Question

A solution contains 0.2 M ${NH}_{4}OH$ and 0.2 ${NH}_{4}CI.$ If 1.0 ml of 0.001 M HCl is added to it. What will be the $\left[{OH}^{-}\right]$ of the resulting solution $[K_b=2\times {10}^{-5}]$:

• A. $2\times {10}^{-5}$
• B. $5\times {10}^{-10}$
• C. $2\times {10}^{-3}$
• D. None of these

Answer 1

The correct option is A $2\times {10}^{-5}$

$N{H}_{4}OH+HCl⟶N{H}_4$
$0.2$
Per $=P{k}_b+\log \frac{\left(Srt\right)}{\left(Brt\right)}$
per $=\log [2\times {10}^{-5}]+\log \frac{0.2}{6.2}$
Per $=-\log \times {10}^{-5}$
$\left[o{n}^{-}\right]=2\times {10}^{-5}$