Answer
a) Initial concentration: $0.515\ g\div 94.11\ g/mol\div 125/1000\ L=0.044\ M$
$K_a=1.3\cdot10^{-10}=x\cdot x/(0.044\ M - x)$
$x=2.39\cdot10^{-6}\ M=[H_3O^+]$
$pH=5.62$
b) With all the hydroxide being consumed:
The initial number of moles of hydroxide: $0.044\ M\times 125/1000=0.0055\ mol$
Volume added: $0.0055\ mol\div 0.123\ M=44.7\ mL$
The concentration of the phenol anion: $0.0055\ mol\div (125+44.7)/1000\ L=0.0324\ M$
Equilibrium:
$K_w/K_a=[OH^-][Hp]/[p^-]=x\cdot x/(0.0323-x)$
$x=0.0015\ M=[OH^-], [C_6H_5O^-]=0.0308\ M$
$[H_3O^+]=6.7\cdot10^{-12}\ M$
Sodium is just diluted: $[Na^+]=0.123\ M\times 44.7/(125+44.7)=0.0323\ M$
c) $pH=11.18$
Work Step by Step
a) Initial concentration: $0.515\ g\div 94.11\ g/mol\div 125/1000\ L=0.044\ M$
$K_a=1.3\cdot10^{-10}=x\cdot x/(0.044\ M - x)$
$x=2.39\cdot10^{-6}\ M=[H_3O^+]$
$pH=5.62$
b) With all the hydroxide being consumed:
The initial number of moles of hydroxide: $0.044\ M\times 125/1000=0.0055\ mol$
Volume added: $0.0055\ mol\div 0.123\ M=44.7\ mL$
The concentration of the phenol anion: $0.0055\ mol\div (125+44.7)/1000\ L=0.0324\ M$
Equilibrium:
$K_w/K_a=[OH^-][Hp]/[p^-]=x\cdot x/(0.0323-x)$
$x=0.0015\ M=[OH^-], [C_6H_5O^-]=0.0308\ M$
$[H_3O^+]=6.7\cdot10^{-12}\ M$
Sodium is just diluted: $[Na^+]=0.123\ M\times 44.7/(125+44.7)=0.0323\ M$
c) $pH=11.18$
