2.5 mL of `(2M)/(5)` weak monoacidic base `(K_(b) = 1xx10^(-12) at 25^(@)C)` is titrated with `(2)/(15)`MHCl in water at `25^(@)C`. The concentration of that equivalence point is:

Options

A. `3.7 xx 10^(-13)M`
B. `3.2 xx 10^(7)M`s
C. `3.2 xx 10^(-2)M`
D. `2.7 xx 10^(-2)M`

Answer

D

Explanation

`BOH + HCl rarr BCl + H_(2)O`
Meq. of BOH = Meq. of Hcl Meq. of BCl
`2.5 xx (2)/(5) xx 1 V xx (2)/(15) xx 1 xx 1 therefore V = 7.5 mL therefore` Total volume `= 2.5 + 7.5 = 10 mL`
Thus, `[BCl] = (1)/(10) = 0.1`, Now for hydrolysis of BCl, `K_(H) = (Ch^(2))/(1-H) = (K_(w))/(K_(b))`
`therefore h = 0.27` `[H^(+)] = c.h. = 0.27 xx 0.1 = 2.7 xx 10^(-2)M`