Answer
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Hint: The solubility product in water as well as buffer solution will be the same. So, by comparing we can find the solubility in buffer solution. Solubility is a property which refers to the ability to dissolve in a solvent for a given substance, the solute. A buffer solution is an aqueous solution composed of a weak acid mixture with its conjugate base, or vice versa. Its pH changes very little when it's added a small amount of strong acid or base.
Complete answer:
- First let’s see what a solubility product is. We know that when a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions.
- The equilibrium constant for the dissolution of such sparingly soluble salts is known as Solubility product ${{K}_{sp}}$.
- The chemical equation of \[Cd{{\left( OH \right)}_{2}}\]in water will be:
\[Cd{{\left( OH \right)}_{2}}\rightleftharpoons C{{d}^{+2}}+2O{{H}^{-}}\]
Solubility: S S 2S
- The equilibrium constant for this solution will be written as:
\[K=\dfrac{\left[ C{{d}^{+2}} \right]{{\left[ O{{H}^{-}} \right]}^{2}}}{\left[ Cd{{\left( OH \right)}_{2}} \right]}\]
Therefore, $\left[ Cd{{\left( OH \right)}_{2}} \right]K={{K}_{sp}}=\left[ C{{d}^{+2}} \right]{{\left[ O{{H}^{-}} \right]}^{2}}$
- So, ${{K}_{sp}}$ = $S\times {{(2S)}^{2}}=4{{S}^{3}}$
Here the power 2 is the stoichiometric constant of $O{{H}^{-}}$ions.
- In the question value of solubility is given as $1.84\times {{10}^{-5}}M$, putting this value of S we get:
${{K}_{sp}}$ = $4\times {{(1.84\times {{10}^{-5}})}^{3}}=2.49\times {{10}^{-14}}$
- Now the pH of the given buffer solution is pH = 12
- Therefore, the pOH will be, pOH = 2
-Through this we can calculate the concentration of $O{{H}^{-}}$ions.
-log[$O{{H}^{-}}$] = 2
[$O{{H}^{-}}$] = 10-2
- Now, let’s see the equation again
\[Cd{{\left( OH \right)}_{2}}\rightleftharpoons C{{d}^{+2}}+2O{{H}^{-}}\]
S ${{10}^{-2}}$
Now, the ${{K}_{sp}}$ will be
${{K}_{sp}}$ = $S\times {{({{10}^{-2}})}^{2}}$
We already calculated the value earlier ${{K}_{sp}}$ = $2.49\times {{10}^{-14}}$
Now put this value and after solving the equation we get
$2.49\times {{10}^{-14}}$= $S\times {{({{10}^{-2}})}^{2}}$
Therefore, S = $2.49\times {{10}^{-10}}M$
So, the expected solubility of \[Cd{{\left( OH \right)}_{2}}\] in a buffer solution of pH = 12 is option D. $2.49\times {{10}^{-10}}M$
Note: Remember that solubility is usually expressed in terms of mass of solute per 100mL of solvent, whereas ${{K}_{sp}}$ is defined in terms of the molar concentrations of the component ions.
While calculating the product of solubility of the ions is raised to the power of their stoichiometric constant. For example:
$\left[ Cd{{\left( OH \right)}_{2}} \right]K={{K}_{sp}}=\left[ C{{d}^{+2}} \right]{{\left[ O{{H}^{-}} \right]}^{2}}$,
The power 2 is the stoichiometric constant of the $O{{H}^{-}}$ions.
Complete answer:
- First let’s see what a solubility product is. We know that when a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions.
- The equilibrium constant for the dissolution of such sparingly soluble salts is known as Solubility product ${{K}_{sp}}$.
- The chemical equation of \[Cd{{\left( OH \right)}_{2}}\]in water will be:
\[Cd{{\left( OH \right)}_{2}}\rightleftharpoons C{{d}^{+2}}+2O{{H}^{-}}\]
Solubility: S S 2S
- The equilibrium constant for this solution will be written as:
\[K=\dfrac{\left[ C{{d}^{+2}} \right]{{\left[ O{{H}^{-}} \right]}^{2}}}{\left[ Cd{{\left( OH \right)}_{2}} \right]}\]
Therefore, $\left[ Cd{{\left( OH \right)}_{2}} \right]K={{K}_{sp}}=\left[ C{{d}^{+2}} \right]{{\left[ O{{H}^{-}} \right]}^{2}}$
- So, ${{K}_{sp}}$ = $S\times {{(2S)}^{2}}=4{{S}^{3}}$
Here the power 2 is the stoichiometric constant of $O{{H}^{-}}$ions.
- In the question value of solubility is given as $1.84\times {{10}^{-5}}M$, putting this value of S we get:
${{K}_{sp}}$ = $4\times {{(1.84\times {{10}^{-5}})}^{3}}=2.49\times {{10}^{-14}}$
- Now the pH of the given buffer solution is pH = 12
- Therefore, the pOH will be, pOH = 2
-Through this we can calculate the concentration of $O{{H}^{-}}$ions.
-log[$O{{H}^{-}}$] = 2
[$O{{H}^{-}}$] = 10-2
- Now, let’s see the equation again
\[Cd{{\left( OH \right)}_{2}}\rightleftharpoons C{{d}^{+2}}+2O{{H}^{-}}\]
S ${{10}^{-2}}$
Now, the ${{K}_{sp}}$ will be
${{K}_{sp}}$ = $S\times {{({{10}^{-2}})}^{2}}$
We already calculated the value earlier ${{K}_{sp}}$ = $2.49\times {{10}^{-14}}$
Now put this value and after solving the equation we get
$2.49\times {{10}^{-14}}$= $S\times {{({{10}^{-2}})}^{2}}$
Therefore, S = $2.49\times {{10}^{-10}}M$
So, the expected solubility of \[Cd{{\left( OH \right)}_{2}}\] in a buffer solution of pH = 12 is option D. $2.49\times {{10}^{-10}}M$
Note: Remember that solubility is usually expressed in terms of mass of solute per 100mL of solvent, whereas ${{K}_{sp}}$ is defined in terms of the molar concentrations of the component ions.
While calculating the product of solubility of the ions is raised to the power of their stoichiometric constant. For example:
$\left[ Cd{{\left( OH \right)}_{2}} \right]K={{K}_{sp}}=\left[ C{{d}^{+2}} \right]{{\left[ O{{H}^{-}} \right]}^{2}}$,
The power 2 is the stoichiometric constant of the $O{{H}^{-}}$ions.