# Question 01170

Jun 8, 2016

The $\text{pH}$ is about $5.1$

#### Explanation:

Because ammonium $\left({\text{NH}}_{4}^{+}\right)$ is a weak acid, you need to determine the amount of ${\text{H}}^{+}$ in the solution by setting an equilibrium problem.

Step 1: find ${K}_{\text{a}}$ of ammonium from the ${K}_{\text{b}}$ of ammonia (acid / base conjugates).

${K}_{\text{a" * K_"b}} = {10}^{-} 14$

${K}_{\text{a}}$ of ${\left({\text{NH}}_{4}\right)}^{+} = {10}^{-} \frac{14}{{K}_{\textrm{b}}} = 5.6 \times {10}^{-} 10$

Step 2: Set up the dissociation of ammonium equilibrium, and find the amount of ${\textrm{H}}^{+}$ in solution at equilibrium.

{: (," ",bb("NH"_4^+),bb->,bb"NH"_3,+,bb("H"^+)), ("initial:",,"0.100 M",,"0 M",,"0 M"), ("change:",,-x,,+x,,+x), ("equilibrium:",,0.1-x,,x,,x) :}

You can approximate that

$0.1 - x \approx 0.1$

b/c ${K}_{\text{a}}$ is very small relative to $\text{0.1 M}$.

K_"a" = (["NH"_3] * ["H"^+])/(["NH"_4^+]) = (x * x) /0.1 = 5.6 xx 10^-10

${x}^{2} = 0.1 \cdot 5.6 \times {10}^{-} 10$

$x = \sqrt{5.6 \times {10}^{-} 11} = 7.5 \times {10}^{-} 6$

$\left[{\text{H}}^{+}\right] = x = 7.5 \times {10}^{-} 6$

Step 3: calculate $\text{pH}$ from the concentration of ${\text{H}}^{+}$ found in step 2.

"pH" = -log["H"^+] = -log(7.5xx10^-6) = 5.1#