Answer
The hydronium ion concentration for that $NH_4Cl$ solution is equal to $1.1 \times 10^{-5}M$. And the pH is equal to 4.98.
Work Step by Step
Analyze the ions of the salt:
$Cl^-$: Conjugate base of $HCl$ (Strong acid). Therefore, it will not affect the pH of the water.
$NH_4^+$: Conjugate acid of a weak base. So it is a weak acid. Calculate the pH for a $0.20M$ $NH_4^+$ solution.
1. Since $NH_4^+$ is the conjugate acid of $NH_3$ , we can calculate its $K_a$ by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 1.8\times 10^{- 5} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_a = 5.56\times 10^{- 10}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$NH_4^+(aq) + H_2O(l) \lt -- \gt NH_3(aq) + H_3O^+(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $NH_3$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M
-$[H_3O^+] = [NH_3] = 0 + x = x$
-$[NH_4^+] = [NH_4^+]_{initial} - x$
For approximation, we are going to consider $[NH_4^+]_{initial} = [NH_4^+]$
3. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][NH_3]}{ [NH_4^+]}$
$Ka = 5.56 \times 10^{- 10}= \frac{x * x}{ 0.20}$
$Ka = 5.56 \times 10^{- 10}= \frac{x^2}{ 0.20}$
$x^2 = 0.20 \times 5.56 \times 10^{-10} $
$x = \sqrt { 0.20 \times 5.56 \times 10^{-10}} = 1.054 \times 10^{-5} $
Percent dissociation: $\frac{ 1.05 \times 10^{- 5}}{ 0.2} \times 100\% = 0.00527\%$
%dissociation < 5% : Right approximation.
Therefore: $[H_3O^+] = [NH_3] = x = 1.054 \times 10^{- 5}M $
And, since 'x' has a very small value (compared to the initial concentration): $[NH_4^+] \approx 0.20M$
4. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.054 \times 10^{- 5})$
$pH = 4.98$
