Answer
a) $pH=7.81$
b) $m=4.324\ g$
Work Step by Step
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 $pH=7.31$:
$pH=pKa+\log([HPO_4^{2-}]/[H_2PO_4^-])$
$7.31=7.21+\log(0.04/V\div n/V)$
$0.1=\log(0.04/n)$
$n=0.032\ mol$
$m=0.032\ mol\dot{} 136.09\ g/mol=4.324\ g$
