Problem 96 - Chemistry: An Atoms-Focused Approach

Question

If only \(0.160 \mathrm{g} \mathrm{Ca}(\mathrm{OH})_{2}\) dissolves in \(0.100 \mathrm{L}\) of water, what is the \(K_{\mathrm{sp}}\) value for calcium hydroxide at that temperature?

Step-by-Step Solution

Answer

The \(K_{sp}\) value for calcium hydroxide is \(4.08 \times 10^{-6}\).

Step 1: Calculate the molarity of \(\mathrm{Ca}(\mathrm{OH})_{2}\) in the solution

First, we need to convert the mass of calcium hydroxide to moles: $$ \text{moles of}\ \mathrm{Ca}(\mathrm{OH})_{2} = \frac{\text{mass of}\ \mathrm{Ca}(\mathrm{OH})_{2}}{\text{molar mass of}\ \mathrm{Ca}(\mathrm{OH})_{2}} $$ The molar mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(40.08 (\mathrm{Ca}) + 2(15.999 (\mathrm{O}) + 1.00784 (\mathrm{H})) = 74.093 \mathrm{g/mol}\). Thus, $$ \text{moles of}\ \mathrm{Ca}(\mathrm{OH})_{2} = \frac{0.160 \mathrm{g}}{74.093 \mathrm{g/mol}} = 0.00216\ \mathrm{mol} $$ Now, we can find the molarity of calcium hydroxide in the solution: $$ \text{molarity of}\ \mathrm{Ca}(\mathrm{OH})_{2} = \frac{\text{moles of}\ \mathrm{Ca}(\mathrm{OH})_{2}}{\text{volume of}\ \mathrm{solution \ (in\ L)}} $$ $$ \text{molarity of}\ \mathrm{Ca}(\mathrm{OH})_{2} = \frac{0.00216\ \mathrm{mol}}{0.100\ \mathrm{L}} = 0.0216 \mathrm{M} $$

Step 2: Write the balanced dissociation equation for \(\mathrm{Ca}(\mathrm{OH})_{2}\)

For calcium hydroxide to dissolve in water, it needs to dissociate into its constituent ions. The balanced dissociation equation is: $$ \mathrm{Ca}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2\mathrm{OH}^{-}(aq) $$

Step 3: Determine the concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions

From the dissociation equation, we can see that for every mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\) that dissolves, it forms 1 mole of \(\mathrm{Ca}^{2+}\) ions and 2 moles of \(\mathrm{OH}^{-}\) ions. Therefore: $$ [\mathrm{Ca}^{2+}] = 0.0216\ \mathrm{M} \quad \text{and} \quad [\mathrm{OH}^{-}] = 2(0.0216) \mathrm{M} = 0.0432\ \mathrm{M} $$

Step 4: Calculate the \(K_{sp}\) value for \(\mathrm{Ca}(\mathrm{OH})_{2}\)

The \(K_{sp}\) value for a compound is the product of its constituent ions raised to the power of their stoichiometric coefficients in the balanced dissociation equation. For \(\mathrm{Ca}(\mathrm{OH})_{2}\), it can be written as: $$ K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{OH}^{-}]^2 $$ Now, we can substitute the values we found earlier: $$ K_{sp} = (0.0216)(0.0432)^2 = 4.08 \times 10^{-6} $$ Therefore, the \(K_{sp}\) value for calcium hydroxide at that temperature is \(\boldsymbol{4.08 \times 10^{-6}}\).

Key Concepts

Calcium Hydroxide Dissolution

Calcium hydroxide, commonly known as slaked lime, dissolves in water to form calcium ions (\(\mathrm{Ca}^{2+}\)) and hydroxide ions (\(\mathrm{OH}^{-}\)). This process is crucial for understanding its solubility and the conditions under which it dissolves.
A key aspect of dissolution is realizing that not all of the solid dissolves in water; instead, an equilibrium is reached between the undissolved solid and the dissolved ions.

• In a solution, the balanced dissociation equation is:\[\mathrm{Ca}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2\mathrm{OH}^{-}(aq)\]
• This means each molecule of calcium hydroxide produces one calcium ion and two hydroxide ions.

Understanding this equilibrium is vital for calculating the solubility product constant (\(K_{sp}\)), which quantifies how well a compound dissolves under specific conditions.

Molarity Calculation

To determine how much calcium hydroxide has dissolved, we need to calculate the molarity of the solution. Molarity is the concentration of a solution expressed as the number of moles of solute per liter of solution.

• First, convert the mass of calcium hydroxide to moles using its molar mass: \[\text{Moles of } \mathrm{Ca}(\mathrm{OH})_{2} = \frac{0.160\ \mathrm{g}}{74.093\ \mathrm{g/mol}=0.00216\ \mathrm{mol}\]
• Next, calculate the molarity by dividing the moles by the volume of the solution:\[\text{Molarity of } \mathrm{Ca}(\mathrm{OH})_{2} = \frac{0.00216\ \mathrm{mol}}{0.100\ \mathrm{L}} = 0.0216\ \mathrm{M}\]

This molarity tells us how concentrated the solution is and is a key factor in calculating ion concentrations.

Ion Concentration

To find out the ion concentrations in the solution, we rely on the balanced dissociation equation of calcium hydroxide. From the equation, we know:

• For every mole of calcium hydroxide that dissolves, one mole of \(\mathrm{Ca}^{2+}\) ions and two moles of \(\mathrm{OH}^{-}\) ions are formed.
• Given a molarity of \(0.0216\ \mathrm{M}\) for \(\mathrm{Ca}(\mathrm{OH})_{2}\), we calculate:
  • The concentration of \(\mathrm{Ca}^{2+}\) ions: \([\mathrm{Ca}^{2+}] = 0.0216\ \mathrm{M}\).
  • The concentration of \(\mathrm{OH}^{-}\) ions: \([\mathrm{OH}^{-}] = 2 \times 0.0216\ \mathrm{M} = 0.0432\ \mathrm{M}\).

These concentrations are crucial because they are used to calculate the solubility product constant, \(K_{sp}\). The formula for \(K_{sp}\) for calcium hydroxide is:\[K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{OH}^{-}]^2\]Substituting the concentrations into the formula allows us to determine the \(K_{sp}\), providing insights into the solubility characteristics of calcium hydroxide.