A monoprotic weak acid when dissolved in water is 0.75% dissociated and produces a solution with a pH of 3.11.Calculate the K_a of the acid. A)7.6 $\times$ 10^-³ B)1.0 $\times$10^-¹ C)5.9 $\times$ 10^-⁶ D)Need to know the initial concentration of the acid. E)None of these.

Question

Quizplus · Accepted Answer

The dissociation constant for a weak acid is given by the expression:
Ka = [H+][A-]/[HA], where [HA], [H+] and [A-] are the concentrations of the weak acid, hydrogen ions, and conjugate base, respectively.

Let's assume that the initial concentration of the weak acid is x mol/L. At equilibrium, the concentration of the weak acid that remains undissociated is (1-0.75%)x = 0.9925x mol/L, and the concentration of the dissociated ions is 0.0075x mol/L.

From the pH of the solution, we know that [H+] = 10^-pH = 10^-3.11 = 7.59x10^-4 mol/L.

Using the fact that the weak acid is monoprotic, we can write the expression for the dissociation constant as:
Ka = [H+][A-]/[HA] = (7.59x10^-4)(0.0075x)/(0.9925x) = 5.94x10^-10

Therefore, the answer is (C) 5.9x10^-10.

A Monoprotic Weak Acid When Dissolved in Water Is 0 ×\times×

A Monoprotic Weak Acid When Dissolved in Water Is 0 $\times$