JEE Chemistry 2023 Question with Solution

Given: $20 \, \text{mL}$ of $0.1 \, \text{M}$ NaOH is added to $50 \, \text{mL}$ of $0.1 \, \text{M}$ acetic acid.

Find: The pH of the resulting solution.

Reaction:

$$\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$$

Moles initially:

$$\text{CH}_3\text{COOH} = 50 \, \text{mL} \times 0.1 \, \text{M} = 5 \, \text{mmol}$$ $$\text{NaOH} = 20 \, \text{mL} \times 0.1 \, \text{M} = 2 \, \text{mmol}$$

After reaction:

$$\text{CH}_3\text{COOH remaining} = 5 - 2 = 3 \, \text{mmol}$$ $$\text{CH}_3\text{COONa formed} = 2 \, \text{mmol}$$

So the final solution is a buffer containing acetic acid and sodium acetate.

Using Henderson-Hasselbalch equation:

$$\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{Salt}]}{[\text{Acid}] }\right)$$ $$\text{pH} = 4.76 + \log_{10}\left(\frac{2}{3}\right)$$

Now,

$$\log_{10}\left(\frac{2}{3}\right) = \log_{10}(2) - \log_{10}(3)$$

Using $\log 2 = 0.30$ and $\log 3 = 0.48$,

$$\log_{10}\left(\frac{2}{3}\right) = 0.30 - 0.48 = -0.18$$

Therefore,

$$\text{pH} = 4.76 - 0.18 = 4.58$$

the solution writes $458 \times 10^{-2}$, which is numerically equal to $4.58$. Therefore, the pH is $4.58$.

Given: Weak acid acetic acid reacts partially with strong base NaOH.

Find: Why the resulting mixture should be treated as a buffer.

NaOH neutralizes an equal amount of acetic acid first. Since NaOH is the limiting reagent, some acetic acid remains unreacted and sodium acetate is produced.

Thus both weak acid and its conjugate base are present together:

• remaining acid = $3 \, \text{mmol}$
• salt formed = $2 \, \text{mmol}$

That is exactly the condition for using the Henderson-Hasselbalch equation. Because both species are in the same final volume, their concentration ratio is the same as their mole ratio:

$$\frac{[\text{Salt}]}{[\text{Acid}]} = \frac{2}{3}$$

Hence,

$$\text{pH} = 4.76 + \log_{10}\left(\frac{2}{3}\right) = 4.58$$

Therefore, the numerical value of the pH is $4.58$.