Experiment 22 Report Sheet Molar solubility. Common-lon Effect Date Lab Sec. Nam

Question

Experiment 22 Report Sheet Molar solubility. Common-lon Effect Date Lab Sec. Name Desk No. A. Molar solubility and solubility Product of Calcium Hydroxide Trial Trial 2 Trial 3 1. volume of saturated Ca(oh), solution ml) 25.0 25.0 2. Concentration of standardized HCI solution (moln. 3. Buret reading, initial (mL) -21 SCen 4. Buret reading, final (mb) 5. Volume of HCl added (mL) 6. Moles of HCI added (mo) 7. Moles of OH in saturated solution (mo) 8. [OH 1, equilibrium (molL) y. [Ca I, equilibrium (mol/L) 10. Molar solubility of Ca(OH)2 (molL) 11. Average molar solubility of Ca(oH): (molL) 12. Kep of Ca(oH. 13. Average K. Analysis, B Analysis, C Standard deviation of K Analysis, D Relative standard deviation of K (%RSD) *Calculations for Trial 1. Experiment 22 26

Explanation / Answer

Concentration of Standard HCl should be Known then only we can proceed with calculation

I am considering Concentration of standard HCl = 0.1 mol/L

Volume of HCl = 21.5 ml = 0.0215

No. of moles of HCl = Concentration * Volume = 0.0215 ml * 0.1 mol/L = 0.00215 moles

No. of HCl will be equal to no. of moles of OH- nuetralized by HCl

No. of moles of OH- = No. of moles of HCl = 0.00215 moles

Volume of Ca(OH)2 = 25 ml = 0.025 L

[OH-] = .00215 moles / 0.025 L = 0.086 mol/L

as the Ca(OH)2 has two moles of OH- for each mole of Ca2+

[Ca2+] = 0.086 mol/L /2 = 0.043 mol/L

Molar solubility of Ca(OH)2 = 0.043 M

Ksp = [OH-]2 [Ca2+] = (0.086)2 (0.043) = 3.18 * 10-4

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