# What is the pH of a 0.85 M NH_4Cl solution?

May 26, 2016

$p H = 4.66$

#### Explanation:

Ammonium chloride is a soluble salt that ionizes completely when dissolved in water.

$\text{ } N {H}_{4} C l \to N {H}_{4}^{+} + C {l}^{-}$

$C {l}^{-}$is a spectator ion that does not affect the pH of the medium.

On the other hand, $N {H}_{4}^{+}$is the conjugate species of the weak base$\setminus N {H}_{3} \setminus$ therefore, it is a weak acid stronger than water. It dissociates in water according to the following equation:

$\text{ } N {H}_{4}^{+} + {H}_{2} O r i g h t \le f t h a r p \infty n s N {H}_{3} + {H}_{3} {O}^{+}$

$\left(N {H}_{4}^{+} / N {H}_{3}\right)$ are conjugate acid base pairs, and for conjugate acid base pairs:

$\text{ } {K}_{a} \times {K}_{b} = {K}_{w}$

Knowing the$\setminus {K}_{b} \setminus$ value for $\setminus N {H}_{3} \left(1.8 \times {10}^{-} 5\right)$, the $\setminus {K}_{a} \setminus$ value for $N {H}_{4}^{+}$ is determined.

${K}_{a} = {K}_{w} / {K}_{b}$

${K}_{a} = \frac{1.00 \times {10}^{-} 14}{1.80 \times {10}^{-} 5}$

${K}_{a} = 5.56 \times {10}^{-} 10$

Use the ICE table to figure out the concentration of each species present at equilibrium.

$\text{ } \underline{N {H}_{4}^{+} + {H}_{2} O r i g h t \le f t h a r p \infty n s N {H}_{3} + {H}_{3} {O}^{+}}$
$\setminus \text{I "0.85 M" " - " } -$
$\text{C " -x" " +x" } + x$
$\text{E " ul( 0.85-x" "x" " x" }$

${K}_{a} = \frac{\left[N {H}_{3}\right] \times \left[{H}_{3} {O}^{+}\right]}{\left[N {H}_{4}^{+}\right]}$

${K}_{a} = \frac{x . x}{\left(0.85 - x\right)} = 5.56 \times {10}^{-} 10$

$\left(0.85 - x\right) \cong 0.85$ since $x$ is too small compared to 0.85

$\frac{{x}^{2}}{0.85} = 5.56 \times {10}^{-} 10$

$x = \sqrt{0.85 \times 5.56 \times {10}^{-} 10}$

$x \cong 2.2 \times {10}^{-} 5 \setminus M$

$x = \left[{H}_{3} {O}^{+}\right]$

$p H = - \log \setminus \left(2.2 \times {10}^{-} 5\right)$

$p H = 4.66$