Problem 130 - A Complete Resource Book in Chemistry for JEE Main

Question

When \(0.1\) mole of \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (ionization constant \(\mathrm{K}_{\mathrm{b}}=5 \times 10^{-4}\) ) is mixed with \(0.08 \mathrm{~mol} \mathrm{HCl}\) and the volume is made up of 1 litre. Find the \(\left[\mathrm{H}^{+}\right]\)of resulting solution. (a) \(8 \times 10^{-2}\) (b) \(2 \times 10^{-11}\) (c) \(1.23 \times 10^{-4}\) (d) \(8 \times 10^{-11}\)

Step-by-Step Solution

Answer

The \([\text{H}^+]\) concentration is approximately \(8 \times 10^{-11}\), option (d).

Step 1: Understand the reaction

When methylamine (\(\text{CH}_3\text{NH}_2\)) is mixed with HCl, they react to form methylammonium chloride (\(\text{CH}_3\text{NH}_3^+\text{Cl}^-\)). Because \(\text{CH}_3\text{NH}_2\) is a base and HCl is a strong acid, HCl will react completely with \(\text{CH}_3\text{NH}_2\).

Step 2: Calculate the moles of remaining species

Initially, we have 0.1 mol \(\text{CH}_3\text{NH}_2\) and 0.08 mol HCl. Due to the reaction, 0.08 mol of \(\text{CH}_3\text{NH}_2\) will react completely with HCl to form 0.08 mol \(\text{CH}_3\text{NH}_3^+\). This leaves 0.02 mol of unreacted \(\text{CH}_3\text{NH}_2\).

Step 3: Set up the equilibrium expression

The equilibrium involves the reaction of the remaining \(\text{CH}_3\text{NH}_2\) with water:\[\text{CH}_3\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{NH}_3^+ + \text{OH}^-\] Given that the mixture is in a 1-liter solution, the concentrations are: \[ [\text{CH}_3\text{NH}_2] = 0.02 \text{ M}\]\[ [\text{CH}_3\text{NH}_3^+] = 0.08 \text{ M}\]

Step 4: Apply the concept of buffer solution

This is a buffer solution, with \(\text{CH}_3\text{NH}_2\) and \(\text{CH}_3\text{NH}_3^+\) acting as a base and its conjugate acid, respectively. Use the Henderson-Hasselbalch equation for a base buffer:\[\text{pOH} = \text{pK}_b + \log\left(\frac{[\text{conjugate acid}]}{[\text{base}]\right)\]Where \(\text{pK}_b = -\log(5 \times 10^{-4})\approx 3.3\). Insert the concentrations:\[\text{pOH} = 3.3 + \log\left(\frac{0.08}{0.02}\right)\]\[\text{pOH} = 3.3 + \log(4)\]\[\text{pOH} = 3.3 + 0.6 = 3.9\]

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

Convert \(\text{pOH}\) to \(\text{pH}\) using the equation:\[\text{pH} = 14 - \text{pOH} = 14 - 3.9 = 10.1\]Calculate the \([\text{H}^+]\) concentration:\[[\text{H}^+] = 10^{-\text{pH}} = 10^{-10.1} \approx 7.94 \times 10^{-11}\]

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a fundamental tool in chemistry for calculating the pH of a buffer solution. This equation provides a simple way to estimate the pH of a solution containing a weak acid and its conjugate base or a weak base and its conjugate acid. In our exercise, let's focus on the base form.
A buffer solution consists of both a weak acid or base and its corresponding salt, which work in tandem to resist changes in pH. The equation is tailored to these systems, and for a base buffer, it reads:

• \( \text{pOH} = \text{pK}_b + \log\left(\frac{[\text{Conjugate Acid}]}{[\text{Base}]}\right) \)

Here, \( \text{pK}_b \) is the negative logarithm of the base ionization constant \( K_b \). For the methylamine example, it’s critical in determining the balance between the remaining amounts of base \( (\text{CH}_3\text{NH}_2) \) and its conjugate acid \( (\text{CH}_3\text{NH}_3^+) \).
This equation allows us to predict the pH or pOH of a buffer by simply plugging in the concentrations of the buffer components. It simplifies the understanding of how the buffer resists changes in pH when small amounts of acid or base are added.

Ionization Constant Kb

The ionization constant for a base, \( K_b \), is a measure of its relative strength as a base. It indicates how well the base accepts protons from water to become its conjugate acid. In the case of methylamine (\( \text{CH}_3\text{NH}_2 \)), the \( K_b \) value tells us how readily it forms \( \text{CH}_3\text{NH}_3^+ \) and \( \text{OH}^- \) ions in water.
For our exercise, the \( K_b \) of methylamine is given as \( 5 \times 10^{-4} \). This relatively high value (for bases) means methylamine is a moderate base, capable of partially ionizing in water.
Understanding \( K_b \) helps us set up the equilibrium expression for methylamine in water. This constant is used to find the pOH using the Henderson-Hasselbalch equation, making it essential for accurately calculating the pH of the solution.

pH and pOH Calculations

Calculating pH and pOH is an essential part of understanding buffer solutions and acid-base equilibrium. Knowing how pH and pOH relate is crucial. The pH is a measure of the acidity of a solution, while pOH measures its basicity. In any aqueous solution at 25°C, the relationship between pH and pOH is given by:

• \( \text{pH} + \text{pOH} = 14 \)

In the exercise, after finding the pOH using the Henderson-Hasselbalch equation, we convert it to pH, helping us solve for the \([\text{H}^+]\) concentration.
To calculate \([\text{H}^+]\):

• \( [\text{H}^+] = 10^{-\text{pH}} \)

In this solution, the resultant pH of 10.1 indicates a basic solution, typical in buffer systems dealing with bases like methylamine. This also confirms the concentration of \([\text{H}^+]\) as found in the step-by-step solution.