Physical and chemical equilibria can respond to a change in their pressure, temperature, and concentration of reactants and products. To describe the change in the equilibrium, we have a principle named Le Chatelier's principle. This we can define in terms of energy, as the free energy change in equilibrium is zero means the system is stable. So if we are doing some changes in equilibrium, then the system having a tendency to reestablish the equilibrium by undoing the effect we brought. Consider the following equilibrium.
Three sparingly soluble salts `A_(2)B, AB`, and `AB_(3)` are given. If all the three having the same value of solubility products `(K_(sp))`, in the saturated solution, the correct order of their solubilites is

Options

A. `AB_(3) gt AB gt A_(2)B`
B. `AB_(3) gt A_(2)B gt AB`
C. `AB gt AB_(3) gt A_(2)B`
D. `AB gt A_(2)B gt AB_(3)`

Explanation

To solve the problem regarding the solubility of the salts \(A_2B\), \(AB\), and \(AB_3\) given that they all have the same solubility product (\(K_{sp}\)), we will analyze the dissociation of each salt and derive their solubility expressions.

### Step-by-Step Solution:

1. **Dissociation of Salt \(AB\)**:
- The dissociation reaction is:
\[ AB(s) \rightleftharpoons A^+(aq) + B^-(aq) \]
- If \(s\) is the solubility of \(AB\), then at equilibrium:
- \([A^+] = s\)
- \([B^-] = s\)
- The solubility product (\(K_{sp}\)) is given by:
\[ K_{sp} = [A^+][B^-] = s \cdot s = s^2 \]
- Therefore, the solubility \(s\) can be expressed as:
\[ s = \sqrt{K_{sp}} \]

2. **Dissociation of Salt \(A_2B\)**:
- The dissociation reaction is:
\[ A_2B(s) \rightleftharpoons 2A^+(aq) + B^{2-}(aq) \]
- If \(s\) is the solubility of \(A_2B\), then at equilibrium:
- \([A^+] = 2s\)
- \([B^{2-}] = s\)
- The solubility product (\(K_{sp}\)) is given by:
\[ K_{sp} = [A^+]^2[B^{2-}] = (2s)^2 \cdot s = 4s^3 \]
- Therefore, the solubility \(s\) can be expressed as:
\[ s = \left(\frac{K_{sp}}{4}\right)^{1/3} \]

3. **Dissociation of Salt \(AB_3\)**:
- The dissociation reaction is:
\[ AB_3(s) \rightleftharpoons A^{3+}(aq) + 3B^-(aq) \]
- If \(s\) is the solubility of \(AB_3\), then at equilibrium:
- \([A^{3+}] = s\)
- \([B^-] = 3s\)
- The solubility product (\(K_{sp}\)) is given by:
\[ K_{sp} = [A^{3+}][B^-]^3 = s \cdot (3s)^3 = 27s^4 \]
- Therefore, the solubility \(s\) can be expressed as:
\[ s = \left(\frac{K_{sp}}{27}\right)^{1/4} \]

4. **Comparing Solubilities**:
- Now we have the expressions for solubility:
- For \(AB\): \(s_{AB} = \sqrt{K_{sp}}\)
- For \(A_2B\): \(s_{A_2B} = \left(\frac{K_{sp}}{4}\right)^{1/3}\)
- For \(AB_3\): \(s_{AB_3} = \left(\frac{K_{sp}}{27}\right)^{1/4}\)
- To determine the order of solubility, we need to compare these values.
- Since \(K_{sp}\) is constant for all three salts, we can compare the coefficients:
- \(\sqrt{K_{sp}} > \left(\frac{K_{sp}}{4}\right)^{1/3} > \left(\frac{K_{sp}}{27}\right)^{1/4}\)

5. **Conclusion**:
- The correct order of solubility from highest to lowest is:
\[ AB > A_2B > AB_3 \]
- Therefore, the correct answer is option 4: \(AB > A_2B > AB_3\).