A solution is prepared by mixing $0.01\mathrm{mol}$ each of ${\mathrm{H}}_2{\mathrm{CO}}_3,{\mathrm{NaHCO}}_3,{\mathrm{Na}}_2{\mathrm{CO}}_3$, and $\mathrm{NaOH}$ in $100 \mathrm{~mL}$ of water. $\mathrm{pH}$ of the resulting solution is
[Given: ${\mathrm{pK}}_{{\mathrm{a}}_1}$ and ${\mathrm{pK}}_{{\mathrm{a}}_2}$ of ${\mathrm{H}}_2{\mathrm{CO}}_3$ are $6.37$ and $10.32$, respectively. $\log 2=0.30$]

First acid base reaction between ${\mathrm{H}}_2{\mathrm{CO}}_3$ and $\mathrm{NaOH}$ takes place.

$\underset{0.01\mathrm{mole}}{{\mathrm{H}}_2{\mathrm{CO}}_3}+\underset{0.01\mathrm{mole}}{\mathrm{NaOH}}⟶\underset{0.01\mathrm{mole}}{\underset{-}{{\mathrm{NaHCO}}_3}}+{\mathrm{H}}_2\mathrm{O}$

After the acid base reaction, we have $0.01\mathrm{mole} {\mathrm{Na}}_2{\mathrm{CO}}_3$  and $0.02$ moles of ${\mathrm{NaHCO}}_3$. Here, This will form an acidic buffer of ${\mathrm{NaHCO}}_3$ and ${\mathrm{Na}}_2{\mathrm{CO}}_3$.

$\therefore \mathrm{pH}={\mathrm{pK}}_{{\mathrm{a}}_2}+\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}$

$=10.32+\log \frac{\left(\frac{0.01}{0.1}\right)}{\left(\frac{0.02}{0.1}\right)}$

$=10.32+\log \frac{1}{2}$

$=10.32-\log 2$

$=10.32-0.3$

$=10.02$

$\therefore \mathrm{pH}=10.02$