Question

What is the `K_a` for a `"1.0-mol L"^(-1)` `"NH"_4^(+)` solution with a `"pH"` of `4.64` ?

Answer 1

`K_a = 5.3 * 10^(-10)`

Explanation:

The ammonium cation will act as a weak acid in aqueous solution, as you can undoubtedly tell from the fact that the `"pH"` of the solution is `<7`.

The balanced chemical equation that describes the partial ionization of the ammonium cation looks like this

`"NH"_ (4(aq))^(+) + "H"_ 2"O"_ ((l)) rightleftharpoons "NH"_ (3(aq)) + "H"_ 3"O"_ ((aq))^(+)`

By definition, the acid dissociation constant, `K_a`, will take the form

`K_a = (["NH"_3] * ["H"_3"O"^(+)])/(["NH"_4^(+)])`

Now, the `"pH"` of the solution will give you the equilibrium concentration of the hydronium cations, since

`["H"_3"O"^(+)] = 10^(-"pH") quad "M"`

Notice that every mole of ammonium cations that ionizes produces `1` mole of ammonia and `1` mole of hydronium cations. This tells you that, at equilibrium, the concentration of ammonia will be equal to the concentration of hydronium cations.

`["NH"_3] = ["H"_3"O"^(+)] = 10^(-"pH") quad "M"`

The equilibrium concentration of the ammonium cations will be equal to

`["NH"_ 4^(+)] = ["NH"_ 4^(+)]_ 0 - 10^(-"pH")`

This is the case because in order for the reaction to produce `["H"_3"O"^(+)] =10^(-"pH") quad "M"`, the initial concentration of the ammoniumc cations, `["NH"_4^(+)]_0`, must decrease by `10^(-"pH") quad "M"`

The acid dissociation constant will thus be equal to--I'll do the calculations without added units!

`K_ a = (10^(-"pH") * 10^(-"pH"))/(["NH"_ 4^(+)]_ 0 - 10^(-"pH"))`

`K_a = 10^(-2"pH")/(["NH"_ 4^(+)]_ 0 - 10^(-"pH")`

In your case, you have

`K_a = 10^(- 2 * 4.64)/(1.0 - 10^(-4.64)) = color(darkgreen)(ul(color(black)(5.3 * 10^(-10))))`

The answer is rounded to two sig figs.