## Chemistry (4th Edition)

(a) KOH is a strong base, therefore: $[OH^-] = [KOH] = 0.066M$ $pOH = -log[OH^-]$ $pOH = -log( 6.6 \times 10^{- 2})$ $pOH = 1.18$ $pH + pOH = 14$ $pH + 1.18 = 14$ $pH = 12.82$ (b) NaOH is a strong base, therefore: $[OH^-] = [NaOH] = 5.43M$ $pOH = -log[OH^-]$ $pOH = -log( 5.43)$ $pOH = -0.734$ $pH + pOH = 14$ $pH + (-0.734) = 14$ $pH = 14.734$ (c) $[Ba(OH)_2]$ is a strong base with 2 $OH^-$ for each molecule, therefore: $[OH^-] = [Ba(OH)_2]*2 = 0.74 * 2 = 1.48M$ $pOH = -log[OH^-]$ $pOH = -log( 1.48)$ $pOH = -0.17$ $pH + pOH = 14$ $pH + (-0.17) = 14$ $pH = 14.17$