## Chemistry (4th Edition)

(a) Since $HCl$ is a strong acid: $[H_3O^+] = [HCl] = 0.12M$ -Find the pH. $pH = -log[H^+]$ $pH = -log(0.12)$ $pH = 0.92$ (b) Since $HNO_3$ is a strong acid: $[H_3O^+] = [HNO_3] = 2.4M$ -Find the pH $pH = -log(2.4)$ $pH = -0.38$ (c) Since $HClO_4$ is a strong acid: $[H_3O^+] = [HClO_4] = 3.2 \times 10^{-4}$ - Find the pH: $pH = -log[H^+]$ $pH = -log( 3.2 \times 10^{- 4})$ $pH = 3.49$