A solution is prepared by mixing 0 . 01mol each of H _ 2 CO _ 3 ,...

First acid base reaction between ${H}_{2}{CO}_{3}$ and $NaOH$ takes place.

$\mathop{{H}_{2}{CO}_{3}}_{0 . 01mole} + \mathop{NaOH}_{0 . 01mole} ⟶ \mathop{\mathop{{NaHCO}_{3}}_{-}}_{0 . 01mole} + {H}_{2}O$

After the acid base reaction, we have $0 . 01mole{Na}_{2}{CO}_{3}$ and $0 . 02$ moles of ${NaHCO}_{3}$. Here, This will form an acidic buffer of ${NaHCO}_{3}$ and ${Na}_{2}{CO}_{3}$.

$∴ pH = {pK}_{{a}_{2}} + log\dfrac{\left[Salt\right]}{\left[Acid\right]}$

$= 10 . 32 + log\dfrac{\left(\dfrac{0 . 01}{0 . 1}\right)}{\left(\dfrac{0 . 02}{0 . 1}\right)}$

$= 10 . 32 + log\dfrac{1}{2}$

$= 10 . 32 - \log 2$

$= 10 . 32 - 0 . 3$

$= 10 . 02$

$∴ pH = 10 . 02$