Calculate the pH of the following aqueous solutions?

Question

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

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Answer 1 · 2018-04-30T17:44:13

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

Explanation:

For a):

Ammonium chloride, $N H_{4} C l$ $NH_4Cl$ dissolves in solution to form ammonium ions $N H_{4}^{+}$ $NH_4^(+)$ which act as a weak acid by protonating water to form ammonia, $N H_{3} (a q)$ $NH_3(aq)$ and hydronium ions $H_{3} O^{+} (a q)$ $H_3O^(+)(aq)$ :

$N H_{4}^{+} (a q) + H_{2} O (l) \to N H_{3} (a q) + H_{3} O^{+} (a q)$ $NH_4^(+)(aq)+H_2O(l) -> NH_3(aq)+H_3O^(+)(aq)$

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

$K_{a} \times K_{b} = 1.0 \times 10^{- 14}$ $K_a times K_b=1.0 times 10^-14$ assuming standard conditions.

So, $K_{a} (N H_{4}^{+}) = \frac{1.0 \times 10^{- 14}}{1.8 \times 10^{- 5}} = 5.56 \times 10^{- 10}$ $K_a(NH_4^(+))=(1.0 times 10^-14)/(1.8 times 10^-5)=5.56 times 10^-10$

Plug in the concentration and the $K_{a}$ $K_a$ value into the expression:

$K_{a} = \frac{[H_{3} O^{+}] \times (N H_{3}]}{[N H_{4}^{+}]}$ $K_a=([H_3O^(+)] times (NH_3])/([NH_4^(+)])$

$5.56 \times 10^{- 10} \approx \frac{[H_{3} O^{+}] \times [N H_{3}]}{[0.1]}$ $5.56 times 10^-10~~([H_3O^(+)] times [NH_3])/([0.1])$

$5.56 \times 10^{- 11} = {[H_{3} O^{+}]}_{3}^{2}$ $5.56 times 10^-11=[H_3O^(+)]^2$

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

$[H_{3} O^{+}] = 7.45 \times 10^{- 6}$ $[H_3O^(+)]=7.45 times 10^-6$

$p H = - log [H_{3} O^{+}]$ $pH=-log[H_3O^(+)]$

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

$p H \approx 5.13$ $pH approx 5.13$

For b):

(i) Determine the species present after mixing.

The equation for the reaction is

$m m m m m {OH}^{-} + {NH}_{4}^{+} \to {NH}_{3} + H_{2} O$ $color(white)(mmmmm)"OH"^"-" + "NH"_4^"+" -> "NH"_3 + "H"_2"O"$
$I/mol : m l l 0.010 m l l 0.010 m o \frac{l}{L} 0$ $"I/mol": color(white)(mll)0.010 color(white)(mll)0.010color(white)(mol/L)0$
$C/mol : m -0.010 m l -0.010 m +0.010$ $"C/mol": color(white)(m)"-0.010"color(white)(ml)"-0.010"color(white)(m)"+0.010"$
$E/mol : m l l 0 m m m m 0 m m m l 0.010$ $"E/mol": color(white)(mll)0color(white)(mmmm)0color(white)(mmml)0.010$

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

${Moles of NH}_{4}^{+} = 0.100 L \times \frac{0.1 mol}{1 L} = 0.010 mol$ $"Moles of NH"_4^"+" = "0.100 L" × "0.1 mol"/"1 L" = "0.010 mol"$

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

$[{NH}_{3}] = \frac{0.010 mol}{0.200 L} = 0.050 mol/L$ $["NH"_3] = "0.010 mol"/"0.200 L" = "0.050 mol/L"$

The chemical equation for the equilibrium is

${NH}_{3} + H_{2} O ⇌ {NH}_{4}^{+} + {OH}^{-}$ $"NH"_3 + "H"_2"O" ⇌ "NH"_4^"+" + "OH"^"-"$

Let's re-write this as

$B + H_{2} O ⇌ {BH}^{+} + {OH}^{-}$ $"B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-"$

We can use an ICE table to do the calculation.

$m m m m m m m l l B + H_{2} O ⇌ {BH}^{+} + {OH}^{-}$ $color(white)(mmmmmmmll)"B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-"$
${I/mol\cdotL}^{-1} : m l l 0.050 m m m m m l l 0 m m m 0$ $"I/mol·L"^"-1":color(white)(mll)0.050color(white)(mmmmmll)0color(white)(mmm)0$
${C/mol\cdotL}^{-1} : m m - x m m m m m l l + x m l l + x$ $"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmll)"+"xcolor(white)(mll)"+"x$
${E/mol\cdotL}^{-1} : m 0.050- x m m m m m x m m m x$ $"E/mol·L"^"-1":color(white)(m)"0.050-"xcolor(white)(mmmmm)xcolor(white)(mmm)x$

$K_{b} = \frac{[{BH}^{+}] [{OH}^{-}]}{[B]} = \frac{x^{2}}{0.050- x} = 1.8 \times 10^{-5}$ $K_text(b) = (["BH"^"+"]["OH"^"-"])/(["B"]) = x^2/("0.050-"x) = 1.8 × 10^"-5"$

Check for negligibility:

$\frac{0.050}{1.8 \times 10^{-5}} = 3 \times 10^{3} > 400$ $0.050/(1.8 × 10^"-5") = 3 × 10^3 > 400$ . ∴ $x ≪ 0.050$ $x ≪ 0.050$

$\frac{x^{2}}{0.050} = 1.8 \times 10^{-5}$ $x^2/0.050 = 1.8 × 10^"-5"$

$x^{2} = 0.050 \times 1.8 \times 10^{-5} = 9.0 \times 10^{-7}$ $x^2 = 0.050 × 1.8 × 10^"-5" = 9.0 × 10^"-7"$

$x = 9.5 \times 10^{-4}$ $x = 9.5 × 10^"-4"$

$[{OH}^{-}] = 9.5 \times 10^{-4} l mol/L$ $["OH"^"-"] = 9.5 × 10^"-4" color(white)(l)"mol/L"$

$pOH = - log (9.5 \times 10^{-4}) = 3.0$ $"pOH" = -log(9.5 × 10^"-4") = 3.0$

$pH = 14.00 - pOH = 14.00 - 3.0 = 11.0$ $"pH = 14.00 - pOH = 14.00 - 3.0" = 11.0$

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