Answer
a) Initial concentration: $0.235\ g\div 122.12\ g/mol\div 100/1000\ L=0.019\ M$
$K_a=6.3\cdot10^{-5}=x\cdot x/(0.019\ M - x)$
$x=1.06\cdot10^{-3}\ M=[H_3O^+]$
$pH=2.97$
b) With all the hydroxide being consumed:
The initial number of moles of hydroxide: $0.019\ M\times 100/1000=0.0019\ mol$
Volume added: $0.0019\ mol\div 0.108\ M=17.6\ mL$
The concentration of benzoate: $0.0019\ mol\div (100+17.6)/1000\ L=0.016\ M$
Equilibrium:
$K_w/K_a=[OH^-][Hb]/[b^-]=x\cdot x/(0.016-x)$
$x=1.6\cdot10^{-6}\ M=[OH^-], [C_6H_5CO_2^-]=0.016\ M$
$[H_3O^+]=6.25\cdot10^{-9}\ M$
Sodium is just diluted: $[Na^+]=0.108\ M\times 17.6/(100+17.6)=0.016\ M$
c) $pH=8.20$
Work Step by Step
a) Initial concentration: $0.235\ g\div 122.12\ g/mol\div 100/1000\ L=0.019\ M$
$K_a=6.3\cdot10^{-5}=x\cdot x/(0.019\ M - x)$
$x=1.06\cdot10^{-3}\ M=[H_3O^+]$
$pH=2.97$
b) With all the hydroxide being consumed:
The initial number of moles of hydroxide: $0.019\ M\times 100/1000=0.0019\ mol$
Volume added: $0.0019\ mol\div 0.108\ M=17.6\ mL$
The concentration of benzoate: $0.0019\ mol\div (100+17.6)/1000\ L=0.016\ M$
Equilibrium:
$K_w/K_a=[OH^-][Hb]/[b^-]=x\cdot x/(0.016-x)$
$x=1.6\cdot10^{-6}\ M=[OH^-], [C_6H_5CO_2^-]=0.016\ M$
$[H_3O^+]=6.25\cdot10^{-9}\ M$
Sodium is just diluted: $[Na^+]=0.108\ M\times 17.6/(100+17.6)=0.016\ M$
c) $pH=8.20$
