## Chemistry (4th Edition)

(a) LiOH is a strong base, therefore: $[OH^-] = [LiOH] = 1.24M$ $pOH = -log[OH^-]$ $pOH = -log( 1.24)$ $pOH = -0.0934$ $pH + pOH = 14$ $pH + (-0.0934) = 14$ $pH = 14.0934$ (b) $Ba(OH)_2$ is a strong base with 2 $OH^-$, therefore: $[OH^-] = 2 * [Ba(OH)_2] = 2 * 0.22 = 0.44M$ $pOH = -log[OH^-]$ $pOH = -log(0.44)$ $pOH = 0.356$ $pH + pOH = 14$ $pH + 0.356 = 14$ $pH = 13.644$ (c) NaOH is a strong base, therefore: $[OH^-] = [NaOH] = 0.085$ $pOH = -log[OH^-]$ $pOH = -log( 8.5 \times 10^{- 2})$ $pOH = 1.07$ $pH + pOH = 14$ $pH + 1.07 = 14$ $pH = 12.93$