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 HCl requires 11.23 \(\mathrm{mL}\) of 0.0983\(M \mathrm{HCl}\) to reach the end point. Calculate \(K_{s p}\) for \(\mathrm{Ca}(\mathrm{OH})_{2}\) . Compare your result with that in Appendix D. Suggest a reason for any differences you find between your value and the one in Appendix D.

Short answer

The calculated Ksp value for Ca(OH)₂, based on the given titration data, is approximately \( 6.7 \times 10^{-6} \). This value should be compared with the one given in Appendix D. Differences between the two values may be due to experimental error, impurities in the chemicals, or temperature variation, as Ksp is dependent on temperature.

Step-by-step solution

Calculate moles of HCl used in titration

To determine the amount of HCl used, we need to use the formula n = M × V, where n is the number of moles, M is the molarity, and V is the volume of the solution. M (HCl) = 0.0983 mol/L V (HCl) = 11.23 mL = 0.01123 L n (HCl) = M × V = 0.0983 mol/L × 0.01123 L = 0.00110103 mol

Use stoichiometry to determine moles of OH⁻ ions neutralized

In the reaction between HCl and Ca(OH)₂: 2 moles of HCl react with 1 mole of Ca(OH)₂. So for every mole of HCl, there will be 0.5 moles of OH⁻ ions consumed. moles of OH⁻ ions neutralized = 0.5 * n (HCl) = 0.5 * 0.00110103 mol = 0.000550515 mol

Calculate the concentration of OH⁻ ions

Since we know the volume of the solution (50.00 mL), we can now determine the concentration of OH⁻ ions in the solution. C (OH⁻) = moles of OH⁻ ions / volume of solution C (OH⁻) = 0.000550515 mol / 0.05000 L = 0.0110103 mol/L

Calculate the concentration of Ca²⁺ ions

The stoichiometry of the dissolution of Ca(OH)₂ is: Ca(OH)₂ → Ca²⁺ + 2 OH⁻ Thus, for each mole of Ca(OH)₂ dissolved, there will be half as many moles of Ca²⁺ ions formed. Using the concentration of OH⁻ ions obtained in Step 3: C (Ca²⁺) = 0.5 × C (OH⁻) = 0.5 × 0.0110103 mol/L = 0.00550515 mol/L

Calculate the Ksp for Ca(OH)₂

Now we can determine the Ksp using the solubility product expression: Ksp = [Ca²⁺] × [OH⁻]² Ksp = (0.00550515 mol/L) × (0.0110103 mol/L)² Ksp = 6.67 × 10⁻⁶ So, the calculated Ksp value for Ca(OH)₂ is 6.67 × 10⁻⁶ (which is approximately \( 6.7 \times 10^{-6} \) ). This should be compared with the value given in Appendix D.

Discuss the differences in the Ksp values

If any differences are found between the calculated value of Ksp and the value given in Appendix D, possible reasons for these differences include: - Experimental error: Measurement inaccuracies and impurities in the chemicals used during the titration may affect the result. - Temperature variation: The Ksp value is dependent on the temperature. If the experiment was not conducted at the same temperature as the Ksp value listed in Appendix D, there would be a difference between the two values.

Key concepts

Titration

Titration is a laboratory technique used to determine the unknown concentration of a substance by reacting it with a solution of known concentration. It involves adding a titrant, which is the solution of known concentration, to a solution containing the unknown concentration until the reaction reaches an endpoint. This endpoint can be detected by a color change, pH change, or a precipitate formation depending on the nature of the reaction.
In the exercise, titration is used to determine how much HCl is required to neutralize the hydroxide ions (OH⁻) in a saturated Ca(OH)₂ solution. By measuring the precise volume of HCl used, we can calculate the moles of OH⁻ (and, by extension, Ca(OH)₂) present in the original solution. This process lays the foundation for finding the solubility product constant, Ksp, of Ca(OH)₂.

Stoichiometry

Stoichiometry is the aspect of chemistry that deals with the quantitative relationships or ratios between reactants and products in chemical reactions. By utilizing the balanced chemical equations, stoichiometry allows chemists to predict the amounts of substances consumed and produced in a given reaction.
In our titration scenario, stoichiometry is critical for interpreting the reaction between hydrochloric acid (HCl) and calcium hydroxide (Ca(OH)₂). The balanced equation tells us that two moles of HCl react with one mole of Ca(OH)₂to form calcium chloride (CaCl₂) and water (H₂O). Using stoichiometry, we can deduce that each mole of HCl will neutralize half a mole of hydroxide ions (OH⁻).

Molarity

Molarity, denoted by M, is a unit of concentration that measures the number of moles of solute per liter of solution. It is a fundamental concept in chemistry for describing the strength of a solution. Knowing the molarity of a solution is essential in titrations, as it allows for precise calculations of reactants' amounts.
For the given exercise, molarity comes into play by helping us calculate the number of moles of hydrochloric acid used (HCl). The molarity of HCl (0.0983 mol/L) and the volume used (11.23 mL) give us the moles of HCl through the formula moles = molarity × volume. This information is then utilized to perform stoichiometric calculations.

Solubility Equilibrium

Solubility equilibrium refers to the dynamic balance between a solute dissolving and precipitating in a solvent. At this equilibrium, the solution becomes saturated and has a constant concentration of the solute under stable conditions. The solubility product constant (Ksp) is a value that represents this equilibrium and depends on the temperature and pressure of the system.
In the textbook exercise, after determining the concentration of OH⁻ and Ca²⁺ ions, we apply the Ksp expression to find the solubility product constant of calcium hydroxide (Ca(OH)₂): Ksp = [Ca²⁺] × [OH⁻]². The calculated value helps portray the solubility of Ca(OH)₂t in water and can be used to understand the precipitate formation in aqueous solutions involving calcium hydroxide.