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

Correct Answer: 10.02

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

\(=10.32+log\frac{1}{2}\)

\(=10.32-log2\)
\(=10.2\)
\(∴ pH = 10.02\)