In the following reaction, $\large AgCl\, + \,KI\,\, \rightleftharpoons \,KCl\, + \,AgI$ . As KI is added, the equilibrium is shifted towards right giving more AgI precipitate, because

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Both AgCl and AgI are sparingly soluble

Both AgCl and AgI have same solubility product.

The K_sp of AgI is lower than K_sp of AgCl

The K_sp of AgI is higher than K_sp of AgCl

answer is B.

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Detailed Solution

$\large AgCl\, + \,KI\,\, \rightleftharpoons \,KCl\, + \,AgI$

On addition of KI to the above equilibrium, the equilibrium shifts forward giving more AgI ,because the solubility of Agcl is more than that of AgI, K_sp of AgCl is more than that of AgI

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