Answer
(a) $pH = 11.28$
(b) $pH = 9.08$
Work Step by Step
(a)
1. We have these concentrations at equilibrium:
-$[OH^-] = [N{H_4}^+] = x$
-$[NH_3] = [NH_3]_{initial} - x = 0.2 - x$
For approximation, we consider: $[NH_3] = 0.2M$
2. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][N{H_4}^+]}{ [NH_3]}$
$Kb = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.2}$
$Kb = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.2}$
$ 3.6 \times 10^{- 6} = x^2$
$x = 1.9 \times 10^{- 3}$
Percent ionization: $\frac{ 1.9 \times 10^{- 3}}{ 0.2} \times 100\% = 0.95\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [N{H_4}^+] = x = 1.9 \times 10^{- 3}M $
$[NH_3] \approx 0.2M$
3. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 1.9 \times 10^{- 3})$
$pOH = 2.72$
4. Find the pH:
$pH + pOH = 14$
$pH + 2.72 = 14$
$pH = 11.28$
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(b)
1. Drawing the ICE table, we get these concentrations at the equilibrium:
$NH_3(aq) + H_2O(l) \lt -- \gt N{H_4}^+(aq) + OH^-(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[NH_3] = 0.2 M - x$
$[N{H_4}^+] = 0.3M + x$
$[OH^-] = 0 + x$
2. Calculate 'x' using the $K_b$ expression.
$ 1.8\times 10^{- 5} = \frac{[N{H_4}^+][OH^-]}{[NH_3]}$
$ 1.8\times 10^{- 5} = \frac{( 0.3 + x )* x}{ 0.2 - x}$
Considering 'x' has a very small value.
$ 1.8\times 10^{- 5} = \frac{ 0.3 * x}{ 0.2}$
$ 1.8\times 10^{- 5} = 1.5x$
$\frac{ 1.8\times 10^{- 5}}{ 1.5} = x$
$x = 1.2\times 10^{- 5}$
Percent ionization: $\frac{ 1.2\times 10^{- 5}}{ 0.2} \times 100\% = 6\times 10^{- 3}\%$
x = $[OH^-]$
3. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 1.2 \times 10^{- 5})$
$pOH = 4.92$
4. Find the pH:
$pH + pOH = 14$
$pH + 4.92 = 14$
$pH = 9.08$
