
If the salts ${M_2}X$, $Q{Y_2}$ and $P{Z_3}$ have the same solubilities, their ${K_{sp}}$ values are related as – (S<1)
A)${K_{sp}}({M_2}X) = {K_{sp}}(Q{Y_2})$
B) ${K_{sp}}({M_2}X) > {K_{sp}}(Q{Y_2}) = {K_{sp}}(P{Z_3})$
C) ${K_{sp}}({M_2}X) = {K_{sp}}(Q{Y_2}) > {K_{sp}}(P{Z_3})$
D) ${K_{sp}}({M_2}X) > {K_{sp}}(Q{Y_2}) > {K_{sp}}(P{Z_3})$
Answer
583.2k+ views
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 $P{Z_3}$.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

