Question

Calculate the concentration of HClO at equilibrium if the initial concentration of HClO is 7.0

Answer 1

HClO + H2O <==> H3O+ + ClO-

At equilibrium you will have formed x M H3O+ and an equal amount of ClO-. The [HClO] will be 0.0075 M - x.

Ka = 3.0 x 10^-8 = (x)(x) / (0.007 - x)
Because Ka is so low we can ignore the -x term
Ka = 3.0 x 10^-8 = x^2 / 0.007
2.3 x 10^-10 = x^2
1.5 x 10^-5 = x = [H3O+] = [ClO-]
[HClO] = 0.007 - 0.00001 = 0.007 M

Answer 2

This is a standard weak acid association problem; just like any other equilibrium problem, and all you need to do

Reaction: HA + H2O <=> H3O+ ClO-
Init: ...........0.0070M.............0...........
Changes,:........-X, +X, +X
Equilib..........0.0070-X ..........X........X

Ka = [H3O+][ClO-]/[HClO]

Substitute in the known value of Ka, and the concentrations from the "Equilibrium" line, and sort out the resulting equation.

As usual in these problems, this will give you a quadratic. However, since Ka is small, you can try assuming that the degree of dissociation is small, so that you can use 0.0070M for [HA], instead of the more exact (0.0070 - X). Try it and see.

Answer 3

The hydrolysis reaction of HClO is: HClO + H2O <==> H3O+ + ClO-.

Molarity . . . .HClO + H2O <==> H3O+ + ClO-
Initial . . . . . 0.0070 . . . . . . . . . . .0 . . . . .0
Change . . . . .-x . . . . . . . . . . . . .x . . . . ..x
Equilibrium 0.0070-x . . . . . . . . . x . . . . . x

Ka = [H3O+][ClO-] / [HClO] = (x)(x) / (0.0070-x) = 3.0 x 10^-8
Because Ka is so small, x is small compared to 0.0070, sodrop it from the 0.0070-x term.

(x^2) / (0.0070) = 3.0 x 10^-8
x = 1.45 x 10^-5 = [H3O+] = [ClO-]

[HClO] = 0.0070 - x = 0.0070 - 0.0000145 = 0.0069

Answer 4

K_a = 3.0