Problem 190 - A Complete Resource Book in Chemistry for JEE Main

Question

When \(0.1\) mol of \(\mathrm{CH}_{3} \mathrm{NH}_{2}\left(\mathrm{~K}_{b}=5 \times 10^{-4}\right)\) is mixedwith \(0.08\) mol of \(\mathrm{HCl}\) and diluted to \(1 \mathrm{~L}\), the \(\mathrm{H}^{+}\)ion concentration in the solution is (a) \(8 \times 10^{-11} \mathrm{M}\) (b) \(6 \times 10^{-5} \mathrm{M}\) (c) \(1.6 \times 10^{-11} \mathrm{M}\) (d) \(8 \times 10^{-2} \mathrm{M}\)

Step-by-Step Solution

Answer

Option (a) with a concentration of \( 8 \times 10^{-11} \mathrm{M} \).

Step 1: Understanding the System

We have a solution that initially contains 0.1 mol of methylamine ( CH_3NH_2), which is a weak base, and 0.08 mol of hydrochloric acid (HCl), a strong acid. When mixed, neutralization between HCl and CH_3NH_2 occurs, leading to the formation of CH_3NH_3Cl and a slight excess of CH_3NH_2.

Step 2: Calculate Neutralization Reaction

The reaction between CH_3NH_2 and HCl is a simple acid-base reaction: \[ CH_3NH_2 + HCl \rightarrow CH_3NH_3^+ + Cl^- \]Since there is 0.08 mol of HCl and 0.1 mol of CH_3NH_2, all of HCl will react with an equal amount of CH_3NH_2, leaving 0.02 mol of CH_3NH_2 unreacted. Thus, the solution now contains 0.02 mol of CH_3NH_2 and 0.08 mol of CH_3NH_3^+.

Step 3: Identify Resulting Buffer System

The mixture behaves as a buffer solution made of CH_3NH_2 (weak base) and CH_3NH_3^+ (conjugate acid). The Henderson–Hasselbalch equation can be used to estimate the pH of the buffer system: \[ \mathrm{pH} = \mathrm{pK}_a + \log \left(\frac{[\text{base}]}{[\text{acid}]}\right) \]Where \( \mathrm{pK}_a = -\log(\frac{K_w}{K_b}) = -\log(\frac{1 \times 10^{-14}}{5 \times 10^{-4}}) \).

Step 4: Calculate pK_a and pH of the Buffer

First, calculate the \( \mathrm{pK}_a \): \[ \mathrm{pK}_a = -\log\left(\frac{1 \times 10^{-14}}{5 \times 10^{-4}} \right) \approx 10.7 \]Now, applying the Henderson–Hasselbalch equation:\[ \mathrm{pH} = 10.7 + \log\left(\frac{0.02}{0.08}\right) \approx 10.7 - 0.6 = 10.1 \]

Step 5: Calculate H^+ Concentration

To find the \( \mathrm{H}^+ \) ion concentration, convert the pH to \( \mathrm{H}^+ \) concentration using the formula:\[ \mathrm{H}^+ = 10^{-\mathrm{pH}} = 10^{-10.1} \approx 7.94 \times 10^{-11} \text{ M} \]This value is closest to option (a) \( 8 \times 10^{-11} \text{ M} \).

Step 6: Conclusion

Thus, the concentration of \( \mathrm{H}^+ \) ions in the solution is \( 7.94 \times 10^{-11} \text{ M} \), rounded to \( 8 \times 10^{-11} \text{ M} \), matching option (a).

Key Concepts

Henderson–Hasselbalch equation

The Henderson–Hasselbalch equation is a fundamental tool in chemistry for estimating the pH of buffer solutions. It relates the pH of a solution to the concentrations of an acid and its conjugate base. This equation is particularly useful when dealing with buffers, which are mixtures that resist changes in pH upon the addition of small amounts of acids or bases. When working with a buffer solution, it contains a conjugate pair like

• an acidic component (\[ \text{HA} \])
• and a basic component (\[ \text{A}^- \]).

For the Henderson–Hasselbalch equation, the formula is:\[pH = \mathrm{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\]The \( \mathrm{pK}_a \) is the negative logarithm of the acid dissociation constant, giving insight into the strength of the acid. When applying this equation to buffer solutions, you consider:

• Balance between the conjugate acid–base pair, this is essential.
• The ratio informs whether the solution is more acidic or basic.

In our original exercise, the equation helped calculate the pH of a buffer formed after the neutralization between methylamine and hydrochloric acid.

Acid-base reaction

Acid-base reactions are fundamental chemical processes where an acid reacts with a base to produce a salt and, often, water. These reactions are central to understanding buffer behavior, as they highlight how acids and bases interact in solutions.In the context of our exercise, methylamine (\( \text{CH}_3\text{NH}_2 \)) acts as a weak base, while hydrochloric acid (\( \text{HCl} \)) is a strong acid. Their reaction can be represented as:\[\text{CH}_3\text{NH}_2 + \text{HCl} \rightarrow \text{CH}_3\text{NH}_3^+ + \text{Cl}^-\]This equation shows:

• Molecules transferring protons from the acid (\( \text{HCl} \)) to the base (\( \text{CH}_3\text{NH}_2 \)), forming a new substance (\( \text{CH}_3\text{NH}_3^+ \)).
• The presence of chloride ions (\( \text{Cl}^- \)), which results from HCl dissociation.

Acid-base reactions are core to developing buffer solutions because they regulate pH by counteracting minor pH shifts upon the introduction of other acidic or basic substances.

pH calculation

Understanding how to calculate pH is crucial for analyzing any aqueous solution, particularly buffer solutions. pH is a measure of the hydrogen ion concentration (\( \text{H}^+ \)) in a solution and indicates how acidic or basic that solution is.The formula to determine the pH is:\[pH = -\log[\text{H}^+]\]A lower pH means a more acidic solution, while a higher pH indicates a more basic one. In the buffer system generated by the reaction in our example, the calculated pH guided us to the hydrogen ion concentration. Using the Henderson-Hasselbalch equation, we first determined the pH of the buffer. From there, the conversion to \( \text{H}^+ \) ion concentration is straightforward:

• 20which is calculated by reversing the logarithmic operation, using: \[\text{H}^+ = 10^{-\text{pH}}\]
• Resulting in \( \text{H}^+ \) concentration.

For our solution, the conversion yieldeda concentration of \[7.94 \times 10^{-11} \text{ M},\]closest to the multiple-choice option of \( \mathbf{8 \times 10^{-11} \text{ M}} \).Accurate pH calculations are vital as they directly impact the buffering capacity and the behavior of chemical reactions in a solution.