Answer
Original base molarity: $0.1547M$
Work Step by Step
1. Find $[OH^-]$
pH + pOH = 14
11.22 + pOH = 14
pOH = 2.78
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 2.78}$
$[OH^-] = 1.66 \times 10^{- 3}$
2. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [N{H_4}^+] = x$
-$[NH_3] = [NH_3]_{initial} - x$
3. Now, use the Kb and x values and equation to find the initial concentration value.
$Kb = \frac{[OH^-][N{H_4}^+]}{ [Initial NH_3] - x}$
$ 1.8\times 10^{- 5}= \frac{[x^2]}{ [Initial NH_3] - x}$
$ 1.8\times 10^{- 5}= \frac{( 1.66\times 10^{- 3})^2}{[Initial NH_3] - 1.66\times 10^{- 3}}$
$[Initial NH_3] - 1.66\times 10^{- 3} = \frac{ 2.754\times 10^{- 6}}{ 1.8\times 10^{- 5}}$
$[Initial NH_3] - 1.66\times 10^{- 3} = 0.153$
$[Initial NH_3] = 0.153 + 1.66\times 10^{- 3}$
$[Initial NH_3] = 0.1547$
