Report Sheet Molar Solubility; Common-Ion Effect Name: Desk No: Date: Lab Sec: Molar Solubility and Solubility Product of Calcium Hydroxide Trial 1 Trial 2 Trial 3 Volume of solid Ca(OH)2 solution (mL): 25.0 25.0 30 Concentration of standardized HCl solution (mol/L): JyY Buret reading: initial (mL): Buret reading: final (mL): Volume of HCl added (mL): 10.40 10.45 10.35 Moles of HCl added (mol): Moles of OH- in Ca(OH)2 solution (mol): [OH-] at equilibrium (mol/L): 0.1080 * 10.40 / 25.00 [Ca2+] at equilibrium (mol/L): Molar solubility of Ca(OH)2 (mol/L): Average molar solubility of Ca(OH)2 (mol/L): Ksp of Ca(OH)2: Average Ksp: Standard deviation of Ksp: Relative standard deviation of Ksp (SRSD): Calculations for Trial:
The Correct Answer and Explanation is:
To complete your Experiment 22 Report Sheet on the Molar Solubility and Solubility Product (Ksp) of Calcium Hydroxide, we need to walk through calculations step-by-step for Trial 1, assuming:
- Volume of Ca(OH)₂ solution = 25.0 mL = 0.02500 L
- Volume of HCl added = 10.40 mL = 0.01040 L
- Concentration of HCl (standardized) = 0.1080 mol/L
We will assume this concentration applies for now since it’s partly shown.
Trial 1 Calculations:
1. Moles of HCl added
mol HCl=MHCl×VHCl=0.1080 mol/L×0.01040 L=1.1232×10−3 mol\text{mol HCl} = M_{\text{HCl}} \times V_{\text{HCl}} = 0.1080 \, \text{mol/L} \times 0.01040 \, \text{L} = 1.1232 \times 10^{-3} \, \text{mol}mol HCl=MHCl×VHCl=0.1080mol/L×0.01040L=1.1232×10−3mol
2. Moles of OH⁻ in Ca(OH)₂ solution
Ca(OH)₂ reacts with HCl as:Ca(OH)2⇌Ca2++2OH−\text{Ca(OH)}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{OH}^-Ca(OH)2⇌Ca2++2OH−HCl+OH−→H2O\text{HCl} + \text{OH}^- \rightarrow \text{H}_2\text{O}HCl+OH−→H2O
Since 1 mol HCl reacts with 1 mol OH⁻:mol OH−=1.1232×10−3 mol\text{mol OH}^- = 1.1232 \times 10^{-3} \, \text{mol}mol OH−=1.1232×10−3mol
3. [OH⁻] at equilibrium
[OH−]=mol OH−volume of Ca(OH)2 solution=1.1232×10−30.02500=0.04493 mol/L[\text{OH}^-] = \frac{\text{mol OH}^-}{\text{volume of Ca(OH)}_2 \text{ solution}} = \frac{1.1232 \times 10^{-3}}{0.02500} = 0.04493 \, \text{mol/L}[OH−]=volume of Ca(OH)2 solutionmol OH−=0.025001.1232×10−3=0.04493mol/L
4. [Ca²⁺] at equilibrium
From stoichiometry:
Each 1 mol of Ca(OH)₂ releases 2 mol OH⁻ and 1 mol Ca²⁺[Ca2+]=[OH−]2=0.044932=0.02247 mol/L[\text{Ca}^{2+}] = \frac{[\text{OH}^-]}{2} = \frac{0.04493}{2} = 0.02247 \, \text{mol/L}[Ca2+]=2[OH−]=20.04493=0.02247mol/L
5. Molar Solubility of Ca(OH)₂
Molar solubility=[Ca2+]=0.02247 mol/L\text{Molar solubility} = [\text{Ca}^{2+}] = 0.02247 \, \text{mol/L}Molar solubility=[Ca2+]=0.02247mol/L
6. Ksp of Ca(OH)₂
Ksp=[Ca2+][OH−]2=(0.02247)(0.04493)2=(0.02247)(0.002018)=4.534×10−5K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 = (0.02247)(0.04493)^2 = (0.02247)(0.002018) = 4.534 \times 10^{-5}Ksp=[Ca2+][OH−]2=(0.02247)(0.04493)2=(0.02247)(0.002018)=4.534×10−5
Explanation
This experiment determines the molar solubility and solubility product constant (Ksp) of calcium hydroxide, Ca(OH)₂, using a titration approach with hydrochloric acid (HCl). When Ca(OH)₂ dissolves in water, it dissociates into calcium (Ca²⁺) and hydroxide (OH⁻) ions. The solubility equilibrium is expressed as:Ca(OH)2(s)⇌Ca2+(aq)+2OH−(aq)\text{Ca(OH)}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{OH}^- (aq)Ca(OH)2(s)⇌Ca2+(aq)+2OH−(aq)
A known volume of the saturated Ca(OH)₂ solution is titrated with a standardized HCl solution. Because each mole of HCl neutralizes one mole of OH⁻, the amount of acid used tells us the total number of moles of OH⁻ in the solution.
In Trial 1, we added 10.40 mL of 0.1080 M HCl, introducing 1.1232 × 10⁻³ moles of HCl, and thus neutralizing the same amount of OH⁻. Dividing by the volume of the Ca(OH)₂ solution gives the [OH⁻] concentration. The [Ca²⁺] is then calculated by dividing [OH⁻] by 2, because each Ca²⁺ corresponds to 2 OH⁻ ions.
This [Ca²⁺] is taken as the molar solubility, since it represents how much Ca(OH)₂ dissociated. Finally, the Ksp is calculated using the equilibrium concentrations in the formula:Ksp=[Ca2+][OH−]2K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2Ksp=[Ca2+][OH−]2
This approach allows for determining the solubility equilibrium of sparingly soluble salts. By performing multiple trials and averaging the Ksp, the reliability of the result increases. Calculating the standard deviation and relative standard deviation (SRSD) further evaluates the precision of the experiment.
