How many grams of Milk of Magnesia, Mg(OH)2(s)(58.3 g / mol) , would be soluble in 200 mL of wat

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How many grams of Milk of Magnesia, Mg(OH)2(s)(58.3 g / mol)...

How many grams of Milk of Magnesia, $\mathrm{Mg}(\mathrm{OH})_{2}(s)(58.3 \mathrm{g} / \mathrm{mol})$ , would be soluble in 200 $\mathrm{mL}$ of water. $K_{\mathrm{sp}}=$ $7.1 \times 10^{-12} .$ Include the ionic reaction and the expression for $K_{\mathrm{sp}}$ in your answer. $\left(K_{\mathrm{w}}=1 \times 10^{-14}=\right.$ $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right] )$

Key Concepts

Stoichiometric Relationships in Solubility Equilibria

The stoichiometry of the dissolution reaction determines the factor by which the molar solubility contributes to the concentration of each ion in solution. For example, if one mole of a salt produces multiple moles of one ion, this must be accounted for in the Ksp expression. This concept allows the proper connection between the solubility (s) and the ion concentrations in the equilibrium expression.

Solubility Product Constant (Ksp)

Ksp is the equilibrium constant for the dissolution of a sparingly soluble ionic compound. It quantifies the product of the concentrations of the dissociated ions, each raised to the power of its coefficient in the balanced dissolution equation. Understanding Ksp is crucial in predicting whether a precipitate will form under given conditions and in calculating the solubility of compounds in solutions.

Ionic Equilibrium and Dissolution Reaction

When a sparingly soluble solid dissolves in water, it establishes an equilibrium between the undissolved solid and its constituent ions in solution. Writing the balanced ionic equation allows one to identify the stoichiometric relationship among the ions, which is key when setting up the expression for Ksp and calculating the solubility.

Molar Solubility

Molar solubility is defined as the number of moles of a substance that can dissolve per liter of solution to reach a saturated state. It is directly tied to the Ksp value through the stoichiometric coefficients of the dissolution reaction. Determining the molar solubility provides the basis for finding the concentration of each ion in equilibrium.

Conversion of Moles to Grams

After calculating the molar solubility, it is necessary to convert moles of the compound to grams using the compound’s molar mass. This step bridges the gap between the theoretical concentration obtained from equilibrium calculations and the practical, measurable mass of the dissolved compound in a given volume of solvent.

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