The acid-dissociation constant at 25.0 °C for hypochlorous acid (HClO) is 3.0 × 10^-8. At equilibrium, the molarity of H₃O⁺ in a 0.030 mol L^-1 solution of HClO is ________. A) 1.52 mol L^-1 B) 3.3 × 10^-10mol L^-1 C) 4.52 mol L^-1 D) 0.030 mol L^-1 E) 3.0 × 10^-5mol L^-1

Question

Quizplus · Accepted Answer

The equilibrium expression for the dissociation of hypochlorous acid is:

HClO ⇌ H+ + ClO-

The acid-dissociation constant (Ka) is given as 3.0 × 10^-8 M.

Ka = [H+][ClO-]/[HClO]

Let the concentration of HClO at equilibrium be x.

Then, [H+] = x M
[ClO-] = x M
[HClO] = (0.030 - x) M

Substituting these values in the expression for Ka, we get:

3.0 × 10^-8 = x^2/(0.030 - x)

Since x is much smaller than 0.030, we can assume that (0.030 - x) ≈ 0.030.

3.0 × 10^-8 = x^2/0.030

x^2 = 9.0 × 10^-10

x = 3.0 × 10^-5 M

Therefore, the molarity of ClO- at equilibrium is also 3.0 × 10^-5 M.

The molarity of H+ at equilibrium is also 3.0 × 10^-5 M, since HClO dissociates to produce one H+ ion and one ClO- ion.

The molarity of HClO at equilibrium is (0.030 - 3.0 × 10^-5) M, which is approximately equal to 0.030 M.

Therefore, the answer is E) 3.0 × 10^-5 mol/L.

The Acid-Dissociation Constant at 25