Question

A solution is prepared by mixing $0.01\, mol$ each of $H _2 CO _3, NaHCO _3, Na _2 CO _3$, and $NaOH$ in $100 \, mL$ of water $pH$ of the resulting solution is ___
[Given : $p K _{ a 1}$ and $p K _{ a 2}$ of $H _2 CO _3$ are $6.37$ and $10.32$, respectively $; \log 2=0.30$ ]

Answer 1

The initial acid-base reaction occurs between $H_2CO_3$ and NaOH.
In the resulting solution, there are 0.01 moles of $Na_2CO_3$ and 0.02 moles of $NaHCO_3.$
In this scenario, a buffer consisting of $NaHCO_3$ and $Na_2CO_3$ is formed.

Therefore, $pH=pK_{a2}+log\frac{[Salt]}{[Acid]}$
$=10.32+log\frac{\frac{0.01}{0.1}}{\frac{0.02}{0.1}}$

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

$=10.32-log2$
$=10.2$
$∴ pH = 10.02$