Chemistry - Ionic Equilibrium Question with Solution | TestHub

Milli.moles of$\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}=30\times 0.2=6$

m.m of$\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}=50\times 0.2=10$

Now:



$\mathrm{p}\mathrm{H}=-\log (2\times 10^{-5})\mathrm{ }+\log \frac{6}{4}$

$=4.87$

Suppose 'v'mL of$\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$is added then

m.mol of${\mathrm{C}\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{N}\mathrm{a}=\left(6+\mathrm{v}\times 0.2\right)$

m.mol of${\mathrm{C}\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}=\left(4-\mathrm{v}\times 0.2\right)$

$\text{pH}={\text{pK}}_{\text{a}}+\log \left[\frac{\text{Salt}}{\text{Acid}}\right]$
$5=-\log (2\times 10^{-5})\mathrm{ }+\log \left(\frac{6+0.2\mathrm{v}}{4-0.2\mathrm{v}}\right)$

$0.3010=\log \left(\frac{6+0.2\mathrm{v}}{4-0.2\mathrm{v}}\right)$

So$\left(\frac{6+0.2\mathrm{v}}{4-0.2\mathrm{v}}\right)=2\Rightarrow \mathrm{v}=3.33 \mathrm{m}\mathrm{L}=\frac{10}{3}\mathrm{m}\mathrm{L}$

So$\mathrm{x}=3$