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 $10.02.$