Problem 115 - Chemistry: The Central Science

Question

Excess \(\mathrm{Ca}(\mathrm{OH})_{2}\) is shaken with water to produce a saturated solution. The solution is filtered, and a \(50.00-\mathrm{mL}\) sample titrated with \(\mathrm{HCl}\) requires \(11.23 \mathrm{~mL}\) of \(0.0983 \mathrm{M} \mathrm{HQ}\) to reach the end point. Calculate \(K_{\mathrm{wp}}\) for \(\mathrm{Ca}(\mathrm{OH})_{2}\). Compare your result with that in Appendix D. \(25^{\circ} \mathrm{C}\). Suggest a reason for any differences you find between your value and the one in Appendix D.

Step-by-Step Solution

Answer

The solubility product constant (\(K_{\mathrm{wp}}\)) for \(\mathrm{Ca}(\mathrm{OH})_{2}\) has been calculated to be \(5.53 \times 10^{-6}\). This result can be compared to the value given in Appendix D. Differences between the two values might be due to experimental error, temperature, or pressure variations.

Step 1: Determine moles of \(\mathrm{HCl}\) used

The titration required \(11.23 \mathrm{~mL}\) of \(0.0983 \mathrm{M}\) \(\mathrm{HCl}\) to reach the end point. To find the moles of \(\mathrm{HCl}\), we use the formula: Moles of \(\mathrm{HCl}\) = Molarity × Volume Since the volume is in mL, we need to convert it to L first: Volume in L = \(\frac{11.23 \mathrm{~mL}}{1000} = 1.123 \times 10^{-2} \mathrm{~L}\) Now, we can find the moles: Moles of \(\mathrm{HCl}\) = \(\left(0.0983 \frac{\text{mol}}{\text{L}}\right) \times \left(1.123 \times 10^{-2} \mathrm{~L}\right) = 1.1037 \times 10^{-3}\text{mol}\)

Step 2: Calculate moles of \(\mathrm{OH}^{-}\) in the saturated sample

The balanced chemical equation for the titration is: \(\mathrm{2HCl} + \mathrm{Ca}(\mathrm{OH})_{2} \longrightarrow \mathrm{CaCl}_2 + 2\mathrm{H}_2\mathrm{O}\) Here, we see that 2 moles of \(\mathrm{HCl}\) react with 1 mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\), releasing 2 moles of \(\mathrm{OH}^{-}\) ions: \(\mathrm{2HCl} + \mathrm{OH}^{-} + \mathrm{OH}^{-} \longrightarrow \mathrm{CaCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{H}_2\mathrm{O}\) So, the moles of \(\mathrm{OH}^-\) in the saturated sample will be equal to the moles of \(\mathrm{HCl}\): Moles of \(\mathrm{OH}^{-}\) = \(1.1037 \times 10^{-3}\text{mol}\)

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

The total volume of the sample is \(50.00 \mathrm{~mL}\). To find the concentration of \(\mathrm{OH}^-\) ions, we use the formula: \(\left[\mathrm{OH}^{-}\right]\) = \(\frac{\text{Moles}}{\text{Volume}}\) First, convert the volume to L: Volume in L = \(\frac{50.00 \mathrm{~mL}}{1000} = 5.000 \times 10^{-2} \mathrm{~L}\) Then, calculate the concentration of \(\mathrm{OH}^{-}\) ions: \(\left[\mathrm{OH}^{-}\right]\) = \(\frac{1.1037 \times 10^{-3}\text{mol}}{5.000 \times 10^{-2}\mathrm{~L}} = 0.022074 \frac{\text{mol}}{\text{L}}\) The dissociation of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is: \(\mathrm{Ca}(\mathrm{OH})_{2} \longrightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^{-}\) From the balanced chemical equation, we know that 1 mole of \(\mathrm{Ca}^{2+}\) ions is released for every 2 moles of \(\mathrm{OH}^{-}\). So, the concentration of \(\mathrm{Ca}^{2+}\) ions will be half the concentration of \(\mathrm{OH}^{-}\) ions: \(\left[\mathrm{Ca}^{2+}\right]\) = \(\frac{\left[\mathrm{OH}^{-}\right]}{2} = \frac{0.022074 \frac{\text{mol}}{\text{L}}}{2} = 0.011037 \frac{\text{mol}}{\text{L}}\)

