Problem 65 - A Complete Resource Book in Chemistry for JEE Main

Question

The pKa pf HCN is \(9.30\). The pH of a solution prepared by mixing \(2.5\) moles of \(\mathrm{KCN}\) and \(2.5\) moles of \(\mathrm{HCN}\) in water and making up the total volume to \(500 \mathrm{ml}\) is (a) \(9.30\) (b) \(8.30\) (c) \(7.30\) (d) \(10.30\)

Step-by-Step Solution

Answer

The pH is 9.30.

Step 1: Identify Given Information

We are given the pKa of HCN as 9.30, and we have 2.5 moles each of KCN and HCN in a 500 ml solution.

Step 2: Recognize the Buffer System

The solution consists of a weak acid (HCN) and its conjugate base (KCN, which provides CN extsuperscript{-}). Thus, it's a buffer solution.

Step 3: Use the Henderson-Hasselbalch Equation

For a buffer solution with equal concentrations of acid and conjugate base, the pH is given by the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \] Since the concentrations are equal (2.5 moles in the same volume), the ratio \( \frac{[\text{A}^-]}{[\text{HA}]} \) is 1.

Step 4: Calculate the pH

Substitute the values into the Henderson-Hasselbalch equation:\[ \text{pH} = 9.30 + \log(1) = 9.30 \] Since \( \log(1) = 0 \), the pH equals the pKa.

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is essential for understanding buffer solutions. It provides a method to calculate the pH of a solution based on the concentrations of an acid and its conjugate base. The equation is written as:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
where:

• \(\text{pH}\) is the measure of the acidity of the solution.
• \(\text{pKa}\) is the acid dissociation constant, providing a measure of the strength of the acid.
• \([\text{A}^-]\) is the concentration of the conjugate base.
• \([\text{HA}]\) is the concentration of the acid.

This equation is incredibly useful for buffer solutions, as it shows how the pH depends on the ratio of the concentrations of the base and the acid. For equal concentrations of the acid and its conjugate base, the \(\log(1) = 0\), and therefore, the pH is equal to the pKa.

pH Calculations

Calculating the pH of buffer solutions using the Henderson-Hasselbalch equation is straightforward when the concentrations of the acid and conjugate base are known. For instance, in our original exercise, we're dealing with a solution containing equal moles of HCN and KCN.
Given the total number of moles (2.5) and the equation:

• Since the concentrations are equal \(\frac{[\text{A}^-]}{[\text{HA}]} = 1\).
• The log of 1 is zero \(\log(1) = 0\).

Thus, the pH of the solution simplifies to:
\[ \text{pH} = \text{pKa} + 0 = \text{pKa}\]In this example, it turns out to be 9.30, clearly showing how the pH reflects the pKa when you're dealing with a perfect balance of components.

Acid-Base Equilibrium

Understanding acid-base equilibrium is key to mastering buffer solutions. When acids and bases interact, they establish an equilibrium that affects the pH.
In the case of a buffer system like HCN/CN⁻, equilibrium is defined by the balance between the weak acid (HCN) donating protons and the conjugate base (CN⁻) accepting protons.

• A buffer solution resists drastic changes in pH. This happens because the weak acid/base combination can neutralize added acids or bases.
• The pKa value is crucial here as it signifies the pH at which the acid is half dissociated, representing a point of stability in the system.
• Reactions in equilibrium,\(\text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^-\), show how the acid and base can shift back and forth, maintaining balance.

This intrinsic resilience against pH change is what makes buffer solutions essential in various chemical and biological processes, such as protein stabilization and enzymatic activities.