Problem 162 - A Complete Resource Book in Chemistry for JEE Main

Question

The \(\mathrm{pH}\) of a solution containing \(0.1 \mathrm{~mol}\) of \(\mathrm{CH}_{3} \mathrm{COOH}\), \(0.2 \mathrm{~mol}\) of \(\mathrm{CH}_{3} \mathrm{COONa}\) and \(0.05 \mathrm{~mol}\) of \(\mathrm{NaOH}\) in \(1 \mathrm{~L}\). \(\left(\mathrm{pK}_{\mathrm{a}}\right.\) of \(\mathrm{CH}_{3} \mathrm{COOH}=4.74\) and \(\left.\log 5=0.7\right)\) (a) \(4.56\) (b) \(5.44\) (c) \(5.04\) (d) \(3.74\)

Step-by-Step Solution

Answer

The pH of the solution is 5.44, so the correct answer is (b).

Step 1: Identify the Solution Type

The solution is a buffer solution consisting of an acetic acid (\[\text{CH}_3\text{COOH}\]) and its conjugate base acetate (\[\text{CH}_3\text{COONa}\]) along with an added base (\[\text{NaOH}\]).

Step 2: Determine Components Affecting pH

NaOH will react with acetic acid (CH\(_3\)COOH) to form more acetate ions (CH\(_3\)COO\(^{-}\)). Calculate the amount of CH\(_3\)COOH that reacts:\[\text{Amount of } \text{CH}_3\text{COOH} \text{ reacting with NaOH} = 0.05 \text{ mol}\]This leaves:\[0.1 - 0.05 = 0.05 \text{ mol of } \text{CH}_3\text{COOH remaining}\]

Step 3: Calculate Moles of Acetate

After the reaction, the acetate ion concentration is increased by the amount of NaOH used.Initial moles of CH\(_3\)COONa:\[0.2\text{ mol}\]Added moles from reaction:\[0.05 \text{ mol}\]Total moles of CH\(_3\)COO\(^{-}\):\[0.2 + 0.05 = 0.25 \text{ mol}\]

Step 4: Apply Henderson-Hasselbalch Equation

Use the Henderson-Hasselbalch equation to find the pH of the buffer solution:\[\text{pH} = \text{pK}_a + \log\left( \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \right)\]Substitute the known values:\[\text{pH} = 4.74 + \log\left( \frac{0.25}{0.05} \right)\]

Step 5: Calculate Logarithmic Value

Simplify the log ratio:\[\frac{0.25}{0.05} = 5\]So,\[\text{pH} = 4.74 + \log(5) = 4.74 + 0.7\]

Step 6: Calculate Final pH

Now calculate the final pH:\[\text{pH} = 5.44\]

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a valuable tool in chemistry for estimating the pH of a buffer solution. A buffer solution contains a weak acid and its conjugate base or a weak base and its conjugate acid. The purpose of a buffer is to maintain a relatively constant pH level even when small amounts of an acid or a base are added.

The equation itself is:\[\text{pH} = \text{pK}_a + \log\left( \frac{[\text{A}^-]}{[\text{HA}]} \right)\]Where:

• \( \text{pH} \) is the measure of acidity or basicity in the solution.
• \( \text{pK}_a \) is the negative log of the acid dissociation constant \( K_a \), reflecting the strength of the weak acid.
• \([\text{A}^-]\) is the concentration of the conjugate base.
• \([\text{HA}]\) is the concentration of the acid.

The equation is useful because it's derived from the equilibrium expression for the dissociation of a weak acid. It simplifies the calculation of pH, especially in systems where a buffer resists drastic changes in pH. This equation assumes that the concentrations of the acid and its conjugate base do not change significantly when the buffer solution is in action.

Understanding how to use this equation is crucial for solving pH problems involving buffers, as demonstrated in the exercise above.

pH calculation

Calculating the pH of a solution involves determining the balance between hydrogen ions \( (\text{H}^+) \) and hydroxide ions \( (\text{OH}^-)\) in that solution. The pH scale runs from 0 to 14, where a low pH below 7 indicates acidity and a high pH above 7 indicates alkalinity.

In buffer solutions, the pH is calculated using the Henderson-Hasselbalch equation, as the exercise demonstrates. With the values provided:

• The weak acid is acetic acid (\(\text{CH}_3\text{COOH}\)), with a known \(\text{pK}_a\) of 4.74.
• The conjugate base is acetate (\(\text{CH}_3\text{COO}^-\)).

We use the Henderson-Hasselbalch equation to arrive at the pH:\[\text{pH} = 4.74 + \log\left( \frac{0.25}{0.05} \right)\]This becomes:\[\text{pH} = 4.74 + 0.7 = 5.44\]Thus, based on the log calculations and formula application, we find that the pH is 5.44. This reflects a slightly basic solution remaining stable due to the buffer created by the weak acid and its conjugate base.

It's important to remember that understanding these calculations helps in predicting how a buffer system would behave when acids or bases are added.

Acid-base reactions

Acid-base reactions are fundamental chemical reactions involving the transfer of hydrogen ions \((\text{H}^+)\) between substances. In the context of buffer solutions, these reactions help maintain a stable pH level in the solution.

For the exercise at hand, it's essential to grasp how the base, sodium hydroxide \((\text{NaOH})\), reacts with the weak acid, acetic acid \((\text{CH}_3\text{COOH})\).

• When \(\text{NaOH}\) is added to the buffer, it reacts with \(\text{CH}_3\text{COOH}\) to form extra acetate ions \((\text{CH}_3\text{COO}^-)\).
• The reaction adjusts the concentrations of the acid and its conjugate base, according to \(\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}\).

After the reaction:

• The amount of \(\text{CH}_3\text{COOH}\) is reduced by the moles of \(\text{NaOH}\) added, while \(\text{CH}_3\text{COO}^-\) correspondingly increases.

These changes are crucial elements in the calculation of pH using the Henderson-Hasselbalch equation.

Understanding acid-base reactions broadens insights into chemical equilibria and the behavior of buffer solutions under various conditions. This comprehension of reaction dynamics enhances prediction of pH shifts, demonstrating control over solution chemistry.