What will be the solubility product of $AX_{3}$?

Question

Accepted Answer

To determine the solubility product of $AX_{3}$, we need to consider the dissociation of the compound in water. The compound $AX_{3}$ dissociates as follows:

$$ AX_{3} \rightleftharpoons A^{3+} + 3X^{-} $$

Let's assume the solubility of $AX_{3}$ in water is $s$ mol/L. This means that at equilibrium, the concentration of $A^{3+}$ ions will be $s$ mol/L, and the concentration of $X^{-}$ ions will be $3s$ mol/L, because for every mole of $AX_{3}$ that dissolves, one mole of $A^{3+}$ and three moles of $X^{-}$ are produced.

The solubility product constant, $K_{sp}$, is given by the expression:

$$ K_{sp} = [A^{3+}][X^{-}]^3 $$

Substituting the equilibrium concentrations, we get:

$$ K_{sp} = (s)(3s)^3 $$

$$ K_{sp} = s \cdot 27s^3 $$

$$ K_{sp} = 27s^4 $$

Therefore, the solubility product of $AX_{3}$ is $27s^4$, which corresponds to Option 1.

Now, let's explain why the other options are incorrect:

- **Option 2: $4s^3$**: This option is incorrect because it does not account for the stoichiometry of the dissociation of $AX_{3}$. The concentration of $X^{-}$ ions is $3s$, and when raised to the third power, it results in a factor of $27$, not $4$.

- **Option 3: $36s^4$**: This option is incorrect because it incorrectly calculates the factor from the concentration of $X^{-}$ ions. The correct factor from $(3s)^3$ is $27$, not $36$.

- **Option 4: $9s^3$**: This option is incorrect because it does not correctly apply the stoichiometry of the dissociation. The concentration of $X^{-}$ ions is $3s$, and the correct factor when raised to the third power is $27$, not $9$. Additionally, the power of $s$ should be 4, not 3, due to the stoichiometry of the dissociation.

In conclusion, the correct solubility product for $AX_{3}$ is $27s^4$, which is why Option 1 is the correct answer.