Step 4: Calculate the solubility product constant \(K_{\mathrm{wp}}\)

The solubility product constant for \(\mathrm{Ca}(\mathrm{OH})_{2}\) is given by: \(K_{\mathrm{wp}} = \left[\mathrm{Ca}^{2+}\right] \times \left[\mathrm{OH}^- \right]^2\) Now, we can plug in the concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^-\) ions: \(K_{\mathrm{wp}} = \left(0.011037 \frac{\text{mol}}{\text{L}}\right) \times \left(0.022074 \frac{\text{mol}}{\text{L}}\right)^2 = 5.53 \times 10^{-6}\) The solubility product constant \(K_{\mathrm{wp}}\) for \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(5.53 \times 10^{-6}\). This result can be compared to the value given in Appendix D. Any differences between the two values might be attributed to experimental error, as well as variations in temperature or pressure.

Key Concepts

Calcium Hydroxide

Calcium hydroxide, also known as slaked lime, is a chemical compound with the formula \( \mathrm{Ca(OH)}_2 \). It is formed when calcium oxide reacts with water. In an aqueous solution, it dissociates to form calcium ions \( \mathrm{Ca^{2+}} \) and hydroxide ions \( \mathrm{OH^-} \).

Calcium hydroxide is slightly soluble in water, and its solubility decreases with an increase in temperature. This property can affect the results of experiments where temperature variations are present. In a saturated solution of calcium hydroxide, there is a dynamic equilibrium between the solid and its ions in the solution.

In laboratory settings, - **Calcium Hydroxide** is commonly used in titration processes to determine concentration levels of acidic solutions.- It reacts with acids, such as hydrochloric acid, during titration and assists in measuring unknown quantities based on reactivity. Its solubility equilibrium is crucial for these calculations.

Understanding the solubility and behavior of calcium hydroxide in water is essential for titration calculations and interpreting chemical equilibria.

Titration Calculations

Titration calculations are vital in chemical experiments to determine unknown concentrations. The specific example here involves determining the solubility product constant \( K_{sp} \) for calcium hydroxide using a titration method. In this process, a strong acid such as hydrochloric acid \( \mathrm{HCl} \) is titrated against the saturated solution of calcium hydroxide.

• Moles of the titrant (\( \mathrm{HCl} \)) are calculated first using its volume and molarity.
• For the reaction \( \text{2HCl} + \mathrm{Ca(OH)}_2 \rightleftharpoons \mathrm{CaCl}_2 + 2\mathrm{H}_2O \), the stoichiometry suggests two moles of \( \mathrm{HCl} \) react with one mole of \( \mathrm{Ca(OH)}_2 \).
• Therefore, moles of \( \mathrm{OH^-} \) can be directly inferred from \( \mathrm{HCl} \) concentration.

The following step is calculating the concentration of hydroxide ions \( [\mathrm{OH^-}] \), and subsequently calcium ions \( [\mathrm{Ca^{2+}}] \) using the known volumes of reactants.
Finally, using these concentrations, the solubility product constant \( K_{sp} \) is determined, reflecting the equilibrium state of the ions in solution. Understanding titration calculations will help in accurately measuring different solubility constants which are pivotal in chemistry.

Chemical Equilibrium

Chemical equilibrium pertains to the state in which the concentrations of reactants and products no longer change with time. For calcium hydroxide, this equilibrium is established in a saturated solution where the solid phase is in balance with its dissolved ions.

The dissociation of calcium hydroxide can be represented as:
\( \mathrm{Ca(OH)}_{2} \rightleftharpoons \mathrm{Ca}^{2+} + 2 \mathrm{OH^-} \)
At equilibrium, the rate of dissolution of calcium hydroxide equals the rate of precipitation from the solution.

The equilibrium concept is applied during titration to calculate the solubility constant \( K_{sp} \), which is specific to the ion concentration in a saturated solution.

• \( K_{sp} \) helps in understanding the solubility behavior of sparingly soluble salts like \( \mathrm{Ca(OH)}_{2} \).
• It shows how the concentration of ions is distributed when equilibrium is reached.

When conducting such experiments, temperature and pressure must be constant as they can shift the equilibrium position. Analyzing chemical equilibrium effectively allows prediction of solubility changes and testing assumptions in experiments against standard values found in tables, like in Appendix D.