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The solubllity of AgCI in water at 298 K is 1.06 x 10-5 mole per litre. Calculate is solubility product at this temperature. |
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Answer» The solubility equilibrium in the saturated solution is AgCl (s) ⇌ Ag+ (aq) + Cl- (aq) The solubility of AgCl is 1.06 x 10-5 mole per litre. [Ag+ (aq)] = 1.06 x 10-5 mol L-1 [Cl- (aq)] = 1.06 x 10-5 mol L-1 Ksp = [Ag+(aq)] [Cl- (aq)] = (1.06 x 10-5 mol L-1) x (1.06 x 10-5 mol L-1) = 1.12 x 10-2 moI-2 L-2 |
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