Correct Answer - A
Weak monoacidic base such as `NH_(4)OH` (aqueous `NH_(3)`) is neutralized by HCI as follows: `NH_(4)OH(aq.)+HCl(aq.)rarrNH_(4)Cl(aq.)+H_(2)O(l)`
At the equivalence point, all `NH_(4)OH` gets converted into `NH_(4)Cl` (salt) which undergoes cationic hydrolysis as follows: `{:(,NH_(4)^(+)(aq.)+H_(2)O(l)hArrNH_(4)OH(aq.)+H^(+)(aq.)),("Initial (M)"," C 0 0"),("Change (M)", " -Ch +Ch +Ch"),("Equilibrium (M)",bar(" C(1-h) Ch Ch")):}`
Thus, the concentration of `H^(+)` ions is due to the hydrolysis of the resulting salt `(NH_(4)Cl)`. To calculate is value, we need the concentration of salt (C ) and the degree of hydrolysis of salt (h).
`"Concentration of salt" = ("Millimoles of base")/("Total volume")`
Millimoles of base `=M.V_(mL)`
`=((2)/(5))(2.5)`
Total volume `=V_("acid")+V_("base")`
`V_("acid")=(M_("base")V_("base"))/(M_("acid"))`
`=((2//5)(2.5))/((2//15))=7.5 mL`
`:. C_("salt")=((2//5)(2.5))/((7.5+2.5))=(1)/(10)=0.1`
For cationic hydrolysis,
`K_(h)=(Ch^(2))/(1-h)=(K_(w))/(K_(b))=(10^(-14))/(10^(-12))=10^(-2)`
We have
`C=0.1`, `K_(w)=1xx10^(-14)`, `K_(b)=1xx10^(-12)`
This gives `h=0.27` (since `K_(h)` is not neglected relative to 1)
`C_(H^(+))=Ch`
`=(0.1)(0.27)`
`=2.7xx10^(-2)M`
If we neglact h relative 1, the answer will be `3.2xx10^(-2)`M.
