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 0.010 mol L^-1 solution of HClO is ________. A)1.7 × 10^-5 B)0.010 C)5.8 × 10^-10 D)4.76 E)2.00

Question

Quizplus · Accepted Answer

The balanced chemical equation for the dissociation of HClO is:

HClO ⇌ H+ + ClO-

From the acid-dissociation constant, we can write:

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

We are given that K_a = 3.0 × 10^-8 and that the initial concentration of HClO is 0.010 M. If we assume that x mol/L of HClO dissociate, then at equilibrium, the concentration of H+ and ClO- will be x mol/L and the concentration of undissociated HClO will be (0.010 - x) mol/L. Substituting these values into the expression for K_a, we get:

3.0 × 10^-8 = (x)(x)/(0.010 - x)

Solving for x using the quadratic equation, we get:

x = 1.7 × 10^-5 M

Therefore, the molarity of HClO that has dissociated is 1.7 × 10^-5 M, and the molarity of H+ and ClO- ions is also 1.7 × 10^-5 M. The molarity of undissociated HClO is:

0.010 M - 1.7 × 10^-5 M = 0.009983 M

So the answer is A) 1.7 × 10^-5 M.

The Acid-Dissociation Constant at 25