## General Chemistry: Principles and Modern Applications (10th Edition)

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.