#### Answer

The initial base concentration should be 0.0273M.

#### Work Step by Step

1. Calculate $[OH^-]:$
pH + pOH = 14
11.12 + pOH = 14
pOH = 2.88
$[OH^-] = 10^{- 2.88}$
$[OH^-] = 1.31 \times 10^{- 3}$
2. Knowing that this solution is pure, we can assume:
$[OH^-] = [(CH_3)_3NH^+] = 1.31 \times 10^{-3}M$
3. Therefore, using the Kb equation we got:
$Kb = \frac{(1.31 \times 10^{-3})(1.31 \times 10^{-3})}{ [(CH_3)_3NH]}$
$6.3 \times 10^{-5} = \frac{1.72 \times 10^{-6}}{[(CH_3)_3NH]}$
$ [(CH_3)_3NH] = \frac{1.72 \times 10^{-6}}{6.3 \times 10^{-5}} = 2.73 \times 10^{-2}M$
- 5% ionization test: $\frac{1.31 \times 10^{-3}}{2.73 \times 10^{-2}} \times 100\% = 4.8\% = OK $
3. Therefore, the initial base concentration should be 0.0273M.