Answer
a) Equivalence: $0.175\ M\times 25.67\ mL = c_{aniline}\times 25.0\ mL$
$c=0.18\ M$
b) Formed aniline cation:
$(0.175\ M\times 25.67\ mL)/(25+25.67)=0.089\ M$
Equilibrium: $K_b=4.0\cdot10^{-10}$
$K_w/K_b=x\cdot x/(0.089-x)$
$x=1.48\cdot10^{-3}\ M=[H_3O^+],\ [C_6H_5NH_3^+]=0.0875\ M$
$[OH^-]=6.76\cdot10^{-12}\ M$
c) $pH=2.83$
Work Step by Step
a) Equivalence: $0.175\ M\times 25.67\ mL = c_{aniline}\times 25.0\ mL$
$c=0.18\ M$
b) Formed aniline cation:
$(0.175\ M\times 25.67\ mL)/(25+25.67)=0.089\ M$
Equilibrium: $K_b=4.0\cdot10^{-10}$
$K_w/K_b=x\cdot x/(0.089-x)$
$x=1.48\cdot10^{-3}\ M=[H_3O^+],\ [C_6H_5NH_3^+]=0.0875\ M$
$[OH^-]=6.76\cdot10^{-12}\ M$
c) $pH=2.83$