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Hydrolysis constant of $NH^+_4$ is $5.55 \times 10^{-10}$. The ionisation constant of $NH^+_4$ is

(a) $1.8 \times 10^9$

(b) $5.55 \times 10^{-10}$

(c) $5.55 \times 10^4$

(d) $1.8 \times 10^{-5}$

Can you answer this question?
 
 

1 Answer

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Answer: $5.55 \times 10^{-10}$
 
$NH^+_4 + H_2O \rightleftharpoons NH_4OH + H^+$
 
$K_h = \frac{[NH_4OH][H^+]}{[NH^+_4]}$
 
Since $[NH_4OH] = [H^+]$
$\therefore K_h = \frac{[H^+]^2}{[NH^+_4]}$
 
$NH^+_4$ ionises as follows:
$NH^+_4 \rightleftharpoons NH_3 + H^+$
 
$K_a = \frac{[NH_3][H^+]}{[NH^+_4]}$
 
Since, $[NH_3] = [H^+]$
$\therefore K_a = \frac{[H^+]^2}{[NH^+_4]}$
 
$\therefore K_a = K_h = 5.55 \times 10^{-10}$

 

answered Nov 29, 2013 by mosymeow_1
 

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