## General Chemistry (4th Edition)

$pOH = 1.40$ $pH = 12.60$
1. $Ba(OH)_2$ is a strong base with 2 $OH^-$ in each molecule: $[OH^-] = 2 * [Ba(OH)_2] = 2* 0.020M = 0.040M$ 2. Calculate pOH and pH: $pOH = -log[OH^-]$ $pOH = -log( 0.040)$ $pOH = 1.40$ $pH + pOH = 14$ $pH + 1.40 = 14$ $pH = 12.60$