When `2.5 mL` of `2//5M` weak monoacidic base `(K_(b) = 1 xx 10^(-12) at 25^(@)C)` is titrated with `2//15 M HCI` in water at `25^(@)C` the concentration of `H^(o+)` at equivalence point is `(K_(w) = 1 xx 10^(-14) at 25^(@)C)`

Options

A. `3.7xx10^(-13)M`
B. `3.2xx10^(-7)M`
C. `3.2xx10^(-2)M`
D. `2.7xx10^(-2)M`

Answer

D

Explanation

`BOH+HCIrarrBCI+H_(2)O`
Meq.of BOH = Meq.of HCI = Meq.of BCI
`2.5xx(2)/(5)xx1=Vxx(2)/(15)xx1=1`
`:. V= 7.5mL`
`:. "Total volume" = 2.5+7.5=10mL`
Thus, for hydrolysis of BCI,
`K_(H) = (Ch^(2))/(1-h)= (K_(w))/(K_(b)) or h= 0.27`
Now, `[H^(+)]= C.h`
`= 0.1xx0.27xx10^(-2)M`