Question

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

Answer 1

`2.5 * 10^(-5)`

Explanation:

The idea here is that the `"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.

`"HClO"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "ClO"_ ((aq))^(-)`

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^(-"pH")` `"M"` of hydronium cations.

In other words, in order for the dissociation of the acid to produce `["H"_3"O"^(+)] = 10^(-"pH")` `"M"`, the concentration of the acid must decrease by `["H"_3"O"^(+)] = 10^(-"pH")` `"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 = (["ClO"^(-)] * ["H"_ 3"O"^(+)])/(["HClO"])`

You will end up with

`K_a = (10^(-"pH") * 10^(-"pH"))/(0.08 - 10^(-"pH"))`

which gets you

`K_a = (10^(-2.85))^2/(0.08 - 10^(-2.85)) = color(darkgreen)(ul(color(black)(2.5 * 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.