Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria - Study Questions - Page 677d: 81

Answer

$pH=9.73$ $[H_3O^+]=1.87\dot{}10^{-10}$

Work Step by Step

Consumption of the $H_3O^+$: $nH_3O^+=0.20\ M\dot{}25.0\ mL\dot{} 10^{-3} L/mL = 0.005 mol$ $nNH_3(before) = 0.40\ M\dot{} 50\ mL \dot{} 10^{-3} L/mL =0.02\ mol$ $[NH_3](after) = (0.02- 0.005)mol \div ((50+25)\ mL\times10^{-3}\ mL/L)=0.2\ M$ $[NH_4^+] = 0.005\ mol \div ((50+25)\ mL\times10^{-3}\ mL/L)=0.067\ M$ By the Henderson-Hasselbalch equation: $pH = pKa + \log ([NH_3]/[NH_4^+])$ $pKb$ of ammonia $4.75$, $pKa = 14-pKb=9.25$ $pH = 9.25 + \log(0.2/0.067)$ $pH = 9.25+0.48$ $pH=9.73$ $[H_3O^+]=10^{-pH}=1.87\dot{}10^{-10}$
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