Problem 107 - 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 \(0.0983 \mathrm{M}\) \(\mathrm{HCl}\) to reach the end point. Calculate \(K_{s p}\) for \(\mathrm{Ca}(\mathrm{OH})_{2}\) Compare your result with that in Appendix D. Do you think the solution was kept at \(25^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

The solubility product constant (\(K_{sp}\)) for \(Ca(OH)_2\) is calculated to be \(5.38 \times 10^{-6}\). This value is close but slightly lower than the \(K_{sp}\) for \(Ca(OH)_2\) at \(25^{\circ} C\) found in Appendix D (\(6.50 \times 10^{-6}\)), suggesting that the solution might not have been kept exactly at \(25^{\circ} C\) or there are some experimental errors. Nevertheless, the obtained value is still reasonably close to the expected value.

Step 1: Calculate the moles of HCl used in titration.

We are given that a \(11.23\: mL\) sample with concentration \(0.0983\: M\) of \(\mathrm{HCl}\) was used to reach the end point. Use the formula: moles = volume × concentration: moles of HCl = \(11.23\: mL \times 0.0983\: M\) = \(0.001104\: moles\)

Step 2: Determine the stoichiometry of reaction between \(Ca(OH)_2\) and HCl.

The balanced chemical equation for the reaction is: \(Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O\) This reaction shows that one mole of \(Ca(OH)_2\) reacts with two moles of HCl.

Step 3: Calculate the moles of \(OH^-\) neutralized by the HCl.

Since one mole of \(Ca(OH)_2\) provides two moles of \(OH^-\) and reacts with two moles of HCl, the number of moles of \(OH^-\) in the \(Ca(OH)_2\) solution is equal to the moles of HCl used: moles of \(OH^-\) = \(0.001104\: moles\)

Step 4: Determine the concentration of \(OH^-\) in the \(Ca(OH)_2\) solution.

Since we started with a \(50.00 \: mL\) sample, to find the concentration of \(OH^-\), we divide the moles by the volume (in liters): \(OH^-\) concentration = \(\frac{0.001104\: moles}{0.050\: L}\) = \(0.02208\: M\)

Step 5: Calculate the concentration of \(Ca^{2+}\) in the \(Ca(OH)_2\) solution.

Since one mole of \(Ca(OH)_2\) provides two moles of \(OH^-\), the concentration of \(Ca^{2+}\) ions in the solution is half the concentration of \(OH^-\) ions: \(Ca^{2+}\) concentration = \(\frac{0.02208}{2} = 0.01104\: M\)

Step 6: Calculate the \(K_{sp}\) for \(Ca(OH)_2\).

The solubility product constant, \(K_{sp}\), is given by the formula: \(K_{sp} = [Ca^{2+}] [OH^-]^2\) where [Ca^{2+}] and [OH^-] are the equilibrium ion concentrations. \(K_{sp} = (0.01104\: M) (0.02208\: M)^2\) = \(5.38 \times 10^{-6}\)

Step 7: Compare the result with the value in Appendix D and deduce if the solution was kept at \(25^{\circ} C\).

The \(K_{sp}\) value for \(Ca(OH)_2\) at \(25^{\circ} C\) in Appendix D is around \(6.50 \times 10^{-6}\) (values may vary slightly depending on the source). The value we obtained is close but slightly lower, which suggests that the solution might not have been kept exactly at \(25^{\circ} C\), or there are some experimental errors. Nevertheless, the obtained value is still reasonably close to the expected value.

Key Concepts

Acid-Base Titration

Acid-base titration is a method used to quantitatively analyze the concentration of an acid or base through a neutralization reaction. In this process, a solution of known concentration, called the titrant, is gradually added to a solution of unknown concentration until the reaction reaches its end point.
The end point is typically indicated by a color change, often facilitated by a pH indicator.

In the exercise given,

• We used hydrochloric acid (HCl) as the titrant to determine the concentration of hydroxide ions in the calcium hydroxide solution.
• The reaction between \( \text{Ca(OH)}_2 \) and \( \text{HCl} \) is: \[ \text{Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O} \]
• The stoichiometry indicates one mole of calcium hydroxide reacts with two moles of hydrochloric acid.

By determining how much HCl is needed to reach the endpoint, we can calculate the original concentration of hydroxide ions, which is crucial for finding the solubility product constant (\( K_{sp} \)).

This technique is widely used due to its accuracy and ability to determine concentrations in a straightforward manner. Understanding titration helps you grasp concepts of molarity and reaction stoichiometry.

Calcium Hydroxide

Calcium hydroxide, \( \text{Ca(OH)}_2 \), is an inorganic compound often used in water treatment and soil stabilization. It is slightly soluble in water, forming a saturated solution where calcium and hydroxide ions are in equilibrium.

In the context of the exercise,

• We focus on the dissolution of \( \text{Ca(OH)}_2 \) in water to produce \( \text{Ca}^{2+} \) and \( \text{OH}^- \) ions. This process can be represented as:
• \[ \text{Ca(OH)}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{OH}^- \]
• The concentration of these ions determines the solubility of \( \text{Ca(OH)}_2 \), which can be further analyzed using its solubility product constant, \( K_{sp} \).

The \( K_{sp} \) calculation involves the equilibrium concentrations of the ions, which are influenced by various factors such as temperature. This means \( K_{sp} \) values can indicate environmental conditions, as suggested by the comparison with Appendix D in the original exercise. Recognizing how slightly soluble compounds behave enhances your understanding of chemical equilibria.

Molar Concentration

Molar concentration, or molarity, is the number of moles of a solute dissolved in one liter of solution. It is expressed as moles per liter (M) and represents the concentration of a substance in a given amount of liquid.

In the exercise, we calculated molar concentrations for both \( \text{OH}^- \) and \( \text{Ca}^{2+} \) ions to find the \( K_{sp} \) value of \( \text{Ca(OH)}_2 \).

To determine molarity:

• Use the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
• Here, moles of \( \text{OH}^- \) are found using the stoichiometry with \( \text{HCl} \).
• Using the volume of the \( \text{Ca(OH)}_2 \) solution, its molarity is established.

Molar concentration handily simplifies the process of quantifying how much of a chemical is present in a solution.
This concept plays a vital role in predicting how substances interact in solutions, which is particularly important in fields such as chemistry, biology, and environmental science.