Chapter 16 - Acids and Bases - Example 16-2 - Relating [H(3)0+],[OH],pH, and pOH - Page 706: Practice Example A

$[H^+] = 1.41 \times 10^{-3}$ $[OH^-] = 7.08 \times 10^{-12}$

Equations that will be used: $[H^+] = 10^{-pH}$ $[OH^-] = 10^{-pOH}$ $pH + pOH = 14$ 1. Calculate the $[H^+]$ $[H^+] = 10^{-2.85} = 1.41 \times 10^{-3}$ 2. Find the pOH $2.85 + pOH = 14$ $pOH = 14 - 2.85 = 11.15$ 3. Calculate $[OH^-]$ $[OH^-] = 10^{-11.15} = 7.08 \times 10^{-12}$