Question

$ {{K}_{sp}} $ of AgCl at $ 18{}^\circ C $ is $ 1.8\times {{10}^{-10}} $ . If $ A{{g}^{+}} $ of solution is $ 4\times {{10}^{-3}} $ mol/litre, the $ C{{l}^{-}} $ that must exceed before AgCl is precipitated would be
(A) $ 4.5\times {{10}^{-8}} $ mol/litre
(B) $ 7.2\times {{10}^{-13}} $ mol/litre
(C) $ 4.0\times {{10}^{-3}} $ mol/litre
(D) $ 4.5\times {{10}^{-7}} $ mol/litre

Answer 1

Hint :Solubility product is a constant or equilibrium constant for the dissociation of solid substances into aqueous solution. It depends on temperature and is represented by Ksp.

Complete Step By Step Answer:
The property of the body by which the solute dissolves in a solvent to form a solution is called solubility.
Solubility product is the equilibrium constant which increases with increase in temperature since solubility increases with temperature.
Solubility product also depends upon lattice energy of salt and salvation enthalpy of ions. Most salts dissociate into ions when they dissolve.
Solubility is different for different salts:
Soluble Solubility $ >0.1M $
Slightly soluble $ 0.01M< $ Solubility $ <0.1M $
 Sparingly soluble Solubility $ <0.1M $
The presence of a common ion lowers the value of the solubility product that is the common-ion effect.
If the ions are uncommon among the solute Ksp will be high due to diverse-ion effects. The presence of ion pairs will also affect Ksp value.
The reaction in the question will take place in the following way:
 $ AgCl\to A{{g}^{+}}+C{{l}^{-}} $
The concentration of $ [A{{g}^{+}}] $ is given to be $ 4\times {{10}^{-3}} $ .
The $ {{K}_{sp}} $ will be:
 $ {{K}_{sp}}=1.8\times {{10}^{-10}} $
 $ \begin{align}
  & [A{{g}^{+}}][C{{l}^{-}}]=1.8\times {{10}^{-10}} \\
 & [C{{l}^{-}}]=\frac{1.8\times {{10}^{10}}}{[A{{g}^{+}}]}=\frac{1.8\times {{10}^{-10}}}{4\times {{10}^{-3}}}=4.5\times {{10}^{-8}} \\
\end{align} $ .
Hence the concentration of $ [C{{l}^{-}}] $ will be $ 4.5\times {{10}^{-8}} $ .
Option A is correct.

Note :
In this reaction it was balanced otherwise to balance the equation before performing any other step. Also we power the stoichiometric coefficient with the concentration in the Ksp value.