# $pK_b$ of $NH_3$ is 4.74. The $pH$ when 100 mL of 0.01 M $NH_3$ solution is 50% neutralised by 0.01 M HCl is,

(a) 4.74

(b) 2.37

(c) 9.26

(d) 9.48

At 50% neutralisation,
$[NH_4^+] = [NH_3]$

The solution is a buffer
$\therefore pOH = pK_b + log \frac{[NH_4^+]}{[NH_3]}$ = 4.74 + log 1 = 4.74

pH = 14 - 4.74 = 9.26