Answer
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Hint: Recall the relationship between solubility product and solubility. Write individually the dissolution reaction of each salt given in the question and then find their ${K_{sp}}$ values. To find ${K_{sp}}$ values, take product of the concentrations of the products, each raised to the power of their stoichiometric coefficient. Also, multiply the concentration by their respective coefficient.
Complete step by step solution:
Consider the general reaction: $aA \rightleftharpoons cC + dD$
Now, if we solve ${K_{sp}}$ for the above reaction, it will be as follows:
${K_{sp}} = {[cC]^c}{[dD]^d}$
Therefore, to solve ${K_{sp}}$ of each salt, we need to take the product of the concentrations of the products, each raise to the power of their stoichiometric constant and also multiply concentration by their respective stoichiometric coefficient.
Now, coming to our given salts ${M_2}X$, $Q{Y_2}$ and $P{Z_3}$. It is given that they have the same solubilities, so let it be $S$. Now their ${K_{sp}}$ values will be as follows:
Dissolution of salt ${M_2}X$:
${M_2}X \rightleftharpoons 2M + X$
${K_{sp}}({M_2}X) = {(2S)^2}S = 4{S^3}$
Dissolution of salt $Q{Y_2}$:
$Q{Y_2} \rightleftharpoons Q + 2Y$
Therefore, ${K_{sp}}(Q{Y_2}) = S{(2S)^2} = 4{S^3}$
Dissolution of salt $P{Z_3}$:
$P{Z_3} \rightleftharpoons P + 3Z$
${K_{sp}}(P{Z_3}) = S{(3S)^3} = 27{S^4}$
Now, find the relationship between above calculated solubility products.
As you can see value of ${K_{sp}}({M_2}X) = {K_{sp}}(Q{Y_2})$ and value of ${K_{sp}}(P{Z_3})$ will be lesser than ${K_{sp}}({M_2}X) = {K_{sp}}(Q{Y_2})$ because a condition is given in the question that, S < 1.
Therefore, the relation between the solubility products is as follows:
${K_{sp}}({M_2}X) = {K_{sp}}(Q{Y_2}) > {K_{sp}}(P{Z_3})$
Thus, option C is the correct answer.
Note: It should be noted that solids are not included while calculating solubility product values. Solubility product (${K_{sp}}$) represents the maximum extent to which a solid can be dissolved in a solution. Higher the value of ${K_{sp}}$ for a substance, more is its solubility. Thus, we can say that salts ${M_2}X$ and $Q{Y_2}$ are more soluble than salt $P{Z_3}$.
Complete step by step solution:
Consider the general reaction: $aA \rightleftharpoons cC + dD$
Now, if we solve ${K_{sp}}$ for the above reaction, it will be as follows:
${K_{sp}} = {[cC]^c}{[dD]^d}$
Therefore, to solve ${K_{sp}}$ of each salt, we need to take the product of the concentrations of the products, each raise to the power of their stoichiometric constant and also multiply concentration by their respective stoichiometric coefficient.
Now, coming to our given salts ${M_2}X$, $Q{Y_2}$ and $P{Z_3}$. It is given that they have the same solubilities, so let it be $S$. Now their ${K_{sp}}$ values will be as follows:
Dissolution of salt ${M_2}X$:
${M_2}X \rightleftharpoons 2M + X$
${K_{sp}}({M_2}X) = {(2S)^2}S = 4{S^3}$
Dissolution of salt $Q{Y_2}$:
$Q{Y_2} \rightleftharpoons Q + 2Y$
Therefore, ${K_{sp}}(Q{Y_2}) = S{(2S)^2} = 4{S^3}$
Dissolution of salt $P{Z_3}$:
$P{Z_3} \rightleftharpoons P + 3Z$
${K_{sp}}(P{Z_3}) = S{(3S)^3} = 27{S^4}$
Now, find the relationship between above calculated solubility products.
As you can see value of ${K_{sp}}({M_2}X) = {K_{sp}}(Q{Y_2})$ and value of ${K_{sp}}(P{Z_3})$ will be lesser than ${K_{sp}}({M_2}X) = {K_{sp}}(Q{Y_2})$ because a condition is given in the question that, S < 1.
Therefore, the relation between the solubility products is as follows:
${K_{sp}}({M_2}X) = {K_{sp}}(Q{Y_2}) > {K_{sp}}(P{Z_3})$
Thus, option C is the correct answer.
Note: It should be noted that solids are not included while calculating solubility product values. Solubility product (${K_{sp}}$) represents the maximum extent to which a solid can be dissolved in a solution. Higher the value of ${K_{sp}}$ for a substance, more is its solubility. Thus, we can say that salts ${M_2}X$ and $Q{Y_2}$ are more soluble than salt $P{Z_3}$.
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