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

Question

The pKa pf HCN is \(9.30 .\) The \(\mathrm{pH}\) of a solution prepared by mixing \(2.5\) moles of \(\mathrm{KCN}\) and \(2.5\) moles of 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 of the solution is 9.30.

Step 1: Understand the Concept

This exercise involves calculating the pH of a solution containing a weak acid (HCN) and its conjugate base (CN-). This is a buffer solution, and the pH can be calculated using the Henderson-Hasselbalch equation, which is applicable for buffer solutions.

Step 2: Write down the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is given by: \[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \] In this case, the pKa of HCN is 9.30, and both the acid (HCN) and the base (KCN) are equal in moles.

Step 3: Determine the Concentrations of Acid and Base

Since we have 2.5 moles of HCN and 2.5 moles of KCN (CN-), and the total volume of the solution is 0.5 L (since 500 mL = 0.5 L), the concentration of both HCN and CN- is:\[ \text{Concentration} = \frac{\text{moles}}{\text{volume in liters}} = \frac{2.5}{0.5} = 5 \text{ M} \] Thus, the [base] = 5 M and [acid] = 5 M.

Step 4: Apply the Henderson-Hasselbalch Equation

Substitute the values into the Henderson-Hasselbalch equation:\[ \text{pH} = 9.30 + \log \left( \frac{5}{5} \right) \] Since \( \log(1) = 0 \), the equation simplifies to:\[ \text{pH} = 9.30 + 0 = 9.30 \]

Key Concepts

pH calculation

When it comes to understanding the acidity or basicity of a solution, pH is a crucial term.
In the context of buffer solutions like the one involving HCN and KCN, pH calculation can become quite straightforward with the right formula.

• The buffer consists of a weak acid and its conjugate base. For the given problem, the weak acid is HCN, and the conjugate base is CN⁻, sourced from KCN.
• To determine the pH, you need to know the concentrations of the acid and the base in the solution and the pKa value of the weak acid.

For this exercise, the presence of equal concentrations of the weak acid and its conjugate base simplifies the pH calculation because the ratio of their concentrations is 1, leading to a log value of 0.

Henderson-Hasselbalch equation

Understanding the Henderson-Hasselbalch equation is key to calculating the pH of buffer solutions. This equation is a useful tool in chemistry for estimating the pH of a buffer solution, which consists of a weak acid and its conjugate base.The Henderson-Hasselbalch equation is expressed as:\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \]Where:

• \(\text{pH}\) is the measure of hydrogen ion concentration in the solution.
• \(\text{pKa}\) is the acid dissociation constant, representing the strength of the weak acid.
• \([\text{base}]\) and \([\text{acid}]\) are the molar concentrations of the conjugate base and the acid, respectively.

In this exercise, since the number of moles of HCN and KCN are equal, their concentrations are the same, leading the ratio \(\frac{[\text{base}]}{[\text{acid}]}\) to be 1.Because the logarithm of 1 is zero, the equation simplifies beautifully in cases like this to show that the \(\text{pH}\) of the solution is equal to the \(\text{pKa}\).

conjugate acid-base pair

A strong grasp of the concept of conjugate acid-base pairs can further enhance your understanding of buffer solutions.

In simple terms, a conjugate acid-base pair consists of two species differing by exactly one proton (H⁺).

When an acid donates a proton, the remaining part forms its conjugate base. Conversely, when a base accepts a proton, it forms its conjugate acid.

• For the given problem, HCN acts as the weak acid, and its loss of a proton leaves behind the conjugate base, CN⁻.
• The presence of both these components in similar proportions in solution forms a buffer, allowing the solution to resist drastic changes in pH when small amounts of acid or base are added.

This balancing act is central to many natural and technological processes, helping to stabilize environments that are sensitive to pH fluctuations.