Problem 230 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{K}\), for \(\mathrm{HCN}\) is \(5 \times 10^{-10}\) at \(25^{\circ} \mathrm{C}\). For maintaining a constant \(\mathrm{pH}\) of 9 , the vol of \(5 \mathrm{M}\) KCN solution required to be added to \(10 \mathrm{ml}\) of \(2 \mathrm{M}\) HCN solution is

Step-by-Step Solution

Answer

Add 2.5 mL of 5 M KCN solution.

Step 1: Establish the Relationship for Buffer Solution

To maintain a constant pH, we use the Henderson-Hasselbalch equation for a buffer system. The equation is given by:\[pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)\]where \(pK_a = -\log K_a\). For HCN, \(K_a = 5 \times 10^{-10}\), which allows us to find \(pK_a\).

Step 2: Calculate pKa of HCN

Calculate \(pK_a\) using \(K_a = 5 \times 10^{-10}\):\[pK_a = -\log(5 \times 10^{-10}) = 9.3\]Now, plug \(pK_a\) into the Henderson-Hasselbalch equation together with the desired pH.

Step 3: Apply Henderson-Hasselbalch Equation

Set the desired \(pH = 9\) in the Henderson-Hasselbalch equation:\[9 = 9.3 + \log\left(\frac{[CN^-]}{[HCN]}\right)\]Solving for the ratio \(\frac{[CN^-]}{[HCN]}\), we have:\[\log\left(\frac{[CN^-]}{[HCN]}\right) = 9 - 9.3 = -0.3\]Therefore,\[\frac{[CN^-]}{[HCN]} = 10^{-0.3}\]

Step 4: Calculate the Required Concentration Ratio

Calculate the ratio:\[\frac{[CN^-]}{[HCN]} = \frac{1}{2} \approx 0.5\]Given the initial concentration of HCN is 2 M in a 10 mL solution, we can determine the required \([CN^-]\).

Step 5: Determine CN- Ions Needed

Since the ratio \(\frac{[CN^-]}{[HCN]} = 0.5\), calculate \([CN^-]\):\[[CN^-] = 0.5 \times 2 = 1 \text{ M}\]This concentration is required for the total volume of the buffer solution. The buffer needs to contain this concentration of \([CN^-]\).

Step 6: Calculate Volume of KCN Solution Required

To find the volume of the 5 M KCN solution needed to provide the 1 M concentration in the final mixture, use the dilution formula:\(V_1C_1 = V_2C_2\)where \(V_1 = ?\) (volume of KCN), \(C_1 = 5 \text{ M}\), \(V_2 = 10 \text{ mL + } V_1\) (total volume), and \(C_2 = 1 \text{ M}\).Rearrange the formula:\(5 \cdot V_1 = 1 \cdot (10 + V_1)\)Solving for \(V_1\) gives us:\[5V_1 = 10 + V_1 \implies 4V_1 = 10 \implies V_1 = 2.5 \text{ mL}\]

Step 7: Conclusion

To maintain a pH of 9, a volume of 2.5 mL of 5 M KCN solution must be added to the 10 mL of 2 M HCN solution.

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a simplified equation used to find the pH of a buffer solution. It connects the pH, the pKa (the negative log of the acid dissociation constant), and the concentration ratio of the buffer’s acid and its conjugate base. The equation is expressed as:
\[pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)\]
In this equation:

• \(pH\) is the measure of acidity or basicity of the solution.
• \(pK_a\) represents the strength of the weak acid component.
• \([A^-]\) is the concentration of the conjugate base.
• \([HA]\) represents the concentration of the weak acid.

For example, to maintain the pH at 9 for the buffer containing HCN (a weak acid) and its conjugate base CN⁻, we start by calculating \(pK_a\). With \(K_a = 5 \times 10^{-10}\), we find \(pK_a\) by taking the negative log:
\[pK_a = -\log(5 \times 10^{-10}) = 9.3\]
Plugging this into our equation, along with the desired pH, helps us find the necessary ratio of the concentrations of CN⁻ and HCN.

pH calculation

Calculating the pH of a buffer solution requires understanding the role of both the weak acid and its conjugate base. Using the Henderson-Hasselbalch equation gives a concise method to determine the solution's pH based on the ratio of these components.
When you have the desired pH (in this example, a pH of 9) and the calculated \(pK_a\) of 9.3, you can find the ratio using:
\[9 = 9.3 + \log\left(\frac{[CN^-]}{[HCN]}\right)\]
Rearranging gives:\[\log\left(\frac{[CN^-]}{[HCN]}\right) = 9 - 9.3 = -0.3\]
Exponentiating both sides with base 10 removes the logarithm:\[\frac{[CN^-]}{[HCN]} = 10^{-0.3} \approx 0.5\]
This means that the concentration of CN⁻ should be half that of HCN to achieve a pH of 9 in this buffer system.

Dilution formula

The dilution formula is instrumental in calculating how much of a concentrated solution is required to achieve a certain concentration in a larger volume. The formula is:
\[V_1C_1 = V_2C_2\]
Here:

• \(V_1\) is the volume of the concentrated solution you need.
• \(C_1\) is the concentration of the original solution.
• \(V_2\) is the total volume of the final solution.
• \(C_2\) is the desired concentration in the final solution.

In our example, to find \(V_1\), the volume of the 5 M KCN solution needed, when the target concentration is 1 M in a solution including 10 mL of 2 M HCN, the equation is set up as:
\[5 \cdot V_1 = 1 \cdot (10 + V_1)\]
Solving gives:\[5V_1 = 10 + V_1 \ 4V_1 = 10 \ V_1 = 2.5 ext{ mL}\]
Thus, to maintain a consistent pH, adding 2.5 mL of the 5 M KCN to the buffer achieves the desired concentration.