Problem 86 - Chemistry

Question

Calculate the mass of manganese hydroxide that dissolves to form 1300 mL of a saturated manganese hydroxide solution. For \(\mathrm{Mn}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=2.0 \times 10^{-13}.\)

Step-by-Step Solution

Answer

The mass of manganese hydroxide that dissolves to form 1300 mL of a saturated manganese hydroxide solution is approximately \(4.25 \times 10^{-6}\) grams.

Step 1: Write the balanced chemical equation and the expression for \(K_{sp}\)

Write the balanced chemical equation for the dissolution of manganese hydroxide in water: \[\mathrm{Mn}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Mn}^{2+}(aq) + 2\mathrm{OH}^-(aq)\] Write the expression for the solubility product constant, \(K_{sp}\), for manganese hydroxide: \[K_{sp} = [\mathrm{Mn}^{2+}][\mathrm{OH}^-]^2\]

Step 2: Define the solubility of manganese hydroxide

Let the solubility of \(\mathrm{Mn}(\mathrm{OH})_{2}\) in water be \(x\) moles per liter, then the equilibrium concentrations of the ions will be: - \([\mathrm{Mn}^{2+}] = x\) moles/L - \([\mathrm{OH}^-] = 2x\) moles/L

Step 3: Substitute the solubilities into the \(K_{sp}\) expression and solve for x

Substitute the solubility expressions into the \(K_{sp}\) expression: \[K_{sp} = (x)(2x)^2\] Solve for x: \[2.0 \times 10^{-13} = x(4x^2)\] \[2.0 \times 10^{-13} = 4x^3\] \[x^3 = \frac{2.0 \times 10^{-13}}{4}\] \[x^3 = 5.0 \times 10^{-14}\] \[x = \sqrt[3]{5.0 \times 10^{-14}}\] Calculate x: \[x \approx 3.68 \times 10^{-5}\ \text{moles/L}\] So, the solubility of \(\mathrm{Mn}(\mathrm{OH})_{2}\) in water is approximately \(3.68 \times 10^{-5}\) moles per liter.

Step 4: Calculate the mass of \(\mathrm{Mn}(\mathrm{OH})_{2}\) dissolved in 1300 mL of the solution

The moles of \(\mathrm{Mn}(\mathrm{OH})_{2}\) dissolved in 1300 mL of the solution is given by: \[n_{\mathrm{Mn}(\mathrm{OH})_{2}} = (\text{solubility}) \times (\text{volume})\] \[n_{\mathrm{Mn}(\mathrm{OH})_{2}} = (3.68 \times 10^{-5}\ \text{moles/L}) \times (0.0013\ \text{L})\] \[n_{\mathrm{Mn}(\mathrm{OH})_{2}} \approx 4.78 \times 10^{-8}\ \text{moles}\] Next, convert the moles to mass: \[m_{\mathrm{Mn}(\mathrm{OH})_{2}} = n_{\mathrm{Mn}(\mathrm{OH})_{2}} \times M_{\mathrm{Mn}(\mathrm{OH})_{2}}\] \[m_{\mathrm{Mn}(\mathrm{OH})_{2}} \approx (4.78 \times 10^{-8}\ \text{moles}) \times (88.94\ \text{g/mol})\] \[m_{\mathrm{Mn}(\mathrm{OH})_{2}} \approx 4.25 \times 10^{-6}\ \text{g}\] The mass of manganese hydroxide that dissolves to form 1300 mL of a saturated manganese hydroxide solution is approximately \(4.25 \times 10^{-6}\) grams.

Key Concepts

Saturated Solution

A saturated solution is an important concept in chemistry, especially when discussing solubility. A solution is termed "saturated" when it contains the maximum amount of a solute that can dissolve at a given temperature and pressure. Beyond this point, any additional solute will not dissolve, and a dynamic equilibrium is established.
In a saturated solution, the rate at which the solute dissolves is equal to the rate at which it precipitates out of the solution. This equilibrium ensures that the solution remains saturated and no net change in concentration occurs.

• Dynamic equilibrium: The equilibrium is dynamic because, on a molecular level, dissolution and precipitation are continuously occurring, although the macroscopic concentration of solute remains unchanged.
• Temperature dependence: The point of saturation can change with temperature—for many solutes, solubility increases with temperature.

This concept is crucial in calculating the maximum quantity of a solute like manganese hydroxide that can dissolve in water to form a saturated solution, as seen in the exercise.

Solubility Product Constant (Ksp)

The solubility product constant, denoted as \(K_{sp}\), is a critical parameter in understanding an ionic compound's solubility in water. It is a specific type of equilibrium constant used for saturated solutions of sparingly soluble ionic compounds.
For manganese hydroxide, \( \mathrm{Mn(OH)}_2 \), the dissociation in water and its corresponding \(K_{sp}\) expression can be written as:
\[\mathrm{Mn(OH)}_2(s) \rightleftharpoons \mathrm{Mn}^{2+}(aq) + 2\mathrm{OH}^-(aq)\]
Thus, the \(K_{sp}\) expression is:
\[K_{sp} = [\mathrm{Mn}^{2+}][\mathrm{OH}^-]^2\]This formula helps you calculate the solubility of a compound under equilibrium conditions. Given that \(K_{sp}\) is a constant under any given condition of temperature, it serves as a predictive tool to determine whether more of a solute can be dissolved or if precipitation will occur.

• Simplifying assumptions: Often, in these calculations, the initial concentration of ions from a solid being dissolved is assumed to be zero when establishing \(K_{sp}\) expressions, as the solute is being added to initially pure water.
• Small \(K_{sp}\) values: Compounds with lower \(K_{sp}\) values are less soluble. This explains why manganese hydroxide has a low solubility in water.

Understanding \(K_{sp}\) is essential for solving problems that involve adding solute to water until saturation.

Equilibrium Calculations

Equilibrium calculations are a fundamental part of understanding chemical reactions and solutions. They enable us to predict the concentrations of reactants and products in a solution once dynamic equilibrium is reached.
In the case of a solubility problem, like that of manganese hydroxide, equilibrium calculations start by writing a balanced chemical equation for the dissolution. This sets the stage for using the \(K_{sp}\) value to determine ion concentrations in a saturated solution:

• Moles per liter conversion: Start by defining the solubility of the compound in terms of moles per liter (mol/L). This is crucial because it helps convert later into grams, connecting the microscopic and macroscopic scales of chemistry.
• Solving for solubility: Substitute the defined solubility values into the \(K_{sp}\) expression and solve for \(x\), which represents solubility in mol/L.
• Mass calculation: Finally, convert the molar solubility to mass using the formula mass = moles × molar mass. This gives the practical amount of the solute that can be dissolved in a given volume of solvent.

These steps show how equilibrium calculations are performed and provide insight into the conditions necessary to reach saturation itself.