# The "pH" of a "0.08 mol dm"^(-3) solution of "HClO" is equal to 2.85. Calculate the ionization constant of the acid?

Jan 7, 2018

$2.5 \cdot {10}^{- 5}$

#### Explanation:

The idea here is that the $\text{pH}$ of the solution will give you the equilibrium concentration of hydronium cations, which, in turn, will give you the equilibrium concentration of the hypochlorite anions and of the hypochlorous acid.

${\text{HClO"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "ClO}}_{\left(a q\right)}^{-}$

Now, you know that when $1$ mole of hypochlorous acid dissociates, it produces $1$ mole of hypochlorite anions and $1$ mole of hydronium cations.

This means that if the solution contains

["H"_ 3"O"^(+)] = 10^(-"pH") quad "M"

then it must also contain

["ClO"^(-)] = 10^(-"pH") quad "M"

and

["HClO"] = (0.08 - 10^(-"pH")) quad "M"

This is what you'd expect the equilibrium concentration of hypochlorous acid to be equal to if its dissociation produced ${10}^{- \text{pH}}$ $\text{M}$ of hydronium cations.

In other words, in order for the dissociation of the acid to produce $\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)$ $\text{M}$, the concentration of the acid must decrease by $\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)$ $\text{M}$.

So now that you know the equilibrium concentrations for all three chemical species that are of interest here, you can plug them into the expression of the acid dissociation constant, ${K}_{a}$.

${K}_{a} = \left(\left[\text{ClO"^(-)] * ["H"_ 3"O"^(+)])/(["HClO}\right]\right)$

You will end up with

${K}_{a} = \left({10}^{- \text{pH") * 10^(-"pH"))/(0.08 - 10^(-"pH}}\right)$

which gets you

${K}_{a} = {\left({10}^{- 2.85}\right)}^{2} / \left(0.08 - {10}^{- 2.85}\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{2.5 \cdot {10}^{- 5}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you have only one significant figure for the initial concentration of the acid.