Answer
a) $pH=4.12$
b) $0.46\ g$ of benzoic acid
Work Step by Step
$[benzoic\ acid]=1.50\ g\div 122.12\ g/mol\div 0.150\ L=0.082\ M$
$[sodium\ benzoate]=1.50\ g\div 144.10\ g/mol\div 0.150\ L=0.069\ M$
a) $pH=pKa+\log([benzoate]/[benzoic\ acid])$
$pH=4.19+\log(0.069/0.082)$
$pH=4.12$
b) More benzoic acid must be added to lower the pH:
$pH=pKa+\log([benzoate]/[benzoic\ acid])$
$4.00=4.19+\log(0.069/[benzoic\ acid])$
$[benzoic\ acid]=0.107\ M$
$m(benzoic\ acid)=0.107M\times0.150\ L\times122.12\ g/mol=1.96\ g$
$m(added)=1.96-1.50=0.46\ g$
c) HCl must be added to lower the pH
$pH=pKa+\log([benzoate]/[benzoic\ acid])$
$4.00=4.19+\log([benzoate]/[benzoic\ acid])$
$[benzoate]/[benzoic\ acid]=0.646$
$[benzoate] = (0.069\ M\times0.150\ L-2.00\ M\times V)\div(0.150\ L+V)$
$[benzoic\ acid]=(0.082\ M\times0.150\ L+2.00\ M\times V)\div(0.150\ L+V)$
$0.646=(0.010-2V)\div(0.012+2V)$
$0.0078+1.292V=0.010-2V$
$3.292V=0.0022$
$V=0.00075\ L=0.75\ mL$
