Problem 115 - Chemistry: The Central Science in SI Units, Global Edition

Question

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

Step-by-Step Solution

Answer

The calculated solubility product, \(K_{sp}\), for \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(5.35 \times 10^{-6}\). Differences between this value and the one provided in Appendix D might be due to experimental errors or variations in temperature, as solubility products are temperature-dependent.

Step 1: Determine the concentration of hydroxide ions from titration

The titration tells us that 50.00 mL of the saturated calcium hydroxide solution was neutralized by 11.23 mL of 0.0983 M HCl. We can use the volume and concentration of HCl to find the number of moles of HCl and then the number of moles of hydroxide ions. Moles of HCl = Volume x Concentration = (11.23 mL) x (0.0983 mol/L) Convert the mL to L: Moles of HCl = (0.01123 L) x (0.0983 mol/L) = 0.00110 mol Since the reaction between HCl and OH- is 1:1, the moles of HCl are equal to the moles of OH- ions. So, moles of OH- = 0.00110 mol

Step 2: Calculate the concentration of hydroxide ions in the saturated solution

To find the concentration of hydroxide ions, we will divide the number of moles by the volume of the solution. Concentration of OH- = moles of OH- / volume of saturated solution = 0.00110 mol / 0.05000 L = 0.0220 mol/L

Step 3: Calculate the concentration of calcium ions

In \(\mathrm{Ca}(\mathrm{OH})_{2}\), for every one calcium ion, there are two hydroxide ions. Therefore, the concentration of calcium ions will be half that of hydroxide ions. Concentration of Ca²⁺ = 1/2 x Concentration of OH- = 1/2 x 0.0220 mol/L = 0.0110 mol/L

Step 4: Calculate the solubility product, \(K_{sp}\)

The \(K_{sp}\) for \(\mathrm{Ca}(\mathrm{OH})_{2}\) is given by the equation: \(K_{sp} = [\mathrm{Ca}^{2+}] [\mathrm{OH}^-]^2\) We can plug in the concentrations we calculated in steps 3 and 2, respectively: \(K_{sp} = (0.0110~\text{mol/L}) (0.0220~\text{mol/L})^2 = 5.35 \times 10^{-6}\)

Step 5: Compare the calculated value with Appendix D and analyze differences

Now, compare the calculated \(K_{sp}\) value with the one given in Appendix D. If the values are different, it could be due to experimental errors, such as inaccuracies in measuring the volume of the saturated solution or the concentration of HCl, or due to variations in temperature. Keep in mind that solubility products are temperature-dependent, and the value in Appendix D is likely to be reported for a standard temperature (such as 25°C), which may not be the same as the temperature at which the experiment was conducted.

Key Concepts

Calcium Hydroxide

Calcium Hydroxide, often referred to as slaked lime, is a chemical compound with the formula \( \text{Ca(OH)}_2 \). It appears as a white powder or in crystal form. When mixed with water, it forms a solution that can become saturated depending on the amount added. This is commonly called lime water. Calcium Hydroxide is slightly soluble in water, meaning only a minimal amount can dissolve to form a saturated solution.

In the context of chemistry, particularly in solubility equilibrium, the saturation point is vital because it determines the maximum concentration of dissolved ions in the solution. These ions are what we measure to find the solubility product constant, \( K_{sp} \). When you create a saturated solution of Calcium Hydroxide, it dissociates into \( \text{Ca}^{2+} \) ions and two \( \text{OH}^- \) ions for every formula unit that dissolves. Understanding this relationship is essential for calculating its \( K_{sp} \).

Titration

Titration is a laboratory technique used to determine the concentration of a solution by reacting it with a solution of known concentration. It typically involves adding a titrant from a burette into a known volume of analyte (solution whose concentration is to be measured) until the reaction reaches its end point.

In this exercise, the titration process involves adding hydrochloric acid (HCl) to the saturated solution of Calcium Hydroxide. The \'end point\' is reached when all the \( \text{OH}^- \) ions from Calcium Hydroxide have reacted with \( \text{HCl} \) to form water. This typically requires precise observation, sometimes with the help of an indicator that changes color at the endpoint.

The data obtained from titration, such as the volume of HCl used, allows one to calculate the number of moles of HCl, which in turn equals the moles of \( \text{OH}^- \) ions neutralized in the solution due to their 1:1 reaction ratio. This step is crucial for determining the concentration of hydroxide ions in the saturated solution, which further aids in calculating the \( K_{sp} \).

Hydroxide Ion Concentration

The concentration of hydroxide ions (\( \text{OH}^- \)) in a solution is an important factor, especially when dealing with bases like Calcium Hydroxide. Hydroxide ions are pivotal in determining the basicity of a solution.

When a saturated solution of Calcium Hydroxide is prepared, the concentration of \( \text{OH}^- \) ions can be determined using titration, as seen in the exercise. During titration, the relationship between \( \text{OH}^- \) and \( \text{HCl} \) is direct; one mole of \( \text{HCl} \) reacts with one mole of \( \text{OH}^- \). Using this stoichiometric relationship, one can calculate the concentration of \( \text{OH}^- \) ions by dividing the moles of \( \text{HCl} \) used by the volume of the solution.

Understanding this process aids in the calculation of the solubility product constant (\( K_{sp} \)) for Calcium Hydroxide, as it helps quantify its solubility under certain conditions. This value is crucial for determining whether a solution remains saturated or if additional calcium hydroxide can dissolve.