The pH of a solution made of 0.100 mol of a weak monoprotic acid HA in 1.000 L of solution is 1.470. Calculate K_a for this acid. A) 29.5 B) 0 C) 0.100 D) 0.0174 E) 0.0339

Question

Quizplus · Accepted Answer

First, we need to use the pH value to find the concentration of H+ ions in the solution:

pH = -log[H+]

1.470 = -log[H+]

[H+] = 2.12 x 10^-2 M

Since HA is a weak acid, we can assume that most of the H+ ions in the solution come from the dissociation of the acid:

HA + H2O ↔ H3O+ + A-

We can use an ICE table to calculate the equilibrium concentrations of the species:

HA      + H2O    ↔ H3O+   + A-

I        0.100 M     0 M        0 M          0 M

C       -x            x           x            x

E       0.100 - x    x           x            x

The equilibrium constant, Ka, is given by:

Ka = [H3O+][A-]/[HA]

We can assume that x is small compared to the initial concentration of HA, so we can simplify the expression:

Ka ≈ x^2/0.100

We need to find x, which is the concentration of H3O+ and A- ions at equilibrium. Since the [H3O+] and [A-] are equal in this case, we can use the expression [H3O+] = x to simplify the calculation:

2.12 x 10^-2 = x

Ka ≈ (2.12 x 10^-2)^2/0.100 = 0.0174

Therefore, the answer is D.

The PH of a Solution Made of 0