# Calculate the pH of the following aqueous solutions?

## Calculate the pH of the following aqueous solutions? a) 0.1 M ammonium chloride. b) 100 ml of 0.1 M solution of ammonium chloride to which 100 ml of 0.1 M solution of sodium hydroxide. Assume that all volumes are additive. Kb (ammonia) = 1.8 x 10-5

Apr 30, 2018

Warning! Long Answer. a) pH = 5.13; b) pH = 11.0

#### Explanation:

For a):

Ammonium chloride, $N {H}_{4} C l$ dissolves in solution to form ammonium ions $N {H}_{4}^{+}$ which act as a weak acid by protonating water to form ammonia, $N {H}_{3} \left(a q\right)$ and hydronium ions ${H}_{3} {O}^{+} \left(a q\right)$:

$N {H}_{4}^{+} \left(a q\right) + {H}_{2} O \left(l\right) \to N {H}_{3} \left(a q\right) + {H}_{3} {O}^{+} \left(a q\right)$

As we know the ${K}_{b}$ for ammonia, we can find the ${K}_{a}$ for the ammonium ion. For a given acid/base pair:

${K}_{a} \times {K}_{b} = 1.0 \times {10}^{-} 14$ assuming standard conditions.

So, ${K}_{a} \left(N {H}_{4}^{+}\right) = \frac{1.0 \times {10}^{-} 14}{1.8 \times {10}^{-} 5} = 5.56 \times {10}^{-} 10$

Plug in the concentration and the ${K}_{a}$ value into the expression:

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

$5.56 \times {10}^{-} 10 \approx \frac{\left[{H}_{3} {O}^{+}\right] \times \left[N {H}_{3}\right]}{\left[0.1\right]}$

$5.56 \times {10}^{-} 11 = {\left[{H}_{3} {O}^{+}\right]}^{2}$

(as we can assume that one molecule hydronium must form for every one of ammonia that forms. Also, ${K}_{a}$ is small, so x ≪ 0.1.)

$\left[{H}_{3} {O}^{+}\right] = 7.45 \times {10}^{-} 6$

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

$p H = - \log \left(7.45 \times {10}^{-} 6\right)$

$p H \approx 5.13$

For b):

(i) Determine the species present after mixing.

The equation for the reaction is

$\textcolor{w h i t e}{m m m m m} \text{OH"^"-" + "NH"_4^"+" -> "NH"_3 + "H"_2"O}$
$\text{I/mol} : \textcolor{w h i t e}{m l l} 0.010 \textcolor{w h i t e}{m l l} 0.010 \textcolor{w h i t e}{m o \frac{l}{L}} 0$
$\text{C/mol": color(white)(m)"-0.010"color(white)(ml)"-0.010"color(white)(m)"+0.010}$
$\text{E/mol} : \textcolor{w h i t e}{m l l} 0 \textcolor{w h i t e}{m m m m} 0 \textcolor{w h i t e}{m m m l} 0.010$

$\text{Moles of OH"^"-" = "0.100 L" × "0.1 mol"/"1 L" = "0.010 mol}$

$\text{Moles of NH"_4^"+" = "0.100 L" × "0.1 mol"/"1 L" = "0.010 mol}$

So, we will have 200 mL of an aqueous solution containing 0.010 mol of ammonia, and the pH should be higher than 7.

(ii) Calculate the pH of the solution

["NH"_3] = "0.010 mol"/"0.200 L" = "0.050 mol/L"

The chemical equation for the equilibrium is

$\text{NH"_3 + "H"_2"O" ⇌ "NH"_4^"+" + "OH"^"-}$

Let's re-write this as

$\text{B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-}$

We can use an ICE table to do the calculation.

$\textcolor{w h i t e}{m m m m m m m l l} \text{B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m l l} 0.050 \textcolor{w h i t e}{m m m m m l l} 0 \textcolor{w h i t e}{m m m} 0$
$\text{C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmll)"+"xcolor(white)(mll)"+} x$
$\text{E/mol·L"^"-1":color(white)(m)"0.050-} x \textcolor{w h i t e}{m m m m m} x \textcolor{w h i t e}{m m m} x$

K_text(b) = (["BH"^"+"]["OH"^"-"])/(["B"]) = x^2/("0.050-"x) = 1.8 × 10^"-5"

Check for negligibility:

0.050/(1.8 × 10^"-5") = 3 × 10^3 > 400. ∴ x ≪ 0.050

x^2/0.050 = 1.8 × 10^"-5"

x^2 = 0.050 × 1.8 × 10^"-5" = 9.0 × 10^"-7"

x = 9.5 × 10^"-4"

["OH"^"-"] = 9.5 × 10^"-4" color(white)(l)"mol/L"

"pOH" = -log(9.5 × 10^"-4") = 3.0

$\text{pH = 14.00 - pOH = 14.00 - 3.0} = 11.0$