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)

2.5 ml of (2)/(5)M weak monoacidic base (K_(b) = 1 xx 10^(-12) at 25^(

1.	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)
Answer» `3.7 xx 10^(-13) M` `3.2 xx 10^(-7) M` `3.2 xx 10^(-2) M` `2.7 xx 10^(-2) M` Solution :`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)`.