Manganese(II) sulfide is one of the compounds found in the nodules on the ocean floor that may eventually be a primary source of many transition metals. The solubility of MnS is \(4.7 \times 10^{-4} \mathrm{~g} / 100 . \mathrm{mL}\) solution. Estimate the \(K_{\mathrm{sp}}\) of \(\mathrm{MnS}\)

Solubility refers to how much of a substance (solute) can dissolve in a solvent to form a homogeneous solution. In this example, the solubility of manganese(II) sulfide (MnS) in water is given as \(4.7 \times 10^{-4} \text{ g} / 100 \text{ mL}\). This value indicates the maximum amount of MnS that can dissolve in 100 mL of water.
Understanding solubility is critical as it impacts how substances interact in solution, which is essential for many chemical reactions and processes.

A dissociation equation shows how a compound splits into its respective ions in water. For MnS, the dissociation can be written as:
\[\text{MnS(s)} \rightleftharpoons \text{Mn}^{2+}(aq) + \text{S}^{2-}(aq)\]
This equation depicts MnS breaking down into manganese ions \(\text{Mn}^{2+}\) and sulfide ions \(\text{S}^{2-}\). Writing the dissociation equation is the first step to solving problems related to solubility because it helps us understand the ion speciation in the solution.

Molarity is a measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution. Let's walk through the core steps:
- **Determine Moles**: Convert the given mass of MnS into moles using its molar mass. The molar masses are Mn = 54.94 g/mol and S = 32.06 g/mol, so MnS = 87.00 g/mol:
\(4.7 \times 10^{-4} \text{g}\times \frac{1 \text{mol}}{87.00 \text{ g}} = 5.40 \times 10^{-6} \text{ mol MnS}\)
- **Convert Volume**: 100 mL is equivalent to 0.100 L.
- **Calculate Molarity**: Find the molarity of the dissolved ions; since 1 mole of MnS dissociates into 1 mole of \(\text{Mn}^{2+}\) and 1 mole of \(\text{S}^{2-}\), the molarity is:
\(\frac{5.40 \times 10^{-6} \text{ mol}}{0.100 \text{ L}} = 5.40 \times 10^{-5} \text{ M}\)
This molarity is used to determine the concentration of each ion in the solution.

For a slightly soluble salt like MnS, the equilibrium expression shows the relationship between the concentrations of the ions at equilibrium. The solubility product constant (\(K_{\text{sp}}\)) represents this equilibrium:
\(K_{\text{sp}} = [\text{Mn}^{2+}][\text{S}^{2-}]\)
Using the calculated molarities of the ions from above:
\(K_{\text{sp}} = (5.40 \times 10^{-5})^2 = 2.92 \times 10^{-9}\)
This small value of \(K_{\text{sp}}\) indicates low solubility, typical for compounds like MnS. Establishing the equilibrium expression and solving for \(K_{\text{sp}}\) is key in understanding the solubility behavior of ionic compounds in water.