Answer
(d) 0.011 M
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ C_2H_5NH_2 ]& [ C_2H_5{NH_3}^{+} ]& [ OH^- ]\\
Initial& 0.200 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.200 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_b$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_b = \frac{[Products]}{[Reactants]} = \frac{[ C_2H_5{NH_3}^{+} ][ H^+ ]}{[ C_2H_5NH_2 ]}$$
$$K_b = \frac{(x)(x)}{[ C_2H_5NH_2 ]_{initial} - x}$$
3. Assuming $ 0.200 \gt\gt x$:
$$K_b = \frac{x^2}{[ C_2H_5NH_2 ]_{initial}}$$
$$x = \sqrt{K_b \times [ C_2H_5NH_2 ]_{initial}} = \sqrt{ 5.6 \times 10^{-4} \times 0.200 }$$
$x = 0.011 $
4. Test if the assumption was correct:
$$\frac{ 0.011 }{ 0.200 } \times 100\% = 5 \%$$
$[OH^-] = x = 0.011 \space M$
