Question

The solubility of \[{\text{CuBr}}\] is $2 \times {10^{ - 4}}{\text{mol/l}}$ at $25^\circ {\text{C}}$ . The ${{\text{K}}_{{\text{sp}}}}$ value for \[{\text{CuBr}}\] is:
A.$4 \times {10^{ - 8}}{\text{mo}}{{\text{l}}^{\text{2}}}{\text{l}}{{\text{t}}^{{\text{ - 2}}}}$
B.$4 \times {10^{ - 11}}{\text{mo}}{{\text{l}}^{\text{2}}}{\text{l}}{{\text{t}}^{{\text{ - 2}}}}$
C.$4 \times {10^{ - 4}}{\text{mo}}{{\text{l}}^{\text{2}}}{\text{l}}{{\text{t}}^{{\text{ - 2}}}}$
D.$4 \times {10^{ - 15}}{\text{mo}}{{\text{l}}^{\text{2}}}{\text{l}}{{\text{t}}^{{\text{ - 2}}}}$

Answer 1

Hint: The product of the concentrations of the ions in a saturated solution of a substance at constant temperature is called the solubility product of that substance.
For a general dissociation of a salt AB into its corresponding ions
\[{{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\left( {\text{s}} \right) \rightleftharpoons {\text{x}}{{\text{A}}^{{\text{ + y}}}}\left( {{\text{aq}}} \right) + {\text{y}}{{\text{B}}^{{\text{ - x}}}}\left( {{\text{aq}}} \right)\]
The solubility product is given by the expression:
${{\text{K}}_{{\text{sp}}}} = {\left[ {{{\text{A}}^{{\text{ + y}}}}} \right]^{\text{x}}}{\left[ {{{\text{B}}^{{\text{ - x}}}}} \right]^{\text{y}}}$

Complete step by step answer:
The maximum number of moles of solute which can be dissolved in a solvent at a constant temperature to obtain 1 litre of solution is called the solubility of the solute.
On the other hand, the solubility product may be defined as the product of the concentration of the ions raised to a power which is equal to the number of ions given by the dissociation of electrolyte at a given temperature when the solution is saturated.
Suppose the concentration of the solute initially before being dissociated is ‘a’. if the solubility of the solute is denoted by ‘S’, then the concentration of the solute after dissociation will be ${\text{a - S}}$ and that of the ions will be ‘xS’ and ‘yS’.
\[{{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\left( {\text{s}} \right) \rightleftharpoons {\text{x}}{{\text{A}}^{{\text{ + y}}}}\left( {{\text{aq}}} \right) + {\text{y}}{{\text{B}}^{{\text{ - x}}}}\left( {{\text{aq}}} \right)\]
  a 0 0
 ${\text{a - S}}$ xS yS
Then, the solubility product in terms of solubility is:
$
  {{\text{K}}_{{\text{sp}}}} = {\left[ {{{\text{A}}^{{\text{ + y}}}}} \right]^{\text{x}}}{\left[ {{{\text{B}}^{{\text{ - x}}}}} \right]^{\text{y}}} \\
   \Rightarrow {{\text{K}}_{{\text{sp}}}} = {\left( {{\text{xS}}} \right)^{\text{x}}}{\left( {{\text{yS}}} \right)^{\text{y}}} \\
   \Rightarrow {{\text{K}}_{{\text{sp}}}} = {{\text{x}}^{\text{x}}}{{\text{y}}^{\text{y}}}{{\text{S}}^{\left( {{\text{x + y}}} \right)}} \\
 $
Now, the given salt \[{\text{CuBr}}\] will be fully dissociated as
           \[{\text{CuBr}} \rightleftharpoons {\text{C}}{{\text{u}}^ + }{\text{ + B}}{{\text{r}}^ - }\]
Initial: S 0 0
Final: 0 S S
Therefore, the solubility product will be:
$
  {{\text{K}}_{{\text{sp}}}} = {1^1} \times {{\text{1}}^1} \times {{\text{S}}^{\left( {{\text{1 + 1}}} \right)}} \\
   \Rightarrow {{\text{K}}_{{\text{sp}}}} = {{\text{S}}^2} \\
 $
According to the given question, the solubility of \[{\text{CuBr}}\] is $2 \times {10^{ - 4}}{\text{mol/l}}$ at $25^\circ {\text{C}}$ .This means S is equal to $2 \times {10^{ - 4}}{\text{mol/l}}$ .
So, from the above expression, we have the solubility product of \[{\text{CuBr}}\] as:
$
  {{\text{K}}_{{\text{sp}}}} = {{\text{S}}^2} \\
   \Rightarrow {{\text{K}}_{{\text{sp}}}} = {\left( {2 \times {{10}^{ - 4}}} \right)^2}{\text{mo}}{{\text{l}}^{\text{2}}}{\text{l}}{{\text{t}}^{{\text{ - 2}}}} \\
   \Rightarrow {{\text{K}}_{{\text{sp}}}} = 4 \times {10^{ - 8}}{\text{mo}}{{\text{l}}^{\text{2}}}{\text{l}}{{\text{t}}^{{\text{ - 2}}}} \\
 $

Hence, the correct option is A.

Note:
If the ionic product is less than the solubility product, the solution is unsaturated and precipitation will not take place. If the ionic product is equal to the solubility product, the solution is saturated and equilibrium exists.
On the other hand, if the ionic product is greater than the solubility product, the solution is supersaturated and precipitation will take place. Hence, the essential condition for precipitation of an electrolyte is that the ionic product, i.e., the product of the concentration of its ions present in a solution should exceed the solubility product of the substance and this is useful in predicting ionic reactions.