Answer
$$[H_3O^+] = 2.1 \times 10^{-2} \space M$$
$$[N{O_2}^-] = 2.1 \times 10^{-2} \space M$$
$$[OH^-] = 4.8 \times 10^{-13} \space M$$
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HNO_2 ]& [ N{O_2}^- ]& [ H_3O^+ ]\\
Initial& 0.60 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.60 -x& x& x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_a = \frac{P_{Products}}{P_{Reactants}} = \frac{[ N{O_2}^- ][ H_3O^+ ]}{[ HNO_2 ]}$$
$$K_a = \frac{(x)(x)}{[ HNO_2 ]_{initial} - x}$$
3. Assuming $ 0.60 \gt\gt x:$
$$K_a = \frac{x^2}{[ HNO_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HNO_2 ]_{initial}} = \sqrt{ 7.1 \times 10^{-4} \times 0.60 }$$
$x = 0.021 $
4. Test if the assumption was correct:
$$\frac{ 0.021 }{ 0.60 } \times 100\% = 3.5 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 0.021 $
6. $$[H_3O^+] = [N{O_2}^-] = x = 0.021 = 2.1 \times 10^{-2}$$
7. Calculate the hydroxide ion concentration:
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 0.021 } = 4.8 \times 10^{-13} \space M$$
