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