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 \( \mathrm{NH_4OH} = 7.75 \))

Hint:

For buffer problems:
Always convert pH to pOH for basic buffers
Use mole ratios (not concentrations) when volumes differ

Answer 1

Correct Answer: 1001

Concept: When a weak base reacts with a strong acid, the resulting solution behaves as a buffer consisting of the weak base and its conjugate acid. 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. Given: \[ \mathrm{pH} = 9.25 \] \[ \mathrm{pOH} = 14 - 9.25 = 4.75 \]
Step 2: Write the buffer equation. \[ 4.75 = 7.75 + \log\left(\frac{\text{salt}}{\text{base}}\right) \] \[ \log\left(\frac{\text{salt}}{\text{base}}\right) = -3 \] \[ \frac{\text{salt}}{\text{base}} = 10^{-3} \]
Step 3: Express salt and base in terms of volumes. 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: \[ \text{Base initially} = 0.1x \] \[ \text{Acid added} = 0.1 \] After reaction: \[ \text{Salt} = 0.1 \] \[ \text{Remaining base} = 0.1x - 0.1 \]
Step 4: Substitute into the ratio. \[ \frac{0.1}{0.1x - 0.1} = 10^{-3} \] \[ \frac{1}{x - 1} = 10^{-3} \] \[ x - 1 = 1000 \] \[ x = 1001 \] \[ \boxed{x = 1001} \]