0.375 g of a monoprotic acid (mm = 245 g/mol) is dissolved in water to produce 25.0 mL of a solution with pH = 3.28. Determine the ionization constant of the acid.

Question

Quizplus · Accepted Answer

Firstly, we need to calculate the concentration of the acid in the solution:

n = m/MM = 0.375 g / 245 g/mol = 0.00153 mol

V = 25.0 mL = 0.0250 L

C = n/V = 0.00153 mol / 0.0250 L = 0.0612 M

Next, we can use the pH to calculate the concentration of H+ ions in the solution:

pH = -log[H+]

[H+] = 10^-pH = 10^-3.28 = 5.28×10^-4 M

When the acid dissociates in water, it produces H+ ions and its conjugate base, A-. We can assume that at equilibrium, the concentration of H+ ions coming from the water dissociation will be negligible in comparison to the concentration coming from the acid dissociation. Consequently, the concentration of acid that dissociates is equal to the initial concentration of the acid minus the concentration of H+ ions that are produced:

[HA] = C - [H+] = 0.0612 M - 5.28×10^-4 M = 0.0607 M

The ionization constant (Ka) can now be calculated using the equation:

Ka = [H+][A-]/[HA]

We know that [A-] = [H+] because the acid is monoprotic and produces only one A- ion per dissociated acid molecule. Therefore,

Ka = [H+]^2 / [HA] = (5.28×10^-4 M)^2 / 0.0607 M = 4.56×10^-6

This value matches with option B.

0375 G of a Monoprotic Acid (Mm = 245 G/mol)