Chapter 18 - Ionic Equilibria I: Acids and Bases - Exercises - The Autoionization of Water - Page 743: 15

(a) $[OH^-] = 1.587 \times 10^{- 11}M$ (b) $[OH^-] = 1.282 \times 10^{- 6}M$

(a) $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $ 6.3 \times 10^{- 4} * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{ 6.3 \times 10^{- 4}}$ $[OH^-] = 1.587 \times 10^{- 11}$ (b) $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $ 7.8 \times 10^{- 9} * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{ 7.8 \times 10^{- 9}}$ $[OH^-] = 1.282 \times 10^{- 6}$