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 \(\left(\mathrm{pK}_{\mathrm{a}}\right.\) of acetic acid is \(4.7\) ) (a) \(7.0\) (b) \(9.35\) (c) \(8.85\) (d) \(6.05\)

Step-by-Step Solution

Answer

The pH of the solution is approximately 8.85 (Option c).

Step 1: Calculate Initial Moles of Reactants

First, calculate the initial moles of \( ext{CH}_3 ext{COOH}\) and \( ext{KOH}\):\[-50 \text{ mL} \times 0.2 \text{ M} = 0.01 \text{ moles of } \text{CH}_3\text{COOH}\] and \[50 \text{ mL} \times 0.2 \text{ M} = 0.01 \text{ moles of } \text{KOH}.\]

Step 2: Analyze Reaction Completion

Both reactants are present in equal molar amounts, meaning they completely neutralize each other: \( ext{CH}_3 ext{COOH} + ext{KOH} \rightarrow ext{CH}_3 ext{COOK} + ext{H}_2 ext{O}\). After the reaction, 0.01 moles of \( ext{CH}_3 ext{COOK}\) (potassium acetate) are formed.

Step 3: Determine Resulting Solution Composition

The solution will contain the acetate ion \( ext{CH}_3 ext{COO}^-\) from potassium acetate. Since potassium acetate is a salt formed from a weak acid and strong base, it can slightly hydrolyze to form \( ext{OH}^-\) ions.

Step 4: Calculate Hydrolysis of Acetate

The equilibrium hydrolysis reaction is \( ext{CH}_3 ext{COO}^- + ext{H}_2 ext{O} \rightleftharpoons ext{CH}_3 ext{COOH} + ext{OH}^-\). The \(K_b\) for acetate can be found using \(K_w = 10^{-14}\) and \(K_a = 10^{-4.7}\) given \(K_b = \frac{K_w}{K_a}\) which calculates to \(K_b \approx 1.79 \times 10^{-10}\).

Step 5: Use Kb to Calculate pH

Using the hydrolysis formula \( [OH^-]^2 = K_b \cdot c\), where \(c = 0.1\) M (total volume is 100 mL), \[ [OH^-] = \sqrt{1.79 \times 10^{-10} \times 0.1} \approx 4.23 \times 10^{-6} \text{ M}.\] Calculate \( ext{pOH} = -\log([OH^-]) \approx 5.37\), thus \(\text{pH} = 14 - \text{pOH} \approx 8.63\).

Step 6: Round to Closest Given Option

The calculated \(\text{pH} \) of 8.63 is closest to option (c) 8.85. Therefore, the answer is \(\boxed{8.85}\).

Key Concepts

Acid-Base Neutralization

When acids and bases come together in a chemical reaction, they neutralize each other. This means that the properties of the acid and the base are cancelled out.The result is typically a salt and water forming. For example, in our original exercise, acetic acid (\(\mathrm{CH}_3\mathrm{COOH}\)) and potassium hydroxide (\(\mathrm{KOH}\)) come together to form potassium acetate (\(\mathrm{CH}_3\mathrm{COOK}\)) and water (\(\mathrm{H}_2\mathrm{O}\)).

• The reaction equation is: \(\mathrm{CH}_3\mathrm{COOH} + \mathrm{KOH} \rightarrow \mathrm{CH}_3\mathrm{COOK} + \mathrm{H}_2\mathrm{O}\)
• Both reactants are present in equal amounts, leading to a complete neutralization.
• No excess of either acid or base remains, making the solution neutral in terms of reactants.

Remember, the goal of neutralization is to find balance, resulting in non-reactive compounds left in the solution.

Hydrolysis of Salts

Hydrolysis is a common process in which a salt reacts with water, altering the pH of the solution. After acid-base neutralization, particularly with weak acids and strong bases, hydrolysis can affect the resulting solution. In this case, potassium acetate, formed from acetic acid and \(\mathrm{KOH}\), can undergo hydrolysis.The equation for this hydrolysis process is:

• \(\mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{CH}_3\mathrm{COOH} + \mathrm{OH}^-\)
• The presence of \(\mathrm{OH}^-\) ions means the solution becomes slightly basic.

This is because the acetate ion, which is a weak conjugate base, partially reacts with water to form \(\mathrm{OH}^-\). Calculating this effect involves finding the equilibrium constant \(\mathrm{K_b}\) for such a process, which can be derived from the known \(\mathrm{K_w}\) and \(\mathrm{K_a}\) values.

Buffer Solution Chemistry

Though not directly a buffer solution, the mixture resulting from the complete neutralization of \(\mathrm{CH}_3\mathrm{COOH}\) and \(\mathrm{KOH}\) contains components that reflect buffer behavior. A true buffer solution consists of a weak acid and its conjugate base or a weak base and its conjugate acid, which helps maintain a stable pH.In this case, although \(\mathrm{CH}_3\mathrm{COOH}\) and \(\mathrm{CH}_3\mathrm{COO}^-\) coexist, the absence of the conjugate acid, in excess amounts, detracts from its buffer properties.

• A buffer resists changes in pH upon adding small amounts of acid or base.
• In our solution, while the presence of \(\mathrm{CH}_3\mathrm{COO}^-\) partially buffers against acid additions, it is less effective due to complete neutralization.

Nonetheless, understanding buffer chemistry assists in grasping how such mixtures shift their pH, considering the interactions of different species.

Equilibrium Constant Calculations

Calculating equilibrium constants is vital in understanding the degree to which reactions occur, especially in hydrolysis and neutralization. For the hydrolysis of salts like acetate, the equilibrium constant \(\mathrm{K_b}\) helps predict \(\mathrm{OH}^-\) ion concentration and thus, the pH.In our example, finding \(\mathrm{K_b}\) involves:

• Using the formula \(K_b = \frac{K_w}{K_a}\)
• This gives \( \mathrm{K_b} \approx 1.79 \times 10^{-10} \)

With this value, realizing the concentration of hydroxide ions becomes manageable.Given the equilibrium reaction \(\mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{CH}_3\mathrm{COOH} + \mathrm{OH}^-\), the \(\mathrm{OH}^-\) concentration can be determined by rearranging the formula: \( \[ \mathrm{[OH^-] = \sqrt{K_b \cdot c}} \] \)Here \(c\) represents the acetate concentration.Thus, knowing these constants simplifies monitoring how the solution behaves, whether it becomes acidic or basic.