Question

A solution is prepared by mixing $0.01$ mole each of ${\text{H}}_2{\text{CO}}_3,{\text{NaHCO}}_3,{\text{Na}}_2{\text{CO}}_3$, and $\text{NaOH}$ in $100\text{mL}$ of water. $\text{pH}$ of the resulting solution is

[Given : ${\text{pKa}}_1$ and ${\text{pKa}}_2$ of ${\text{H}}_2{\text{CO}}_3$ are $6.37$ and 10.32, respectively; $\log 2=0.30$]

Answer 1

${\text{n}}_{{\text{H}}_2{\text{CO}}_3}={\text{n}}_{{\text{NaHCO}}_3}={\text{n}}_{{\text{Na}}_2{\text{CO}}_3}={\text{n}}_{\text{NaOH}}=0.01$ mol

$\begin{array}{ccccc}{H}_2{CO}_3& +& \text{NaOH}& ⟶& {NaHCO}_3\\ \text{0.01 mol}& +& \text{0.01 mol}& & \text{0.01 mol}\\ 0& & 0& & \text{0.02 mol}\\ {NaHCO}_3& +& {\text{Na}}_2{\text{CO}}_3& & \\ \text{0.02 mol}& +& \text{0.01 mol}& & \end{array}$

So the resulting solution acts as a buffer solution of $N{a}_{2}C{O}_3$ and $NaHC{O}_3$.

$\text{pH}={\text{pKa}}_2+\log \frac{\left[{\text{Na}}_2{\text{CO}}_3\right]}{\left[{\text{NaHCO}}_3\right]}$

$=10.32+\log \left(\frac{0.01}{0.02}\right)$
$=10.32-0.3$
$\text{pH}=10.02$