AIIMS 2000 Chemistry Solubility Product Constant (Ksp) MCQ Question

The solubility product (Ksp) of a salt is expressed in terms of the concentrations of its ions raised to the power of their stoichiometric coefficients. For barium phosphate, the dissociation can be represented as:

$\text{Ba}_3\text{(PO}_4\text{)}_2 (s) \rightleftharpoons 3\text{Ba}^{2+} (aq) + 2\text{PO}_4^{3-} (aq)$

The solubility product expression is given by:

$K_{sp} = [\text{Ba}^{2+}]^3 [\text{PO}_4^{3-}]^2$

If we let the solubility of barium phosphate be 's' mol/dm³, then:

• The concentration of $\text{Ba}^{2+}$ will be $3s$ mol/dm³.
• The concentration of $\text{PO}_4^{3-}$ will be $2s$ mol/dm³.

Substituting these values into the Ksp expression gives:

$K_{sp} = (3s)^3 (2s)^2 = 27s^3 \cdot 4s^2 = 108s^5$

Thus, the unit of Ksp is:

$\text{mol}^5 \text{dm}^{-15}$

This means the correct answer is D) mol⁵ dm⁻¹⁵.

The other options are incorrect because they do not match the derived units from the Ksp calculation for barium phosphate.