Answer
a) Equivalence: $0.0105\ M\times 36.78\ mL = c_{NH_3}\times 25.0\ mL$
$c=0.0154\ M$
b) Formed ammonium: $(0.0105\ M\times 36.78\ mL)/(25+36.78)=0.00625\ M$
Equilibrium:
$K_a=x\cdot x/(0.00625-x)$
$x=1.87\cdot10^{-6}\ M=[H_3O^+],\ [NH_4^+]=0.00624\ M$
$[OH^-]=5.35\cdot10^{-9}\ M$
c) $pH=5.73$
Work Step by Step
a) Equivalence: $0.0105\ M\times 36.78\ mL = c_{NH_3}\times 25.0\ mL$
$c=0.0154\ M$
b) Formed ammonium: $(0.0105\ M\times 36.78\ mL)/(25+36.78)=0.00625\ M$
Equilibrium:
$K_a=x\cdot x/(0.00625-x)$
$x=1.87\cdot10^{-6}\ M=[H_3O^+],\ [NH_4^+]=0.00624\ M$
$[OH^-]=5.35\cdot10^{-9}\ M$
c) $pH=5.73$
