## Chemistry and Chemical Reactivity (9th Edition)

pH = 10.079 Hydronium concentration: $8.333 \times 10^{-11}M$
- KOH is a strong base, therefore: $[OH^-] = [KOH] = 1.2 \times 10^{-4}M$ 1. Calculate the pOH, then find the pH. $pOH = -log[OH^-]$ $pOH = -log( 1.2 \times 10^{- 4})$ $pOH = 3.921$ $pH + pOH = 14$ $pH + 3.921 = 14$ $pH = 10.079$ 2. Calculate the $[H_3O^+]$ $[OH^-] * [H_3O^+] = Kw = 10^{-14}$ $1.2 \times 10^{- 4} * [H_3O^+] = 10^{-14}$ $[H_3O^+] = \frac{10^{-14}}{ 1.2 \times 10^{- 4}}$ $[H_3O^+] = 8.333 \times 10^{- 11}$