1.

The solubility of AgCl in water at 250 C is found to be 1.06*10-5 moles per litre. Calculate the solubility product of AgCl at this temperature. 

Answer»

AgCl ionizes completely in the solution as AgCl ----> Ag+ +Cl 

One mole of AgCl in the solution gives 1 mole of Ag+ ions and 1 mole of Clions.

Now as the solubility of AgCl= 1.06*10—5 moles per litre. 

Therefore [Ag+]=1.06*10—5 moles per litre and [Cl--] = 1.06*10—5 moles per litre. 

Ksp for AgCl=[Ag+] [Cl--]=1.06*10—5 *1.06*10—5 =1.1*10—10



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