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 theequivalence point is reached when \(40 \mathrm{ml}\) of the acid have been added. The \(\mathrm{pH}\) at the equivalence point is (Given : \(\mathrm{K}_{\mathrm{a}}\) for \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\)is \(\left.4 / 3 \times 10^{-11}\right)\).

Step-by-Step Solution

Answer

The pH at the equivalence point is 6.

Step 1: Identify the Reaction

The titration involves methylamine ( \(\text{CH}_3\text{NH}_2\) ) which is a weak base being titrated with a strong acid, \(\text{HCl}\). The reaction at equivalence point is: \(\text{CH}_3\text{NH}_2 + \text{HCl} \rightarrow \text{CH}_3\text{NH}_3^+ + \text{Cl}^-\). At the equivalence point, all the methylamine has reacted to form \(\text{CH}_3\text{NH}_3^+\).

Step 2: Calculate Moles at Equivalence Point

At equivalence point, the moles of \(\text{HCl}\) added are equal to the initial moles of \(\text{CH}_3\text{NH}_2\) present. The moles of \(\text{HCl}\) is calculated by multiplying its concentration with volume in liters: \(0.15 \text{ M} \times 0.040 \text{ L} = 0.006 \text{ moles}\). Therefore, the moles of \(\text{CH}_3\text{NH}_3^+\) at equivalence point is 0.006 moles.

Step 3: Calculate Concentration of \(\text{CH}_3\text{NH}_3^+\)

The concentration of \(\text{CH}_3\text{NH}_3^+\) at equivalence point is given by dividing the moles by total volume of the solution. The total volume is the sum of volume of both solutions: \(40 \text{ ml} + 40 \text{ ml} = 80 \text{ ml} = 0.080 \text{ L}\). Thus, concentration is \(\frac{0.006}{0.080} = 0.075 \text{ M}\).

Step 4: Use \(\mathrm{K}_a\) to Find \([\text{H}^+]\) and \(\text{pH}\)

The acidic ion \(\text{CH}_3\text{NH}_3^+\) partially dissociates in solution: \(\text{CH}_3\text{NH}_3^+ \leftrightarrow \text{CH}_3\text{NH}_2 + \text{H}^+\). Ka is given as \(4/3 \times 10^{-11}\). For weak acids, \([\text{H}^+] = \sqrt{K_a \times c}\), where \(c\) is the concentration of the acid. Thus, \([\text{H}^+] = \sqrt{\frac{4}{3} \times 10^{-11} \times 0.075}\).

Step 5: Calculate \([\text{H}^+]\) and \(\text{pH}\)

\([\text{H}^+] = \sqrt{\frac{4}{3} \times 10^{-11} \times 0.075} = \sqrt{10^{-11} \times 0.1} \approx \sqrt{10^{-12}} = 10^{-6}\). Thus, \(\text{pH} = -\log(10^{-6}) = 6\).

Key Concepts

Equivalence Point

In titration, the equivalence point is a crucial moment in the process where the amount of titrant added equals the amount of substance initially present. This means that all original molecules have reacted completely.

In our exercise, methylamine ( \(\text{CH}_3\text{NH}_2\) ), a weak base, reacts with hydrochloric acid (\(\text{HCl}\)), a strong acid. The equivalence point occurs when \(40 \text{ ml}\) of \(\text{HCl}\) has been added to the \(40 \text{ ml}\) solution of methylamine. When this happens, all the methylamine converts to \(\text{CH}_3\text{NH}_3^+\), resulting in a solution that no longer has the original weak base but its conjugate acid.

The significance of the equivalence point is that it allows us to measure either the concentration of an unknown solution or to determine the \(\text{pH}\) of the solution after complete neutralization. It marks the transition phase in the chemical reaction used during titration—switching from a reactant state to a product state.

pH Calculations

Calculating \(\text{pH}\) at the equivalence point for a weak base and strong acid titration involves several steps. This process incorporates understanding the acid dissociation constant, \(\text{K}_a\), to determine the concentration of hydronium ions, \([\text{H}^+]\).

Once the equivalence point is reached, the solution consists of water and the conjugate acid formed from the original weak base. This conjugate acid slightly dissociates in water, producing \(\text{H}^+\) ions.

• Use the dissociation reaction, \(\text{CH}_3\text{NH}_3^+ \leftrightarrow \text{CH}_3\text{NH}_2 + \text{H}^+\), where \(\text{K}_a\) is given as \(\frac{4}{3} \times 10^{-11}\).
• Plug the given \(\text{K}_a\) value and the concentration of \(\text{CH}_3\text{NH}_3^+\) into the hydronium ion concentration formula: \([\text{H}^+] = \sqrt{\text{K}_a \times c}\).
• Through this calculation, find \([\text{H}^+]\) to then calculate \(\text{pH} = -\log([\text{H}^+])\).

These calculations bridge the understanding between chemical reactions and their numerical presentation on the \(\text{pH}\) scale.

Weak Base and Strong Acid Reaction

Reactions between weak bases and strong acids form the bedrock for understanding fundamental chemical equilibria in titrations. The exercise presented involves a classic example where methylamine, a weak base, reacts with a strong acid, \(\text{HCl}\).

When mixed, these two substances undergo a chemical reaction forming \(\text{CH}_3\text{NH}_3^+\) along with \(\text{Cl}^-\).

• The reaction is represented by: \(\text{CH}_3\text{NH}_2 + \text{HCl} \rightarrow \text{CH}_3\text{NH}_3^+ + \text{Cl}^-\). This signifies complete conversion of the weak base into its conjugate acid.
• This new product, \(\text{CH}_3\text{NH}_3^+\), slightly ionizes in water, which impacts the solution's overall \(\text{pH}\) level. The degree of ionization is dependent on the acid dissociation constant \(\text{K}_a\).

Understanding this reaction process helps in determining the composition and properties of the resulting solution at the equivalence point. It highlights how acid-base reactions modify the state of the original compounds, and the new forms influence the solution's acidity.