Question

Which of the following mixture gives a buffer solution with \( \text{pH} = 9.25 \)?
Given: \( \mathrm{p}K_b(\mathrm{NH_4OH}) = 4.75 \)

• A. \(0.2\,\text{M } \mathrm{NH_4OH}\,(0.5\,\text{L}) + 0.1\,\text{M } \mathrm{HCl}\,(0.5\,\text{L})\)
• B. \(0.4\,\text{M } \mathrm{NH_4OH}\,(1\,\text{L}) + 0.1\,\text{M } \mathrm{HCl}\,(1\,\text{L})\)
• C. \(0.2\,\text{M } \mathrm{NH_4OH}\,(0.4\,\text{L}) + 0.1\,\text{M } \mathrm{HCl}\,(1\,\text{L})\)
• D. \(0.5\,\text{M } \mathrm{NH_4OH}\,(0.2\,\text{L}) + 0.2\,\text{M } \mathrm{HCl}\,(0.5\,\text{L})\)

Hint:

For a basic buffer, when \([\text{salt}] = [\text{base}]\), the pH equals \(14 - \text{p}K_b\).

Answer 1

The Correct Option is A

To determine which mixture gives a buffer solution with a pH of 9.25, we need to consider the properties of buffer solutions, which are commonly made from a weak base and its conjugate acid.

In this question, the weak base is ammonium hydroxide (\( \mathrm{NH_4OH} \)) and its conjugate acid is ammonium chloride (\( \text{NH}_4^+ \)), produced from the reaction with hydrochloric acid (\( \mathrm{HCl} \)). The pH of a buffer solution is determined by the Henderson-Hasselbalch equation:

\[ \text{pH} = \text{p}K_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]

Since we have been given \( \text{p}K_b(\mathrm{NH_4OH}) = 4.75 \), we can find \( \text{p}K_a \) using the relation between \( \text{p}K_w \), \( \text{p}K_a \) and \( \text{p}K_b \):

\[ \text{p}K_w = \text{p}K_a + \text{p}K_b = 14 \]

Therefore,

\[ \text{p}K_a = 14 - 4.75 = 9.25 \]

The desired pH of the buffer solution is equal to the calculated \( \text{p}K_a \), therefore the concentrations of the weak base and its conjugate acid must be equal in order for the log term to be zero in the Henderson-Hasselbalch equation.

Let's analyze the given options:

1. 0.2 M \( \mathrm{NH_4OH}\) (0.5 L) + 0.1 M \( \mathrm{HCl} \) (0.5 L)
   • Moles of \(\mathrm{NH_4OH}\) = 0.2 × 0.5 = 0.1 moles
   • Moles of \(\mathrm{HCl}\) = 0.1 × 0.5 = 0.05 moles
   • Excess \(\mathrm{NH_4OH}\) after \(\mathrm{HCl}\) neutralization = 0.1 - 0.05 = 0.05 moles (for remaining weak base)
   • Moles of \(\mathrm{NH_4Cl}\) formed = 0.05 moles (for conjugate acid)
2. 0.4 M \( \mathrm{NH_4OH} \) (1 L) + 0.1 M \( \mathrm{HCl} \) (1 L)
   • Excess \(\mathrm{NH_4OH}\) after \(\mathrm{HCl}\) neutralization is high, altering the required 1:1 ratio.
3. 0.2 M \( \mathrm{NH_4OH} \) (0.4 L) + 0.1 M \( \mathrm{HCl} \) (1 L)
   • Excess \(\mathrm{HCl}\) is present leading to no buffer solution formation as excess acid remains.
4. 0.5 M \( \mathrm{NH_4OH} \) (0.2 L) + 0.2 M \( \mathrm{HCl} \) (0.5 L)
   • Again, excessive \(\mathrm{HCl}\) leads to neutralization of all \(\mathrm{NH_4OH}\) and excess acid remains.

Thus, the correct choice is:

0.2 M \( \mathrm{NH_4OH}\,(0.5\,\text{L}) + 0.1\,\text{M } \mathrm{HCl}\,(0.5\, \text{L})\), as it achieves the equal moles of \(\mathrm{NH_4OH}\) and \(\mathrm{NH_4Cl}\) required for a pH of 9.25.