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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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A weak base, B has basicity constant, $K_b = 2 \times 10^{-5}$. The $pH$ of any solution in which $[B] = [BH^+]$ is

(a) 4.7

(b) 7.9

(c) 9.3

(d) 9.7

Can you answer this question?
 
 

1 Answer

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Answer: 9.3010
 
$B + H_2O \rightleftharpoons BH^+ + OH^-$
 
$K_b = \frac{[BH^+][OH^-]}{[B]}$
Given, $[BH^+] = [B]$
$\therefore K_b = [OH^-]$ = $2 \times 10^{-5}$
 
$[H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{2 \times 10^{-5}} = \frac{10^{-9}}{2}$
 
$pH = - log [H^+] = - log \frac{10^{-9}}{2} = 9 + log 2 = 9 + 0.3010 = 9.3010$

 

answered Nov 29, 2013 by mosymeow_1
 

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