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 $637$ and $1032$, respectively $; \log 2=030$ ]

Answer 1

We have a solution containing:

• \(𝐻_2𝐶𝑂_3\)
• \(𝑁𝑎𝐻𝐶𝑂_3\)
• \(𝑁𝑎_2𝐶𝑂_3\)
• \(𝑁𝑎𝑂𝐻\) Each with a concentration of 0.01 mol.

When \(𝐻_2𝐶𝑂_3\) reacts with \(𝑁𝑎𝑂𝐻\), it forms \(𝑁𝑎𝐻𝐶𝑂_3\)

Then, \(𝑁𝑎𝐻𝐶𝑂_3\)​ reacts with \(𝑁𝑎_2𝐶𝑂_3\)​, resulting in \(𝑁𝑎_2𝐶𝑂_3\)​ and \(𝑁𝑎𝐻𝐶𝑂_3\)​ both having a concentration of 0.02 mol.

This solution acts as a buffer solution of \(𝑁𝑎_2𝐶𝑂_3\)​ and \(𝑁𝑎𝐻𝐶𝑂_3​.\)

To find the pH of this solution, we use the Henderson-Hasselbalch equation: \(p_H = p_{K_a2} + \log\left(\frac{[Na_2CO_3]}{[NaHCO_3]}\right)\)

Given: \(𝑝𝐾_{𝑎2}=10.32\)
 \(\frac{[Na_2CO_3]}{[NaHCO_3]} = \frac{0.01}{0.02} = 0.5\)

Plugging in the values: \(p_H = 10.32 + \log(0.5) = 10.32 - 0.3 = 10.02\)

So, the pH of the resulting solution is \(10.02.\)