Answer
(a) $$pH = 2.37 $$
(b) $$pOH = 11.53$$
Work Step by Step
(a)
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HZ ]& [ Z^- ]& [ H_3O^+ ]\\
Initial& 0.075 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.075 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_a = \frac{P_{Products}}{P_{Reactants}} = \frac{[ Z^- ][ H_3O^+ ]}{[ HZ ]}$$
$$K_a = \frac{(x)(x)}{[ HZ ]_{initial} - x}$$
3. Assuming $ 0.075 \gt\gt x:$
$$K_a = \frac{x^2}{[ HZ ]_{initial}}$$
$$x = \sqrt{K_a \times [ HZ ]_{initial}} = \sqrt{ 2.55 \times 10^{-4} \times 0.075 }$$
$x = 4.4 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 4.4 \times 10^{-3} }{ 0.075 } \times 100\% = 5.9 \%$$
The percent is greater than 5%, therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ HZ ]_{initial} - x}$$
$$K_a [ HZ ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ HZ ] = 0$$
- Using bhaskara:
$$x_1 = \frac{- 2.55 \times 10^{-4} + \sqrt{( 2.55 \times 10^{-4} )^2 - 4 (1) (- 2.55 \times 10^{-4} ) ( 0.075 )} }{2 (1)}$$
$$x_1 = 4.25 \times 10^{-3} $$
$$x_2 = \frac{- 2.55 \times 10^{-4} - \sqrt{( 2.55 \times 10^{-4} )^2 - 4 (1) (- 2.55 \times 10^{-4} )( 0.075 )} }{2 (1)}$$
$$x_2 = -4.5 \times 10^{-3} $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 4.25 \times 10^{-3} $$
6. $$[H_3O^+] = x = 4.25 \times 10^{-3} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 4.25 \times 10^{-3} ) = 2.37 $$
(b)
3. Assuming $ 0.045 \gt\gt x:$
$$K_a = \frac{x^2}{[ HZ ]_{initial}}$$
$$x = \sqrt{K_a \times [ HZ ]_{initial}} = \sqrt{ 2.55 \times 10^{-4} \times 0.045 }$$
$x = 3.4 \times 10^{-3} $
6. $$[H_3O^+] = x = 3.4 \times 10^{-3} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 3.4 \times 10^{-3} ) = 2.47 $$
$$pOH = 14.00 - 2.47 = 11.53$$
