Question

Consider the following equilibria in aqueous solution (1) Ag++ Cl-ーAgCl(aq) (2) AgCl(aq)CAgC12 (3) AgCls)Ag*CI K = 20-103 K = 93 K= 1.8.10-10 (a) Find K for the reaction AgCI(s)AgCl(aq). The species AgCl(aq) is an ion pair consisting of Ag and Cl associated with each other in solution. (b) Find [AgCl(aq)] in equilibrium with excess AgCl(s).

Answer 1

Explanation:

(a)  Chemical reaction equation is given as follows.

           $Ag^{+} + Cl^{-} \rightarrow AgCl(aq) \rightarrow K_{1}$

Also,  

         $AgCl(s) \rightarrow Ag^{+} + Cl^{-} \rightarrow K_{3}$

Therefore, the net reaction equation is as follows.

         $AgCl(s) \rightarrow AgCl(aq)$

Now, we will calculate the value of K for this reaction as follows.

          K = $K_{1} \rightarrow K_{2}$

             = $2.0 \times 10^{3} \times 1.8 \times 10^{-10}$

             = $3.6 \times 10^{-7 }$

Hence, the value of K for the given reaction is  $3.6 \times 10^{-7 }$.

(b)  As the reaction  is given as follows.

              $AgCl(s) \rightarrow AgCl(aq)$

Therefore, when excess of AgCl(s) is added then the amount of [AgCl(aq)] present in equilibrium is as follows.

                   K = [AgCl(aq)] = $3.6 \times 10^{-7 }$

Thus, the value of [AgCl(aq)] in equilibrium with excess AgCl(s) is $3.6 \times 10^{-7 }$.