Question

Calculate the pKa of hypochlorous acid. The pH of a 0.015 M solution of hypochlorous acid has a pH of 4.64 ?

Answer 1

`"pK"_a = -log(K_a) = ???`

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Well, you can work backwards. The `"pH"` gives the `["H"_3"O"^(+)]`:

`"pH" = -log["H"_3"O"^(+)]`

`10^(-"pH") = ["H"_3"O"^(+)] = 10^(-4.64) = 2.29 xx 10^(-5) "M"`

But hypochlorous acid is a monoprotic acid, and hence,

`"HClO"(aq) + "H"_2"O"(l) rightleftharpoons "ClO"^(-)(aq) + "H"_3"O"^(+)(aq)`

we find that `["H"_3"O"^(+)] = ["ClO"^(-)] = x` at equilibrium. The amount `x` that is generated at equilibrium was dissociated from `"HClO"`, i.e. the concentration dropped from `"0.015 M"` to `"0.015 M" - x`.

Writing the `K_a` expression,

`K_a = (["ClO"^(-)]["H"_3"O"^(+)])/(["HClO"])`

`= x^2/("0.015 M" - x)`

But we know `x`. Therefore,

`color(blue)(K_a) = (2.29 xx 10^(-5) "M")^2/("0.015 M" - 2.29 xx 10^(-5) "M")`

`= color(blue)(3.52 xx 10^(-8))`

How then do you get the `"pK"_a` from here, if `hatp(" ") = -log(" ")`?