Chapter 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria - Study Questions - Page 677e: 90

$400.0\ mL$

$pH=pKa+\log([NH_3]/[NH_4^+])$ $9.00=9.25+\log([NH_3]/[NH_4^+])$ $0.562=[NH_3]/[NH_4^+]$ $[NH_3]=(0.250\ M\times0.5\ L-0.200\ M\times V)\div(0.5+V)$ $[NH_4^+]=0.200\ M\times V\div(0.5+V)$ $[NH_3]/[NH_4^+] = (0.250\ M\times0.5\ L-0.200\ M\times V)\div(0.200\ M\times V)$ $0.562=(0.125-0.2V)/0.2V$ $0.112V=0.125-0.2V$ $V=0.125/0.312$ $V=0.40\ L = 400.0\ mL$