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Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Equilibrium

The $pH$ of 0.1 M solution of $NH_4OH$ (having dissociation constant $K_b = 1.0 \times 10^{-5})$ is equal to

(a) 10

(b) 6

(c) 11

(d) 12

1 Answer

Answer: 11
 
$[OH^-] = \sqrt{K_b \times (NH_4OH)} = (10^{-5} \times 10^{-1})^{\frac{1}{2}} = 1 \times 10^{-3}$
 
$[H^+] = \frac{10^{-14}}{10^{-3}} = 10^{-11}$
 
Hence, $pH = - log 10^{-11} = 11.0$

 

answered Nov 29, 2013 by mosymeow_1
 

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