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