Question

Volume ratio of decimolar \( \mathrm{NH_4OH} \) and decimolar \( \mathrm{HCl} \) to give a solution of pH \(= 9.25\) at \(25^\circ\mathrm{C}\) is \(x : 1\). Find \(x\). (Given: \( \mathrm{p}K_b \) of \( \mathrm{NH_4OH} = 4.75\))

Hint:

When \(\mathrm{pOH} = \mathrm{p}K_b\):
Concentration of base = concentration of salt
This condition simplifies buffer problems greatly

Answer 1

Correct Answer: 2

Concept: When a weak base reacts partially with a strong acid, a buffer solution is formed. For a basic buffer: \[ \mathrm{pOH} = \mathrm{p}K_b + \log \left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \] Also, \[ \mathrm{pH} + \mathrm{pOH} = 14 \]
Step 1: Calculate pOH of the solution. \[ \mathrm{pOH} = 14 - 9.25 = 4.75 \]
Step 2: Apply the Henderson equation. \[ \mathrm{pOH} = \mathrm{p}K_b + \log \left(\frac{[\mathrm{NH_4Cl}]}{[\mathrm{NH_4OH}]}\right) \] Substitute values: \[ 4.75 = 4.75 + \log \left(\frac{[\mathrm{NH_4Cl}]}{[\mathrm{NH_4OH}]}\right) \] \[ \log \left(\frac{[\mathrm{NH_4Cl}]}{[\mathrm{NH_4OH}]}\right) = 0 \] \[ \frac{[\mathrm{NH_4Cl}]}{[\mathrm{NH_4OH}]} = 1 \]
Step 3: Use stoichiometry to find volume ratio. Let volume of decimolar \( \mathrm{NH_4OH} = x \) L Let volume of decimolar \( \mathrm{HCl} = 1 \) L Reaction: \[ \mathrm{NH_4OH + HCl \rightarrow NH_4Cl + H_2O} \] Moles: \[ \text{Moles of } \mathrm{NH_4OH} = 0.1x \] \[ \text{Moles of } \mathrm{HCl} = 0.1 \] After reaction: \[ \text{Moles of } \mathrm{NH_4Cl} = 0.1 \] \[ \text{Remaining } \mathrm{NH_4OH} = 0.1x - 0.1 \] Given ratio: \[ \frac{0.1}{0.1x - 0.1} = 1 \] \[ 0.1x - 0.1 = 0.1 \] \[ x = 2 \] \[ \boxed{x = 2} \]