# Question #71432

May 10, 2015

The pH of that solution will be 4.82.

To solve this problem you're going to need the acid dissociation constant, ${K}_{a}$, for the ammonium ion, $N {H}_{4}^{+}$, which is listed as being equal to $5.6 \cdot {10}^{- 10}$.

The ammonium chloride will dissociate in aqueous solution to give ammonium, $N {H}_{4}^{+}$, and chloride, $C {l}^{-}$, ions.

$N {H}_{4} C {l}_{\left(s\right)} \to N {H}_{4 \left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$

The ammonium ion will act as a weak acid, reacting with water to produce ammonia and hydronium ions. Use an ICE table to solve for the concentration of the hydroxide ions in solution.

$\text{ } N {H}_{4 \left(a q\right)}^{+} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s N {H}_{3 \left(a q\right)} + {H}_{3} {O}_{\left(a q\right)}^{+}$
I.......0.42......................................0.................0
C.......(-x).......................................(+x)............(+x)
E....0.42-x......................................x.................x

By definition, the acid dissociation constant will be equal to

${K}_{a} = \frac{\left[N {H}_{3}\right] \cdot \left[{H}_{3} {O}^{+}\right]}{\left[N {H}_{4}^{+}\right]} = \frac{x \cdot x}{0.42 - x} = {x}^{2} / \left(0.42 - x\right)$

Because ${K}_{a}$ is so small, you can approximate (0.42 - x) with 0.42. This will result in

${x}^{2} / 0.42 = 5.6 \cdot {10}^{- 10} \implies x = 1.53 \cdot {10}^{- 5}$

This will be equal to the concentration of the hydronium ions

$\left[{H}_{3} {O}^{+}\right] = 1.53 \cdot {10}^{- 5} \text{M}$

As a result, the pH of the solution will be

$p {H}_{\text{sol}} = - \log \left(\left[{H}_{3} {O}^{+}\right]\right)$

$p {H}_{\text{sol}} = - \log \left(1.53 \cdot {10}^{- 15}\right) = \textcolor{g r e e n}{4.82}$