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

We can approach this problem by determining which protonation state of alanine dominates at the given pH, and then using the corresponding dissociation constants to calculate the concentrations of the relevant species.

First, let's consider the effect of adding NaOH to the solution. NaOH is a strong base and will react with the acidic protons on alanine to form water and the corresponding salts. In particular, the first protonation state (HS+) will react first:

HS+ + OH- → H2O + S + Na+

The reaction consumes one mole of OH- for each mole of HS+, so after adding 25.00 mL of 0.100 M NaOH, we have:

0.100 mol/L × 0.02500 L = 0.00250 mol OH-

Since alanine is diprotic, we need to consider the reaction with the second proton as well. However, the second dissociation constant is much smaller than the first one, meaning that the second protonation state (H+) will be negligible at pH values where we expect to see significant amounts of alanine. Therefore, we can assume that all alanine molecules are either in the HS+ or the S form.

Next, we can use the Henderson-Hasselbalch equation to relate the pH to the concentrations of the acid and its conjugate base:

pH = pKa + log([S]/[HS+])

For the given dissociation constants, the pKa values are:

pKa1 = 3.34
pKa2 = 8.69

Since we only have one strong acid-base reaction, we can use the simplified version of the Henderson-Hasselbalch equation:

pH = pKa1 + log([S]/[HS+])

We know the initial concentration of alanine is 0.0500 M, but we need to determine the concentrations of HS+ and S after the addition of NaOH. Let x be the amount (in moles) of HS+ that reacts with NaOH, then the amount of S that forms is also x (since the reaction consumes and produces equal amounts of acid and conjugate base). The equilibrium concentrations of HS+ and S are then:

[HS+] = 0.0500 - x
[S] = x

The value of x can be determined from the stoichiometry of the reaction with NaOH:

1 mol HS+ reacts with 1 mol OH-

0.00250 mol OH- was added

Therefore, x = 0.00250 mol or 2.50 × 10^-3 mol

Plugging these values into the Henderson-Hasselbalch equation gives:

pH = 3.34 + log(2.50 × 10^-3 / (0.0500 - 2.50 × 10^-3)) = 5.59

Therefore, the pH of the solution after adding NaOH is 5.59.

The correct answer is (C).

The Dissociation Equilibrium Constants for the Protonated Form of Alanine