Question

The values of ${K_{sp}}$for sparingly soluble $AB$ and $M{B_2}$ are each equal to $4.0 \times {10^{ - 18}}$.Which salt is more soluble?

Answer 1

Hint : For any salt, its solubility product is the mathematical product of the concentrations of all the ions present in the solution expressed in Molarity. If salt is made up of three ions then, it will dissociate into three ions when it will be dissolved and if salt is made up of two ions then it will give two ions when dissolved.

Complete step by step solution :
We are given that Solubility product (${K_{sp}}$) of $AB$ and $M{B_2}$ is equal to $4.0 \times {10^{ - 18}}$. We know that solubility product is the product of concentration of ions of the salt in the solution.
From the value of ${K_{sp}}$, we can find out the concentration of the ions and then we can determine which salt has more solubility. First let’s determine the concentration of all the ions.
When AB salt gets dissolved, it will dissociate into two ions.
$AB \to {A^ + }_{(aq)} + {B^{^ - }}_{(aq)}$
So, according to definition of Solubility product, its formula is ${K_{sp}} = \left[ {{A^ + }} \right]\left[ {{B^ - }} \right]$
So, we can put the value of the solubility product because it is given to us.
$4.0 \times {10^{ - 18}} = \left[ {{A^ + }} \right]\left[ {{B^ - }} \right]$
Now, according to the law of conservation of mass, concentration of $\left[ {{A^ + }} \right]$ and $\left[ {{B^ - }} \right]$ must be equal.
so, we can write that
$4.0 \times {10^{ - 18}} = {\left[ {{A^ + }} \right]^2} = {\left[ {{B^ - }} \right]^2}$
So, $\left[ {{A^ + }} \right] = \left[ {{B^ - }} \right] = 2.0 \times {10^{ - 9}}M$
From the reaction, we can say that the amount of salt that is dissolved must be equal to concentration of ions according to the law of mass conservation law.
So, we can say that $2.0 \times {10^{ - 9}}M$ of salt $AB$ would have been dissolved in the solution.
Let’s find the same for $M{B_2}$ salt. We can write its dissociation reaction as below that it will dissociate into three ions, one positive and two negative.
$M{B_2} \to {M^{2 + }}_{(aq)} + 2{B^{^ - }}_{(aq)}$
Now, its solubility product can be expressed as
${K_{sp}} = \left[ {{M^{2 + }}} \right] \times {\left[ {{B^ - }} \right]^2}$
We are given the value of solubility product, so
$4.0 \times {10^{ - 18}} = \left[ {{M^ + }} \right] \times {\left[ {{B^ - }} \right]^2}$………………(1)
Now, if we say that $\left[ {{M^{2 + }}} \right]$ has concentration equal to $x$ then we can write that concentration of $\left[ {{B^ - }} \right]$ is equal to $2x$, putting that into the equation (1), we get
$4.0 \times {10^{ - 18}} = \left[ x \right]{\left[ {2x} \right]^2}$

$4.0 \times {10^{ - 18}} = 4{x^3}$ , so
$1.0 \times {10^{ - 18}} = {x^3}$
$x = {10^{ - 6}}M$
That means that ${10^{ - 6}}M$of $M{B_2}$ salt would have been dissolved in the solution.
- Based on the values obtained, we can say that $M{B_2}$ has higher solubility in the solution compared to $AB$.

Additional Information:
- Solubility product is the Equilibrium constant itself in the case of dissolution of salts in the solution. Suppose we have a reaction shown as below.
$a{A_{(s)}} \to b{B_{(aq)}} + c{C_{(aq)}}$
Now,
$K = \dfrac{{{{\left[ B \right]}^b}{{\left[ C \right]}^c}}}{{{{\left[ A \right]}^a}}}$
But as A is solid, we take its concentration equal to 1 because its concentration does not affect the solubility product. So, we can write that
${K_{sp}} = {\left[ B \right]^b}{\left[ C \right]^c}$

Note :
- Keep in mind that when power of 10 is in minus sign then higher the power, lesser the value of the number. Sometimes when in a hurry, these types of mistakes may happen.
- When we are trying to find a solubility product of any salt like $M{B_2}$ , then do not forget that it gives three ions when dissolved.