Answer
$pH = 10.08$
Work Step by Step
1. Drawing the ICE table we get these concentrations at the equilibrium:
$CH_3NH_2(aq) + H_2O(l) \lt -- \gt CH_3N{H_3}^+(aq) + OH^-(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[CH_3NH_2] = 0.2 M - x$
$[CH_3N{H_3}^+] = 0.6M + x$
$[OH^-] = 0 + x$
2. Calculate 'x' using the $K_b$ expression.
$ 3.7\times 10^{- 4} = \frac{[CH_3N{H_3}^+][OH^-]}{[CH_3NH_2]}$
$ 3.7\times 10^{- 4} = \frac{( 0.6 + x )* x}{ 0.2 - x}$
Considering 'x' has a very small value.
$ 3.7\times 10^{- 4} = \frac{ 0.6 * x}{ 0.2}$
$ 3.7\times 10^{- 4} = 3x$
$\frac{ 3.7\times 10^{- 4}}{ 3} = x$
$x = 1.2\times 10^{- 4}$
Percent ionization: $\frac{ 1.2\times 10^{- 4}}{ 0.2} \times 100\% = 0.062\%$
x = $[OH^-]$
3. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 1.2 \times 10^{- 4})$
$pOH = 3.92$
4. Find the pH:
$pH + pOH = 14$
$pH + 3.92 = 14$
$pH = 10.08$
