## Chemistry (7th Edition)

a) $pH = 0.32$ b) $pH = 11.40$ c) $pH = 12.88$
a) 1. Since $HCl$ is a strong acid: $[H_3O^+] = [HCl] = 0.48$ 2. Calculate the pH value: $pH = -log[H_3O^+]$ $pH = -log( 0.48)$ $pH = 0.3188 = 0.32$ b) 1. Since $Ba(OH)_2$ is a strong base, and it has 2 $OH^-$ in each molecules: $[OH^-] = 2 * [Ba(OH)_2] = 2 * 2.5\times 10^{- 3} = 5\times 10^{- 3}M$ 2. Calculate the pOH, and then the pH value: $pOH = -log[OH^-]$ $pOH = -log( 2.5 \times 10^{- 3})$ $pOH = 2.602$ $pH + pOH = 14$ $pH + 2.602 = 14$ $pH = 11.398 = 11.40$ c) 1. Since $NaOH$ is a strong base: $[OH^-] = [NaOH] = 0.075$ 2. Calculate the pOH, and then the pH value: $pOH = -log[OH^-]$ $pOH = -log( 0.075)$ $pOH = 1.125$ $pH + pOH = 14$ $pH + 1.125 = 14$ $pH = 12.875 = 12.88$