Given: 200 ml of 0.005 M AgNo3 reacts with 300 ml of 0.01 M KCl.

To find: The maximum concentration of AG+ in mixture?

Solution:

• Now we have number of moles of:

          Ag+ ions = 200 x 0.005 / 1000 = 10^-3 moles

          Cl- ions = 300 x 0.01 / 100 = 3 x 10^-3 moles.

• Now the ionic equation is:

          Ag+(aq)   +  Cl-(aq)   ---------->   AgCl (s)

• But Cl-  ions are more that Ag+ ions, so it will consume all Ag+  ions and FORM AgCl.

• Then the amount of Cl- ions left will be:

          3 x 10^-3  - 10^-3

          2 x 10^-3 moles.

          [ Cl-] (left ) = 2 x 10^-3 / 500 x 1000

                            = 4 x 10^-3 M

                            = 0.004 M

• Now dissociation of AgCl into its ions will PRODUCE an equal amount of each ion. Let the produced ions be S.

• Then [ Ag+ ] = S and [ a-  ] = S + 0.004 ≃ 0.004

                 K(cp) =  [ Ag+ ] [ Cl- ] = 1.8 x 10^-10

                 [ Ag+ ] = 1.8 x 10^-10 / [ Cl- ]

                 [ Ag+ ] = 1.8 x 10^-10 / 0.004

• So, maximum concentration of Ag+  in the mixture is 4.5×10^-8 M.

Answer:

            So, maximum concentration of Ag+  in the mixture is 4.5×10^-8 M.