Question

The solubility products of three sparingly soluble salt ${M_2}X$ , $MX$ and $M{X_3}$ are identical. What will be the order of their solubilities?
A. $M{X_3} > {M_2}X > MX$
B.$M{X_3} > MX > {M_2}X$
C.$MX > {M_2}X > M{X_3}$
D.$MX > M{X_3} > {M_2}X$
What volume of $C{O_2}$ at STP is obtained by $5g$?

Answer 1

First, think about solubility definition. Then think about the formula of solubility and the way in which the salts will break down in the ionization process. Write the equations for the same and using the equations calculate the solubility.

Step by step answer: Let’s define solubility first. Solubility is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. In simple words, Solubility indicates the maximum amount of substance that can be dissolved in a solvent at a given temperature. It is calculated by gram of solute dissolved per $100g$ of solvent and also by number of moles dissolved per 1 litre of solvent. The formula for solubility is as follows:

$$$$ $S = \sqrt {{K_{sp}}}$

Here ${K_{SP}}$ is the solubility product constant. Now, let’s break down each of our soluble salts into their ion forms and find their solubility. They are ionized as follows:
For ${M_2}X$

${l}{K_{sp}} = {[{M^ + }]^2}[{X^{2 - }}]\\\\{K_{sp}} = [{(2s)^2}][(s)] = 4{s^3}$ ${K_{sp}} = {[{M^ + }]^2}[{X^{2 - }}]$
${K_{sp}} = [{(2s)^2}][(s)] = 4{s^3}$
Here, the ${M_2}X$ is broken down into two ions of $M$ and one ion of $X$ . These are then replaced with the solubility s. Now let’s calculate for other salts.

For $MX$

${K_{sp}} = [{M^ + }][{X^ - }]$
${K_{sp}} = [s][(s)] = {s^2}$

For $M{X_3}$

${K_{sp}} = [{M^{3 + }}]{[{X^ - }]^3}$
${K_{sp}} = [s][{(3s)^3}] = 27{s^4}$
Here we can see the solubilities of the three salts. We get that $M{X_3}$ has highest solubility and $MX$ has lowest solubility. Therefore the order of solubility is as follows
$MX < {M_2}X < M{X_3}$
Therefore, Option that is correct is A.

Now, we will move on to the second part of the question where we have to find the volume of $C{O_2}$. So, we have to find the volume of $5g$ of $C{O_2}$ at standard temperature and pressure. STP is ${0^ \circ }C$ or 273.15 Kelvin. Here we will use ideal gas law to answer this question. The formula is:
$PV = nRT$
Where,
$P$ is pressure
$V$ is volume
$n$ is moles
$R$ is gas constant
$T$ is temperature in kelvins
Now, we require moles of $C{O_2}$ but we have the mass of $C{O_2}$. So we will calculate the moles from mass by multiplying given mass by the inverse of molar mass of $C{O_2}$ which is 44.009 g/mol.
$0.5g \times \dfrac{{1mol}}{{44.009g}} = 0.1136mol$
We get 0.1136 moles of $C{O_2}$. Now we will calculate using ideal gas law to find volume which will be as follows:
$V = \dfrac{{nRT}}{P}$
$V = \dfrac{{0.1136 \times 8.3145 \times 273.15}}{{100}} = 2.6L$

Therefore, we get the volume of carbon dioxide $2.6L$.

Note: When calculating the solubility we need to take care when we ionize the salts as we might make mistakes. Now, for the second part of the question, we need to have the knowledge of STP, the gas constant and the formula. Units of the constant should be written in every step.