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

Feb 15, 2018

${K}_{a} = 5.3 \cdot {10}^{- 10}$

#### Explanation:

The ammonium cation will act as a weak acid in aqueous solution, as you can undoubtedly tell from the fact that the $\text{pH}$ of the solution is $< 7$.

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

${\text{NH"_ (4(aq))^(+) + "H"_ 2"O"_ ((l)) rightleftharpoons "NH"_ (3(aq)) + "H"_ 3"O}}_{\left(a q\right)}^{+}$

By definition, the acid dissociation constant, ${K}_{a}$, will take the form

${K}_{a} = \left(\left[{\text{NH"_3] * ["H"_3"O"^(+)])/(["NH}}_{4}^{+}\right]\right)$

Now, the $\text{pH}$ 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

$\left[\text{NH"_ 4^(+)] = ["NH"_ 4^(+)]_ 0 - 10^(-"pH}\right)$

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, ${\left[{\text{NH}}_{4}^{+}\right]}_{0}$, must decrease by 10^(-"pH") quad "M"

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

${K}_{a} = \left({10}^{- \text{pH") * 10^(-"pH"))/(["NH"_ 4^(+)]_ 0 - 10^(-"pH}}\right)$

${K}_{a} = {10}^{- 2 \text{pH")/(["NH"_ 4^(+)]_ 0 - 10^(-"pH}}$

${K}_{a} = {10}^{- 2 \cdot 4.64} / \left(1.0 - {10}^{- 4.64}\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{5.3 \cdot {10}^{- 10}}}}$