Answer
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Hint: Solubility product, ${ K }_{ sp }$, is the product of a substance’s dissolved ion concentrations raised to the power of their stoichiometric coefficients.
Complete step by step solution:
> It is given that,
S = ${ 1\times 10 }^{ -4 }$
${ K }_{ sp }$ =?
> The salt ${ A }_{ 2 }{ B }_{ 3 }$ will dissociates as shown below;
${ A }_{ 2 }{ B }_{ 3 }{ \rightarrow 2Al }^{ +3 }{ +3B }^{ -2 }$
> Let, the solubility of ${ A }_{ 2 }{ B }_{ 3 }$ is S, then the solubility of the products shown above.
> As we know, solubility products are the products of substances raised to their power of coefficients.
So, ${ K }_{ sp }$ = ${ (Al }^{ +3 }{ ) }^{ 2 }{ \times (B }^{ -2 }{ ) }^{ 3 }$
= ${ (2s) }^{ 2 }{ \times (3s) }^{ 3 }$
= ${ (4s) }^{ 2 }{ (27s) }^{ 3 }$
= ${ 108s }^{ 5 }$
As we know, S = ${ 1\times 10 }^{ -4 }$
Now, put the value of S, we get
${ K }_{ sp }$ = ${ 108\times (10 }^{ -4 }{ ) }^{ 5 }$
${ K }_{ sp }$ = ${ 1.08\times 10 }^{ -18 }$
The correct option is D.
> Ionic product is the product of the concentrations of ions, each raised to a power equal to the stoichiometric number for a solution of salt at a specified concentration.
Ionic product equals the solubility product for a saturated solution.
> Common ion effect: When a salt containing a common ion (an ion which is already in the given ionic solution)is added to an ionic equilibrium, then the equilibrium shifts to the backward direction. This effect caused by the added ion is called a common ion effect.
> According to Le Chatelier's Principle, if the equilibrium is disturbed by any of the factors, on which it depends, such as concentration, temperature, and pressure, then the equilibrium will shift in the direction, to balance the induced change.
Note: The possibility to make a mistake is that you may forget to multiply with their powers as the solubility product is the product of the substances raised to their stoichiometric coefficients.
Complete step by step solution:
> It is given that,
S = ${ 1\times 10 }^{ -4 }$
${ K }_{ sp }$ =?
> The salt ${ A }_{ 2 }{ B }_{ 3 }$ will dissociates as shown below;
${ A }_{ 2 }{ B }_{ 3 }{ \rightarrow 2Al }^{ +3 }{ +3B }^{ -2 }$
> Let, the solubility of ${ A }_{ 2 }{ B }_{ 3 }$ is S, then the solubility of the products shown above.
> As we know, solubility products are the products of substances raised to their power of coefficients.
So, ${ K }_{ sp }$ = ${ (Al }^{ +3 }{ ) }^{ 2 }{ \times (B }^{ -2 }{ ) }^{ 3 }$
= ${ (2s) }^{ 2 }{ \times (3s) }^{ 3 }$
= ${ (4s) }^{ 2 }{ (27s) }^{ 3 }$
= ${ 108s }^{ 5 }$
As we know, S = ${ 1\times 10 }^{ -4 }$
Now, put the value of S, we get
${ K }_{ sp }$ = ${ 108\times (10 }^{ -4 }{ ) }^{ 5 }$
${ K }_{ sp }$ = ${ 1.08\times 10 }^{ -18 }$
The correct option is D.
> Ionic product is the product of the concentrations of ions, each raised to a power equal to the stoichiometric number for a solution of salt at a specified concentration.
Ionic product equals the solubility product for a saturated solution.
> Common ion effect: When a salt containing a common ion (an ion which is already in the given ionic solution)is added to an ionic equilibrium, then the equilibrium shifts to the backward direction. This effect caused by the added ion is called a common ion effect.
> According to Le Chatelier's Principle, if the equilibrium is disturbed by any of the factors, on which it depends, such as concentration, temperature, and pressure, then the equilibrium will shift in the direction, to balance the induced change.
Note: The possibility to make a mistake is that you may forget to multiply with their powers as the solubility product is the product of the substances raised to their stoichiometric coefficients.
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