IIT JEE 2008 P1 Q52

let V be the volume of HCl used...

2.5 * 2/5 = V * 2/15

V= 7.5 ml.....

total volume = 10 ml...
Conc. of salt produced (c) = (2.5 * 2/5)/10
= 0.1 M

K_h= K_w/K_b

= 10^-14/10^-12

=10^-2

for hydrolysis of base..

B⁺ +H₂O -> BOH + H⁺

c 0 0

c(1-h) ch ch

K_h= ch²/(1-h)

Putting the values of K_h = 10^-2 and c= 0.1 M

We get h= 0.27

[H⁺] = 0.1 * 0.27 = 2.7 * 10^-2 M

that will be when in equation ch²/1-h u take 1-h ~ 1

bhaiya what is the correct answer.....

bhaiya iska answer toh do.....

Ans D

http://www.iitkgp.ernet.in/jee/paper12008.pdf