Answer
$1 BaCrO_4(s) \lt -- \gt 1Ba^{2+}(aq) + 1Cr{O_4}^{2-}(aq)$
$K_{sp} (BaCrO_4) = [Ba^{2+}][Cr{O_4}^{2-}]$
Work Step by Step
1. Write the dissociation equation for this salt:
- Identify the ions of the salt: $(Ba^{2+})$ and $(Cr{O_4}^{2-})$, these are the products, and the reactant is the solid salt.
$BaCrO_4(s) \lt -- \gt Ba^{2+}(aq) + Cr{O_4}^{2-}(aq)$
- Balance the equation:
$1 BaCrO_4(s) \lt -- \gt 1Ba^{2+}(aq) + 1Cr{O_4}^{2-}(aq)$
2. Now, write the $K_{sp}$ expression.
- Multiply the concentrations of the ions;
- The equilibrium coefficients represent the exponent of these concentrations:
$K_{sp} (BaCrO_4) = [Ba^{2+}]^1 \times [Cr{O_4}^{2-}]^1$
