A. Molar solubility and Solubility Product of Calcium Hydroxide 1. Volume of sat

Question

A. Molar solubility and Solubility Product of Calcium Hydroxide 1. Volume of saturated Ca(OH)2 solution (mL) 2. concentration of standardized HCI solution (mol/L) 3. Buret reading, initial (mL) 4. Buret reading, final (mL) 5. Volume of HCI added (mL) 6. Moles of HCI added (mol) 7. Moles of OH^- in saturated solution (mol 8. [OH^-] equilibrium (mol/L) 9. [Ca^2+], equilibrium (mol/L) 10. Molar solubility of Ca(OH)2 (mol/L) 11. Average molar solubility of Ca(OH)2 (mol/L) 12. Ksp of Ca(OH)2 13. Average Ksp 14. Standard deviation of Ksp 15. Relative standard deviation of Ksp(%RSD) Calculations for Trial 1.

Explanation / Answer

Solution:

The volume of saturated Ca(OH)2 solution (mL) is 25.00

Step 1) The chemical equation:

Ca(OH)2 <===> Ca2+ + 2OH¯

Step 2) The Ksp expression:

Ksp = [Ca2+] [OH¯]2

Step 3) Use titration data to determine moles of OH¯ in the 25.0 mL sample (Remember, every one H+ neutralizes one OH¯.):

molarity = moles ÷ volume (in liters)

0.0250 mol/L = x ÷ 0.02250 L

x = 0.0005625 mol

Step 4) Use moles of OH¯ and sample volume to determine [OH¯]:

0.0005625 mol / 0.02500 L = 0.0225 mol/L

Step 5) Determine [Ca2+]:

it is exactly half the [OH¯]. This is because of the 1:2 molar ratio from the balanced equation.

Step 6) Calculate the Ksp for Ca(OH)2

Ksp = (0.01125) (0.0225)2 = 5.70 x 10¯6 mol-3 dm

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