Answer
$1Sr_3(PO_4)_2(s) \lt -- \gt 3Sr^{2+}(aq) + 2P{O_4}^{3-}(aq)$
$K_{sp} (Sr_3(PO_4)_2) = [Sr^{2+}]^3 \times [P{O_4}^{3-}]^2$
Work Step by Step
- Strontium $(Sr^{2+})$ phosphate $(P{O_4}^{3-})$: $Sr_3(PO_4)_2$
1. Write the dissociation equation for this salt:
- Identify the ions of the salt: $(Sr^{2+})$ and $(P{O_4}^{3-})$, these are the products, and the reactant is the solid salt.
$Sr_3(PO_4)_2(s) \lt -- \gt Sr^{2+}(aq) + P{O_4}^{3-}(aq)$
- Balance the equation:
$1Sr_3(PO_4)_2(s) \lt -- \gt 3Sr^{2+}(aq) + 2P{O_4}^{3-}(aq)$
2. Now, write the $K_{sp}$ expression.
- Multiply the concentrations of the ions;
- The equilibrium coefficients represent the exponent of these concentrations:
$K_{sp} (Sr_3(PO_4)_2) = [Sr^{2+}]^3 \times [P{O_4}^{3-}]^2$
