Question

The salt $ {\text{Al}}{\left( {{\text{OH}}} \right)_3} $ is involved in the following two equilibria.
 $ {\text{Al}}{\left( {{\text{OH}}} \right)_3}\left( {\text{s}} \right) \rightleftharpoons {\text{A}}{{\text{l}}^{ + 3}}\left( {{\text{aq}}} \right) + 3{\text{O}}{{\text{H}}^ - }\left( {{\text{aq}}} \right);{{\text{K}}_{{\text{sp}}}} $
 $ {\text{Al}}{\left( {{\text{OH}}} \right)_3} + {\text{OH}}\left( {{\text{aq}}} \right) \rightleftharpoons {\text{Al}}\left( {{\text{OH}}} \right)_4^ - \left( {{\text{aq}}} \right);{{\text{K}}_{\text{c}}} $
Which of the following relationships is correct when the solubility is minimum?
(A) $ \left[ {{\text{O}}{{\text{H}}^ - }} \right] = {\left( {\dfrac{{{{\text{K}}_{{\text{SP}}}}}}{{{{\text{K}}_{\text{c}}}}}} \right)^{1/3}} $
(B) $ \left[ {{\text{O}}{{\text{H}}^ - }} \right] = {\left( {\dfrac{{{{\text{K}}_{\text{c}}}}}{{{{\text{K}}_{{\text{SP}}}}}}} \right)^{1/4}} $
(C) $ \left[ {{\text{O}}{{\text{H}}^ - }} \right] = \sqrt {{{\left( {\dfrac{{{{\text{K}}_{{\text{SP}}}}}}{{{{\text{K}}_{\text{c}}}}}} \right)}^{1/4}}} $
(D) None of these

Answer 1

Hint: The total solubility will be equal to the solubility of the ions produced. The derivative with respect to the concentration of hydroxide ion is zero at minimum and maximum concentration.

Complete step by step solution:
The salt that is $ {\text{Al}}{\left( {{\text{OH}}} \right)_3} $ is present in two forms in dissolved form that is $ {\text{A}}{{\text{l}}^{ + 3}} $ and $ {\text{Al}}\left( {{\text{OH}}} \right)_4^ - $ , so the total solubility will be equal to the sum of solubility of these ions:
 $ {\text{solubility}} = \left[ {{\text{A}}{{\text{l}}^{ + 3}}} \right] + \left[ {{\text{Al}}\left( {{\text{OH}}} \right)_4^ - } \right] $
The concentration of aluminium ion will be calculated from the equation where this appear that is the first equation and the value of $ {\text{Al}}\left( {{\text{OH}}} \right)_4^ - $ will be calculated from the equation second.
 $ {\text{Al}}{\left( {{\text{OH}}} \right)_3}\left( {\text{s}} \right) \rightleftharpoons {\text{A}}{{\text{l}}^{ + 3}}\left( {{\text{aq}}} \right) + 3{\text{O}}{{\text{H}}^ - }\left( {{\text{aq}}} \right);{{\text{K}}_{{\text{sp}}}} $
The solubility product will be calculated as:
 $ {{\text{K}}_{{\text{sp}}}} = \left[ {{\text{A}}{{\text{l}}^{ + 3}}} \right]{\left[ {{\text{O}}{{\text{H}}^ - }} \right]^3} $
Now we will calculate the concentration of aluminium ion which will be:
 $ \dfrac{{{{\text{K}}_{{\text{sp}}}}}}{{{{\left[ {{\text{O}}{{\text{H}}^ - }} \right]}^3}}} = \left[ {{\text{A}}{{\text{l}}^{ + 3}}} \right] $
In the same way we will calculate the concentration of the other ion from the following equation:
 $ {\text{Al}}{\left( {{\text{OH}}} \right)_3} + {\text{OH}}\left( {{\text{aq}}} \right) \rightleftharpoons {\text{Al}}\left( {{\text{OH}}} \right)_4^ - \left( {{\text{aq}}} \right);{{\text{K}}_{\text{c}}} $
The value of equilibrium constant will be:
 $ {{\text{K}}_{\text{C}}} = \dfrac{{{\text{Al}}\left( {{\text{OH}}} \right)_4^ - }}{{{\text{O}}{{\text{H}}^ - }}} $
Hence the concentration of the ion will be:
 $ {{\text{K}}_{\text{C}}}\left[ {{\text{O}}{{\text{H}}^ - }} \right] = {\text{Al}}\left( {{\text{OH}}} \right)_4^ - $
We will now substitute the value in solubility equation:
 $ {\text{S}} = \dfrac{{{{\text{K}}_{{\text{sp}}}}}}{{{{\left[ {{\text{O}}{{\text{H}}^ - }} \right]}^3}}} + {{\text{K}}_{\text{C}}}\left[ {{\text{O}}{{\text{H}}^ - }} \right] $
For the minimum solubility the derivative will be zero:
 $ \dfrac{{{\text{dS}}}}{{{\text{d}}{{\left[ {{\text{O}}{{\text{H}}^ - }} \right]}^3}}} = 0 $
Doing the derivative we will get:
  $ \dfrac{{ - 3{{\text{K}}_{{\text{sp}}}}}}{{{{\left[ {{\text{O}}{{\text{H}}^ - }} \right]}^4}}} + {{\text{K}}_{\text{C}}} = 0 $
Both the solubility product and equilibrium constant are constants and hence we will not take their derivative.
Rearranging the above equation we will get:
 $ {\left[ {{\text{O}}{{\text{H}}^ - }} \right]^4} = \dfrac{{3{{\text{K}}_{{\text{SP}}}}}}{{{{\text{K}}_{\text{c}}}}} $
Taking fourth power on both sides we will get:
  $ \left[ {{\text{O}}{{\text{H}}^ - }} \right] = {\left[ {\dfrac{{3{{\text{K}}_{{\text{SP}}}}}}{{{{\text{K}}_{\text{c}}}}}} \right]^{\dfrac{1}{4}}} $
Thus, the correct option is D, that is, none of the above.

Note:
The solubility product is defined for the sparingly soluble salts which are not completely dissociated in solution. Both equilibrium and solubility constant are defined for the equilibrium reaction. They both are the function of temperature.