Question

Calculate amount of ammonium chloride required to dissolve in 500 ml water to make pH = 4.5
$(K_{b}forN{H}_3=1.8\times {10}^{-5})$

• A. 48.15 gm
• B. 47.15 gm
• C. 48.12 gm
• D. 48.18 gm

Answer 1

The correct option is A 48.15 gm

$\left[H^{+}\right]={10}^{-pH}={10}^{-4.5}=3.162\times {10}^{-5}M$
$N{H}^{+}\rightleftharpoons N{H}_3+H^{+}$
$C(1-h)$ $chch$
$K_h=C{h}^2[\text{assuming h}\ll 1]$
$K_h=\frac{Kw}{Kb}=\frac{{10}^{-4}}{1.8\times {10}^{-5}}=5.5\times {10}^{-10}$
$\therefore h=\frac{K_h}{ch}=\frac{K_n}{\left[H^{+}\right]}=\frac{5.5\times {10}^{-10}}{3.162\times {10}^{-5}}=1.74\times {10}^{-5}$
$\therefore C=\frac{\left[H^{+}\right]}{h}=\frac{3.162\times {10}^{-5}}{1.74\times {10}^{-5}}=1.8mol/L$
500 ml of $H_{2}O$ contain $\frac{1.8}{2}=0.9$
mole Mass in gm $=0.9\times 53.5=48.15$ gm