Answer
$1PbCO_3(s) \lt -- \gt 1Pb^{2+}(aq) + 1C{O_3}^{2-}(aq)$
$K_{sp} (PbCO_3) = [Pb^{2+}][C{O_3}^{2-}]$
Work Step by Step
- Lead (II) $(Pb^{2+})$ carbonate $(C{O_3}^{2-})$: $PbCO_3$
1. Write the dissociation equation for this salt:
- Identify the ions of the salt: $(Pb^{2+})$ and $(C{O_3}^{2-})$, these are the products, and the reactant is the solid salt.
$PbCO_3(s) \lt -- \gt Pb^{2+}(aq) + C{O_3}^{2-}(aq)$
- Balance the equation:
$1PbCO_3(s) \lt -- \gt 1Pb^{2+}(aq) + 1C{O_3}^{2-}(aq)$
2. Now, write the $K_{sp}$ expression.
- Multiply the concentrations of the ions;
- The equilibrium coefficients represent the exponent of these concentrations:
$K_{sp} (PbCO_3) = [Pb^{2+}]^1 \times [C{O_3}^{2-}]^1$
