Which of the followmg mixture gives a buffer solution with pH=9.25?
Given : $$pK_{b}$$ $$(NH_{4}OH)$$ = 4.75

We need to find which mixture gives a buffer with pH = 9.25.

Key Formula: For a basic buffer: $$pOH = pK_b + \log\frac{[\text{salt}]}{[\text{base}]}$$

Since pH = 9.25 and pH + pOH = 14: $$pOH = 14 - 9.25 = 4.75 = pK_b$$

This means $$\log\frac{[\text{salt}]}{[\text{base}]} = 0$$, so $$[\text{salt}] = [\text{base}]$$, i.e., equal moles of $$NH_4OH$$ and $$NH_4Cl$$ after the reaction.

Checking Option 4: 0.2 M $$NH_4OH$$ (0.5 L) + 0.1 M HCl (0.5 L)

Moles $$NH_4OH$$ = $$0.2 \times 0.5 = 0.10$$ mol

Moles HCl = $$0.1 \times 0.5 = 0.05$$ mol

After reaction: $$NH_4OH$$ remaining = $$0.10 - 0.05 = 0.05$$ mol, $$NH_4Cl$$ formed = 0.05 mol.

Moles of salt = moles of base = 0.05. Equal! ✓

$$pOH = 4.75 + \log(1) = 4.75$$, so $$pH = 9.25$$ ✓

The correct answer is Option 4.