## Chemistry and Chemical Reactivity (9th Edition)

a) $pH=7.81$ b) $m=4.324\ g$
a) $n(KH_2PO_4)=1.360\ g\div136.09\ g/mol=0.01\ mol$ $n(Na_2HPO_4)=5.677\ g\div 141.96\ g/mol=0.04\ mol$ By the Henderson-Hasselbach equation: $pH=pKa+\log([HPO_4^{2-}]/[H_2PO_4^-])$ $pH=7.21+\log(0.04/V\div 0.01/V)$ $pH=7.81$ b) For $pH=7.31$: $pH=pKa+\log([HPO_4^{2-}]/[H_2PO_4^-])$ $7.31=7.21+\log(0.04/V\div n/V)$ $0.1=\log(0.04/n)$ $n=0.032\ mol$ $m=0.032\ mol\dot{} 136.09\ g/mol=4.324\ g$