Answer
The pH of a 0.42 M $NH_4Cl$ solution is equal to 4.82
Work Step by Step
$Cl^-$ does not hydrolyze, since it is the conjugate base of a strong acid.
But, $N{H_4}^+$ does hydrolyze in water, acting like an acid, following this equation:
$$N{H_4}^+(aq) \leftrightharpoons NH_3(aq) + H^+(aq)$$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ NH_4^{+} ]& [ NH_3 ]& [ H^+ ]\\
Initial& 0.42 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.42 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ NH_3 ][ H^+ ]}{[ NH_4^{+} ]}$$
$$K_a = \frac{(x)(x)}{[ NH_4^{+} ]_{initial} - x}$$
3. Solve for x:
$$K_a [ NH_4^{+} ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ NH_4^{+} ] = 0$$
- Thus:
$$x_1 = \frac{- 5.6 \times 10^{-10} + \sqrt{( 5.6 \times 10^{-10} )^2 - 4 (1) (- 5.6 \times 10^{-10} ) ( 0.42 )} }{2 (1)}$$
$$x_1 = 1.5 \times 10^{-5} $$
$$x_2 = \frac{- 5.6 \times 10^{-10} - \sqrt{( 5.6 \times 10^{-10} )^2 - 4 (1) (- 5.6 \times 10^{-10} )( 0.42 )} }{2 (1)}$$
$$x_2 = -1.5 \times 10^{-5} $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 1.5 \times 10^{-5} $$
4. $$[H^+] = x = 1.5 \times 10^{-5} $$
5. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 1.5 \times 10^{-5} ) = 4.82 $$
