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

Question

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

Step-by-Step Solution

Answer

2 mL of 5 M KCN is needed.

Step 1: Understanding the Components

The problem involves a weak acid, HCN, and its conjugate base, CN⁻, present in the KCN solution. The goal is to find the volume of 5 M KCN needed to maintain a pH of 9.

Step 2: Applying the Henderson-Hasselbalch Equation

Use the Henderson-Hasselbalch equation for the buffer solution:\[\text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]\right)\]Here, \(\text{A}^-\) is CN⁻ and \(\text{HA}\) is HCN. Calculate \(\text{pK}_a\) using \(\text{pK}_a = -\log(\text{K}_a)\). For \(\text{K}_a = 5 \times 10^{-10}\), \(\text{pK}_a = 9.3\). The target pH is 9.

Step 3: Solving the Buffer Equation

Set up the equation for pH 9:\[9 = 9.3 + \log\left(\frac{[\text{CN}^-]}{[\text{HCN}]\right)\]Rearrange to find the ratio:\[\log\left(\frac{[\text{CN}^-]}{[\text{HCN}]\right) = 9 - 9.3 = -0.3\]Thus, \[\frac{[\text{CN}^-]}{[\text{HCN}]} = 10^{-0.3} \approx 0.5\]

Step 4: Calculate Initial Moles of HCN

Calculate the initial moles of HCN in 10 mL of 2 M solution:\[\text{moles of HCN} = 2 \, \text{M} \times 0.01 \, \text{L} = 0.02 \, \text{moles}\]

Step 5: Determine Required Moles of CN⁻

According to the ratio already calculated,\[\frac{[\text{CN}^-]}{0.02} = 0.5 \Rightarrow [\text{CN}^-] = 0.01 \, \text{moles}\]

Step 6: Calculate Volume of KCN Solution

To find the required volume of 5 M KCN:\[0.01 \, \text{moles} = 5 \, \text{M} \times V \]Solve for V:\[V = \frac{0.01}{5} = 0.002 \, \text{L} = 2 \, \text{mL}\]

Step 7: Conclusion

The volume of 5 M KCN solution required to maintain a pH of 9 is 2 mL.

Key Concepts

Understanding Buffer Solutions

Buffer solutions are incredibly important in both chemistry and biology because they help maintain a stable pH environment. They resist changes in pH when small amounts of an acid or base are added. This stability is achieved through the presence of a weak acid and its conjugate base. The weak acid can donate protons to neutralize added bases, while the conjugate base can accept protons to neutralize added acids.
This characteristic makes buffer solutions very practical. For instance, blood in our bodies acts as a buffer to maintain a pH range around 7.4, which is vital for proper cellular function. In the exercise above, the buffer solution is created using hydrocyanic acid (HCN) as the weak acid and its conjugate base, cyanide ion (CN⁻), from potassium cyanide (KCN). The pH balance of this solution is crucial as it enables the precise control needed when conducting various chemical reactions or experiments.

Steps in pH Calculation Using the Henderson-Hasselbalch Equation

The pH calculation for a buffer solution is typically done using the Henderson-Hasselbalch equation. This critical formula is

• \[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Conjugate Base}]}{[\text{Weak Acid}]} \right) \]

In our scenario, we are using the components HCN and CN⁻. To find the pH:

• First, calculate \( \text{pK}_a \) using the given \( \text{K}_a \) of the weak acid. For HCN, \( \text{K}_a = 5 \times 10^{-10} \), so \( \text{pK}_a = -\log(5 \times 10^{-10}) = 9.3 \).
• Then, apply the formula by setting the target \( \text{pH} \) to 9 and solve for the ratio of conjugate base to weak acid, which dictated the needed amounts of KCN.

The calculation of this specific pH is critical for determining how much of the conjugate base (CN⁻) is required, which allows us to see how much KCN must be added to our buffer to maintain stability.

Interaction Between Weak Acid and Conjugate Base in Buffers

A weak acid and its conjugate base form the cornerstone of a buffer solution. Their interaction is what allows the solution to resist changes in pH. The weak acid partially dissociates in water, establishing an equilibrium between the acid and its conjugate base. When a small quantity of acid or base is added to the solution:

• The weak acid donates a proton (\( H^+ \)), thus neutralizing any added bases.
• Conversely, the conjugate base accepts (\( H^+ \)) protons, neutralizing any added acids.

In our exercise, the weak acid is hydrocyanic acid (HCN), and the conjugate base is cyanide ion (CN⁻). The Henderson-Hasselbalch equation reflects this balance by incorporating concentrations of both species. This balance is essential for maintaining the pH of 9. For a successful buffer system, it's crucial to have significant amounts of both the weak acid and its conjugate base, ensuring it can adequately counteract added acids and bases.