Problem 227 - A Complete Resource Book in Chemistry for JEE Main

Question

A \(40 \mathrm{ml}\) sample of an aqueous solution of the methylamine at \(25^{\circ} \mathrm{C}\) is titrated with \(0.15 \mathrm{M} \mathrm{HCl}\) and the equivalence point is reached when \(40 \mathrm{ml}\) of the acid have been added. The \(\mathrm{pH}\) at the equivalence point is (Given: \(\mathrm{K}\), for \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\)is \(\left.4 / 3 \times 10^{-11}\right)\).

Step-by-Step Solution

Answer

The \(\mathrm{pH}\) at the equivalence point is approximately 5.55.

Step 1: Understanding the Equivalence Point

At the equivalence point, the number of moles of the titrant (HCl in this case) is equal to the number of moles of the analyte (methylamine, \( \mathrm{CH}_{3}\mathrm{NH}_{2} \)). This means the solution at this point contains methylammonium chloride (\( \mathrm{CH}_{3}\mathrm{NH}_{3}\mathrm{Cl} \)). The reaction is: \( \mathrm{CH}_{3}\mathrm{NH}_{2} + \mathrm{HCl} \rightarrow \mathrm{CH}_{3}\mathrm{NH}_{3}^+ + \mathrm{Cl}^- \).

Step 2: Calculating Moles of HCl and Methylamine

The volume of HCl added is \(40 \mathrm{ml} = 0.04 \mathrm{L}\). Thus, moles of \( \mathrm{HCl} \) = \( 0.15 \mathrm{M} \times 0.04 \mathrm{L} = 0.006 \mathrm{mol} \) (0.15 moles/L \( \times \) 0.04 L). Since we reach the equivalence point, we also have \(0.006 \mathrm{mol}\) of methylamine.

Step 3: Concentration of \(\mathrm{CH}_{3}\mathrm{NH}_{3}^+\)

The total volume of the solution at the equivalence point is \( 40 \mathrm{ml} + 40 \mathrm{ml} = 80 \mathrm{ml} = 0.08 \mathrm{L} \). The concentration of \( \mathrm{CH}_{3}\mathrm{NH}_{3}^+ \) is therefore \( \frac{0.006 \mathrm{mol}}{0.08 \mathrm{L}} = 0.075 \mathrm{M} \).

Step 4: Calculating the \(\mathrm{pH}\) at Equivalence Point using \(\mathrm{K}_a\)

\(\mathrm{pH}\) at the equivalence point is governed by the hydrolysis of \( \mathrm{CH}_{3}\mathrm{NH}_{3}^+ \). The equilibrium expression is \( \mathrm{CH}_{3}\mathrm{NH}_{3}^+ + \mathrm{H}_2\mathrm{O} \leftrightarrow \mathrm{CH}_{3}\mathrm{NH}_{2} + \mathrm{H}_3\mathrm{O}^+ \) with \(K_a = \frac{K_w}{K_b(\mathrm{CH}_{3}\mathrm{NH}_2)} \). Given \(K_a = 4/3 \times 10^{-11}\).

Step 5: Using Ice Table for Hydrolysis

Set up an ICE table for \(\mathrm{CH}_{3}\mathrm{NH}_{3}^+\) hydrolysis. Initial concentration = 0.075. Change = \(-x\) for \(\mathrm{CH}_{3}\mathrm{NH}_{3}^+\), and \(+x\) for \(\mathrm{H}_3\mathrm{O}^+\). Solve \(K_a = \frac{x^2}{0.075-x}\), assuming \(x \ll 0.075\).

Step 6: Solve for \(x\) which is \([\mathrm{H}_3\mathrm{O}^+]\)

Since \(x \) is small, \(x^2 \approx 4/3 \times 10^{-11} \times 0.075 \). We can solve \(x = \sqrt{4/3 \times 10^{-11} \times 0.075} \approx 2.82 \times 10^{-6}\). At \(25^\circ C\), \([\mathrm{H}_3\mathrm{O}^+] = x\).

Step 7: Calculate \(\mathrm{pH}\)

\(\mathrm{pH} = -\log([\mathrm{H}_3\mathrm{O}^+]) = -\log(2.82 \times 10^{-6}) \approx 5.55\).

Key Concepts

Equivalence Point

In an acid-base titration, the equivalence point is a critical concept. It's the stage where the number of moles of the acid equals the number of moles of the base. For our exercise, this means the moles of hydrochloric acid (HCl) added are equal to the moles of the methylamine in the original solution. This balance signifies that all the original base has reacted with the acid.

At the equivalence point in this titration, the solution contains primarily the salt formed from the reaction. Here, that's methylammonium chloride (\( \mathrm{CH}_{3}\mathrm{NH}_{3}^+ \)). This is because the reaction between methylamine (\( \mathrm{CH}_{3}\mathrm{NH}_{2} \)) and the added HCl produces this compound. The major components of the solution are now the conjugate acid of the original base and the chloride ion. Understanding this conversion is vital because it sets the stage for calculating the resulting pH of the solution at this point.

• The equivalence point marks the completion of neutralization.
• Identifying when this point occurs helps us determine the properties of the resulting solution.

pH Calculation

Calculating the pH at the equivalence point requires an understanding of both the nature of the solution and its constituents. After the reaction reaches the equivalence point, the resulting solution's pH depends largely on the characteristics of the methylammonium ion (\( \mathrm{CH}_{3}\mathrm{NH}_{3}^+ \)). This ion is acidic due to its inclination to donate a proton in water, forming hydronium ions (\( \mathrm{H}_3\mathrm{O}^+ \)).

To find the pH, set up an equation incorporating the acid dissociation constant (\( K_a \)) of the methylammonium ion. The formula \( K_a = \frac{x^2}{[\mathrm{CH}_{3}\mathrm{NH}_{3}^+] - x} \) can be used, where x is the hydronium ion concentration (\( [\mathrm{H}_3\mathrm{O}^+] \)). Given the problem's parameters, simplify to assume \( x \ll [\mathrm{CH}_{3}\mathrm{NH}_{3}^+] \) and solve for x. This approximation helps make calculations more manageable.

• Use the formula \( x^2 = K_a \cdot [\mathrm{CH}_{3}\mathrm{NH}_{3}^+] \).
• Popping x into the pH formula \( \mathrm{pH} = -\log([\mathrm{H}_3\mathrm{O}^+]) \) gives us the final pH.

Methylamine Hydrolysis

Understanding methylamine hydrolysis is crucial in acid-base chemistry, especially when predicting the behavior of its conjugate acid. Methylammonium chloride, the product of the titration reaction, undergoes hydrolysis in water. The process involves the dissociation of the methylammonium ion (\( \mathrm{CH}_{3}\mathrm{NH}_{3}^+ \)) into a neutral methylamine molecule (\( \mathrm{CH}_{3}\mathrm{NH}_{2} \)) and a hydronium ion (\( \mathrm{H}_3\mathrm{O}^+ \)), thus contributing to the acidity of the solution.

This hydrolysis reaction hinges on the concepts of equilibrium and acid dissociation constants. As we consider this reaction, remember the role of water in shifting the balance toward forming more hydronium ions. The presence of \( \mathrm{H}_3\mathrm{O}^+ \) is what decreases the pH, indicating that the solution becomes more acidic when hydrolysis occurs.

• Hydrolysis affects the pH and the acidity of the resulting solution.
• The direction of the reaction is determined by the relative strengths of the acids and bases involved.