07500 L.
Therefore, the new concentration of H+ ions is:
(x - 2.5 × 10S1U1P1-S1S1P0 mol) / 0.07500 L = (x - 3.33 × 10S1U1P1-S1S1P0) M
pH = -log[H+] = -log(x - 3.33 × 10S1U1P1-S1S1P0)
pH = -log(1.15 × 10S1U1P1-S1S1P0 - 3.33 × 10S1U1P1-S1S1P0) = 5.59. 
Therefore, the final answer is choice C, 5.59.

The Dissociation Equilibrium Constants for the Protonated Form of Alanine