Problem 217 - A Complete Resource Book in Chemistry for JEE Main

Question

\(50 \mathrm{~mL}\) of \(0.2 \mathrm{M}\) aqueous \(\mathrm{CH}_{3} \mathrm{COOH}\) is mixed with \(50 \mathrm{~mL}\) of \(0.2 \mathrm{M}\) aqueous \(\mathrm{KOH}\) solution. The \(\mathrm{pH}\) of resulting solution is \(\left(\mathrm{pK}_{2}\right.\) of acetic acid is \(4.7\) ) (a) \(7.0\) (b) \(9.35\) (c) \(8.85\) (d) \(6.05\)

Step-by-Step Solution

Answer

(b) 9.35

Step 1: Determine moles of acetic acid and KOH

Calculate the moles of acetic acid (\(\text{CH}_3\text{COOH}\)) and KOH separately. Use the formula \(\text{moles} = \text{molarity} \times \text{volume (L)}\). For both solutions, since the molarity is \(0.2\ \text{M}\) and the volume is \(50\ \text{mL} = 0.050\ \text{L}\), the moles are \(0.2 \times 0.050 = 0.01\) moles each.

Step 2: Analyze the reaction between acetic acid and KOH

The equation for the reaction is \(\text{CH}_3\text{COOH} + \text{KOH} \rightarrow \text{CH}_3\text{COOK} + \text{H}_2\text{O}\). Since the moles of both \(\text{CH}_3\text{COOH}\) and \(\text{KOH}\) are equal, they completely neutralize each other, forming \(\text{CH}_3\text{COOK}\), a salt.

Step 3: Assess the nature of the salt

\(\text{CH}_3\text{COOK}\) is a salt derived from a weak acid (acetic acid) and a strong base (KOH). In such cases, the solution will be basic, and the pH will be greater than 7.

Step 4: Calculate pH using the hydrolysis of the salt

The pH of the solution can be calculated using the formula for hydrolysis of a salt from a weak acid and strong base: \[\text{pH} = \frac{1}{2} (\text{pK}_a + 14)\] where \(\text{pK}_a = 4.7\). Thus, \[\text{pH} = \frac{1}{2} (4.7 + 14) = \frac{18.7}{2} = 9.35\].

Key Concepts

Acid-Base Neutralization

Acid-base neutralization describes the reaction between an acid and a base, resulting in the formation of a salt and water. When acetic acid (\(\text{CH}_3\text{COOH}\)) is mixed with potassium hydroxide (\(\text{KOH}\)), an interesting interaction occurs.
At the molecular level, the hydrogen ions (\(\text{H}^+\)) from the acetic acid combined with the hydroxide ions (\(\text{OH}^-\)) from KOH to create water (\(\text{H}_2 ext{O}\)).
The remaining ions, acetate (\(\text{CH}_3\text{COO}^-\)) from the acetic acid and potassium (\(\text{K}^+\)) from KOH, combine to form the salt, potassium acetate (\(\text{CH}_3\text{COOK}\)). This is a classic example of neutralization where a weak acid is neutralized by a strong base.

• The reaction is: \( \text{CH}_3\text{COOH} + \text{KOH} \rightarrow \text{CH}_3\text{COOK} + \text{H}_2\text{O} \)
• Complete neutralization occurs when moles of acid equal moles of base.

Salt Hydrolysis

When a salt forms from the neutralization of a weak acid and a strong base, an interesting process called salt hydrolysis can occur. Salt hydrolysis involves the interaction of ions with water to form a solution that isn't necessarily neutral. In this case, potassium acetate (\(\text{CH}_3\text{COOK}\)) is the resulting salt.
Potassium (\(\text{K}^+\)) does not interact much with water but the acetate ions (\(\text{CH}_3\text{COO}^-\)) can react with water forming acetic acid and hydroxide ions (\(\text{OH}^-\)).

• As a result, the solution becomes slightly basic.
• This process changes the pH of the solution.

Thus, salt hydrolysis, in this example, shifts the pH above 7, demonstrating the basic nature of the solution.

Buffer Solution

A buffer solution is a special type of solution that maintains its pH relatively constant when small amounts of acid or base are added. Although the solution in this exercise is not a buffer, understanding buffers can help clarify why the pH changes after complete neutralization.
Buffs are usually made from a weak acid and its conjugate base or vice versa. In our exercise, the acetic acid and acetate ions do have buffering potential.

• Even after neutralizing, the remaining acetate ions can buffer changes in pH.

However, in this case, since there are no excess acetic acid ions remaining, the solution acts more like a basic salt solution. This lack of excess acid is why the solution isn't functioning as a true buffer here.

pKa

The \( \text{pKa} \) value is a crucial concept in chemistry, particularly when dealing with acids and bases. It is the negative base-10 logarithm of the acid dissociation constant \( \text{Ka} \), which gives a measure of the strength of an acid.

• Lower \( \text{pKa} \) indicates a stronger acid.
• In the context of the problem, acetic acid's \( \text{pKa} \) is 4.7, showing it is a weak acid.

These concepts help us estimate the degree of dissociation of the acid in water and predict the behavior of the resulting solutions.
This is crucial for calculations like determining pH through hydrolysis formulas used in this exercise. The half-neutralization formula, \( \text{pH} = \frac{1}{2} (\text{pKa} + 14) \), helps to further calculate the pH when linked with a strong base.