Chapter 14 - Acids and Bases - Questions for Review and Thought - Topical Questions - Page 652b: 34d

$pH = 1.602$ $[OH^-] = 4 \times 10^{- 13}M$ Acidic

$pH = -log[H_3O^+]$ $pH = -log( 0.025)$ $pH = 1.602$ $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $ 2.5 \times 10^{- 2} * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{ 2.5 \times 10^{- 2}}$ $[OH^-] = 4 \times 10^{- 13}M$ $[H_3O^+] > [OH^-]$: Acidic solution; Assuming $25^{\circ} C$: pH < 7: Acidic solution;