Step 1: Understanding the Concept:
For a binary sparingly soluble salt like \(\text{NiS}\), the solubility product (\(K_{sp}\)) is the product of the concentrations of its component ions at equilibrium. Step 2: Key Formula or Approach:
For \(\text{NiS(s)} \rightleftharpoons \text{Ni}^{2+}(\text{aq}) + \text{S}^{2-}(\text{aq})\):
\[ K_{sp} = [\text{Ni}^{2+}][\text{S}^{2-}] = S \times S = S^2 \]
Therefore, solubility \(S = \sqrt{K_{sp}}\). Step 3: Detailed Explanation:
1. Given \(K_{sp} = 4.9 \times 10^{-5}\).
2. \(S = \sqrt{4.9 \times 10^{-5}} = \sqrt{49 \times 10^{-6}}\).
3. \(S = 7.0 \times 10^{-3} \text{ mol/dm}^3\). Step 4: Final Answer:
The solubility is \(7.0 \times 10^{-3} \text{ moldm}^{-3}\).