# 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

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