JEE Main 2017 — Ionic Equilibrium Question with Solution

NH₃ + HCl $$\to$$ NH₄Cl

moles of HCl = 0.2 M × 25 × 10^–3 L = 0.005 moles HCl (total consumed)
moles of NH₃ = 0.2 M × 50 × 10^–3 L = 0.01 moles HCl
excess NH₃ = 0.01 – 0.005 = 0.005 moles

From reaction, 1 mole ammonia = 1 mole NH4Cl

$$\therefore$$ 0.005 NH₃ = 0.005 NH₄Cl

Total Volume = V_HCl + V_NH3 = 25 + 50 = 75 mL

[NH₃] = [NH₄Cl] = $$\frac{0.005}{75\times {10}^{-3}}$$ = 0.066 M

pOH = pK_b + log $$\frac{\left[salt\right]}{\left[base\right]}$$

= pK_b + log$$\frac{\left[N{H}_{4}Cl\right]}{\left[N{H}_3\right]}$$

= 4.75 + log$$\frac{\left[0.066\right]}{\left[0.066\right]}$$

$$\Rightarrow$$ pOH = 4.75

$$\therefore$$ pH = 14 – 4.75 = 9.25