## Chemistry (4th Edition)

(a) 1. HCl is a strong acid, therefore: $[H_3O^+] = [HCl] = 0.0010M$ $pH = -log[H_3O^+]$ $pH = -log( 1 \times 10^{- 3})$ $pH = 3$ (b) 1.KOH is a strong base, therefore: $[OH^-] = [KOH] = 0.76M$ $pOH = -log[OH^-]$ $pOH = -log( 0.76)$ $pOH = 0.119$ $pH + pOH = 14$ $pH + 0.119 = 14$ $pH = 13.88$