Answer
(a) $$[H_3O^+] = 6.7 \times 10^{-4} $$
(b) $$[OH^-] = 3.4 \times 10^{-12} \space M$$
Work Step by Step
$$K_a = 10^{-4.89} = 1.29 \times 10^{-5}$$
(a)
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HQ ]& [ Q^- ]& [ H_3O^+ ]\\
Initial& 3.5 \times 10^{-2} & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 3.5 \times 10^{-2} -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_a = \frac{P_{Products}}{P_{Reactants}} = \frac{[ Q^- ][ H_3O^+ ]}{[ HQ ]}$$
$$K_a = \frac{(x)(x)}{[ HQ ]_{initial} - x}$$
3. Assuming $ 3.5 x 10^{-2} \gt\gt x:$
$$K_a = \frac{x^2}{[ HQ ]_{initial}}$$
$$x = \sqrt{K_a \times [ HQ ]_{initial}} = \sqrt{ 1.29 \times 10^{-5} \times 3.5 \times 10^{-2} }$$
$x = 6.7 \times 10^{-4} $
4. Test if the assumption was correct:
$$\frac{ 6.7 \times 10^{-4} }{ 3.5 \times 10^{-2} } \times 100\% = 1.9 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 6.7 x 10^{-4} $
6. $$[H_3O^+] = x = 6.7 \times 10^{-4} $$
(b)
3. Assuming $ 0.65 \gt\gt x:$
$$K_a = \frac{x^2}{[ HQ ]_{initial}}$$
$$x = \sqrt{K_a \times [ HQ ]_{initial}} = \sqrt{ 1.29 \times 10^{-5} \times 0.65 }$$
$x = 2.9 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 2.9 \times 10^{-3} }{ 0.65 } \times 100\% = 0.45 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 2.9 \times 10^{-3} $
6. $$[H_3O^+] = x = 2.9 \times 10^{-3} $$
7. Calculate the hydroxide ion concentration:
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 2.9 \times 10^{-3} } = 3.4 \times 10^{-12} \space M$$
