Which of the following graph truly represents the titration of solution against solution? — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Understand conductometric titration of HCl vs NH₄OH.** In this titration, a strong acid (HCl) is titrated against a weak base (NH₄OH). The conductance of the solution changes as the base is added. **Step 2: Analyse the initial stages (before equivalence point).** Initially, HCl is fully dissociated: $$\ce{HCl -> H+ + Cl-}$$ $\text{H}^+$ has very high molar conductance ($\lambda^\circ \approx 350\ \text{S cm}^2\text{mol}^{-1}$). As NH₄OH is added: $$\ce{H+ + NH4OH -> NH4+ + H2O}$$ $\text{H}^+$ ions (high conductance) are replaced by $\text{NH}_4^+$ ions (low conductance, $\lambda^\circ \approx 73\ \text{S cm}^2\text{mol}^{-1}$). Conductance **decreases sharply**. **Step 3: Analyse the region after equivalence point.** After the equivalence point, excess $\text{NH}_4\text{OH}$ is added. Since NH₄OH is a weak base and the solution already contains $\text{NH}_4^+$ ions (common ion effect), further ionisation of $\text{NH}_4\text{OH}$ is suppressed: $$\ce{NH4OH <=> NH4+ + OH-} \quad \text{(equilibrium suppressed)}$$ The conductance remains **nearly constant** (very slight increase, practically a plateau). **Step 4: Identify the correct graph description.** The conductance–volume graph shows: - A **sharp decrease** before the equivalence point (as $\text{H}^+$ is replaced by $\text{NH}_4^+$) - A **plateau** after the equivalence point (due to suppression of NH₄OH ionisation) $$\boxed{\text{Answer: B}}$$