## Chemistry (7th Edition)

Published by Pearson

# Chapter 15 - Aqueous Equilibria: Acids and Bases - Worked Example - Page 621: 15

#### Answer

(a) $pH = 1.301$ (b) $pH = -0.7781$ (c) $pH = 14.602$ (d) $pH = 12.301$

#### Work Step by Step

(a) $HClO_4$: Monoprotic (One $H^+$) strong acid: $[H_3O^+] = [HClO_4] = 0.050M$ $pH = -log[H_3O^+]$ $pH = -log( 0.05)$ $pH = 1.301$ (b) $HCl$: Monoprotic (One $H^+$) strong acid: $[H_3O^+] = [HCl] = 6.0M$ $pH = -log[H_3O^+]$ $pH = -log( 6)$ $pH = -0.7781$ (c) $KOH$: Strong base with one $OH^-$: $[OH^-] = 1 * [KOH] = 4.0M$ $pOH = -log[OH^-]$ $pOH = -log( 4)$ $pOH = -0.602$ $pH + pOH = 14$ $pH + -0.602 = 14$ $pH = 14.602$ (d) $Ba(OH)_2$: Strong base with 2 $OH^-$: $[OH^-] = 2 * [Ba(OH)_2] = 0.010 * 2 = 0.020M$ $pOH = -log[OH^-]$ $pOH = -log( 0.02)$ $pOH = 1.699$ $pH + pOH = 14$ $pH + 1.699 = 14$ $pH = 12.301$

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