Question

2.5 ml of `(2)/(5)M` weak monoacidic base (`K_(b) = 1 xx 10^(-12)` at `25^(@)C`) is titrated with `(2)/(15)` M HCl in water at `25^(@)C`. The concentration of `h^(+)` at equivalence point is (`K_(w) = 1xx 10^(-14)` at `25^(@)C`)
A. `3.7 xx 10^(-13) M`
B. `3.2 xx 10^(-7) M`
C. `3.2 xx 10^(-2) M`
D. `2.7 xx 10^(-2) M`

Answer 1

Correct Answer - C
`BOH + HCl rarr BCl + H_(2)O`
At equivalence, `M_(1)V_(1) = M_(2)V_(2)`
`(2)/(5) xx 2.5 = (2)/(15) xx V_(2), V_(2) = 7.5 ml HCl = 7.5 ml`
Moles of BOH in 2.5 ml = 0.01
Moles of HCl in 7.5 ml = 0.001
`:.` Moles of salt formed = 0.001 [salt of S.A.F. W. B]
Total value = 2.5 + 7.5 = 10 ml. = 0.01
Conc. of salt `= (0.001)/(0.01) = 0.1` mole/litre
Hydrolysis of salt take place `pH = 7 - (1)/(2) (pK_(b) + log C)`
`pK_(b) = -log 10^(-12) = 12 pH = 7 - (1)/(2)(12-1)`
`[because log C = -1 log.1 = -1]`
`pH = 1.5, -log H^(+) = 1.5 log H^(+) = -1.5, log H^(+) = 2.5`
`H^(+) = 3.2 xx 10^(-2)`.