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Answer the following · Q7

Q.viii. Solubility of a sparingly soluble salt get affected in presence of a soluble salt having one common ion. Explain.

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Step 1. Set up the base equilibrium. Consider sparingly soluble AgCl in equilibrium with its own saturated solution: AgCl(s)⇌Ag+(aq)+Cl−(aq)AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq), Ksp=[Ag+][Cl−]K_{sp}=[Ag^+][Cl^-].

Step 2. Add a common-ion salt. Suppose AgNO3, a soluble strong electrolyte, is added to this saturated solution. It dissociates completely, AgNO3(aq)→Ag+(aq)+NO3−(aq)AgNO_3(aq)\rightarrow Ag^+(aq)+NO_3^-(aq), supplying a large EXTRA amount of Ag+ -- the ion AgCl and AgNO3 have in common.

Step 3. Apply Le Chatelier's principle. The extra Ag+ increases the concentration of one of the PRODUCTS of AgCl's own dissolution equilibrium. The system responds by shifting toward the reactant (solid AgCl) side -- i.e. the reverse (precipitation) reaction is favoured, until equilibrium is re-established.

Step 4. Interpret the shift. This shift means MORE Ag+ and Cl- combine back into solid AgCl than before -- so the total amount of AgCl that stays dissolved (its solubility) DECREASES in the presence of the common Ag+ ion, compared with its solubility in pure water.

Step 5. Note what does NOT change. Ksp itself is a genuine equilibrium constant at a given temperature, so it stays exactly the same value throughout -- what changes is only the individual ion concentrations (and hence how much solid stays undissolved), not Ksp. …

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