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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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To prepare a buffer of $pH$ 8.26 amount of $(NH_4)_2SO_4$ to be added into 50 mL of 0.01 M $NH_4OH$ solution is [$pK_a (NH_4^+)$ = 9.26]

(a) 0.05 mol

(b) 0.025 mol

(c) 0.10 mol

(d) 0.005 mol

Can you answer this question?
 
 

1 Answer

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Answer: 0.025 mol
 
Let $(NH_4)_2SO_4$ to be added to 500 mL solution = x mol
 
$\therefore [NH^+_4] = 4x M$
$ pH = 8.26$
$ pOH = 14 - 8.26 = 5.74$
$pK_b = 14 - 9.26 = 4.74$
 
$pOH = pK_b + log \frac{[NH^+_4]}{[NH_4OH]}$
$\Rightarrow 5.74 = 4.74 + log \frac{4x}{0.01}$
 
$log 4x \times 10^2$ = 1
$\Rightarrow 4x \times 10^2 = 1$
$\Rightarrow x = \frac{1}{40}$ mol = 0.025 mol

 

answered Nov 29, 2013 by mosymeow_1
 

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