Question

Find the h3o+ concentration of a 0.210 m hypochlorous acid solution (whose acid dissociation constant is ka=2.9×10−8).

Answer 1

Given that:
Ka = 2.9 x 10⁻⁸      and [HClO] = 0.210 M
Ka = $\frac{[H_{3}O^+][ClO^-]}{[HClO]}$
                    HClO + H₂O → H₃O⁺ + ClO⁻
Initial           0.210                   0           0
Change         -x                      +x          +x
Equilibrium  (0.210-x)             x             x
so:
2.9 x 10⁻⁸ = $\frac{(x)(x)}{(0.210-x)}$
solving x:
x = 7.8 x 10⁻⁵
so [H₃O⁺] = 7.8 x 10⁻⁵ M

Answer 2

The $\rm [H_3O^+]$ concentration of a 0.210 m hypochlorous acid solution is $7.8 \times 10^-^5M$

What is hypochlorous acid?

Hypochlorous acid is a weak acid formed by the dissolving of chlorine in water.

Given that, Ka = $2.9 \times 10^-^8$

[HClO] = 0.210 M

$Ka = \dfrac{[H_3O^+][ClO^-]}{[HCLO]}$

$\rm HClO + H_2O = H_3O^+ + ClO^-$

The table

Initial           0.210                   0           0

Change         -x                      +x          +x

Equilibrium  (0.210-x)             x             x

$2.9 \times 10^-^8 = \dfrac{[x][x]}{(0.210-x)} \\\\\\x = 7.8 \times 10^-^5$

Thus, the $\rm [H_3O^+]$ = $7.8 \times 10^-^5M$

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