The solubility of AgCl (s) with solubility product 1.6 x 10-10 in 0.1 M NaCl solution would be

  • Option 1)

    1.26 x 10-5 M

  • Option 2)

    1.6 x 10-9 M

  • Option 3)

    1.6 x 10-11 M

  • Option 4)

    zero

 

Answers (1)

As learnt in

Solubility product constant -

e.g.

BaSO_{4}\rightleftharpoons Ba^{2+}(aq)+SO_{4}^{2-}(aq)


K=\frac{[Ba^{2+}]\:[SO_{4}^{2-}]}{[BaSO_{4}]}

- wherein

For a pure solid substance the concentration remain constant

\therefore \:K_{sp}=K[BaSO_{4}]
 

               =[Ba^{2+}]\:[SO_{4}^{2-}]
 

    K_{sp}=solubility\:product

 

 AgCl \left ( s \right )\rightleftharpoons Ag^{+}+ Cl ^{-}\:\:,Ksp = 1.6 \times 10^{-10}

Now [Cl-] = 0.1 M, as NaCl is a strong electrolyte.

\therefore [Ag+] = s

    [Cl-] = 0.1

\therefore  s x 0.1 = 1.6 x 10-10  as s << 1

     D s = 1.6 x 10-9


Option 1)

1.26 x 10-5 M

This option is incorrect

Option 2)

1.6 x 10-9 M

This option is correct

Option 3)

1.6 x 10-11 M

This option is incorrect

Option 4)

zero

This option is incorrect

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