Question

Need help on this calculation sheet for my pre lab!

A determination of the molar solubility and the K_sp calcium hydroxide was completed according to the experimental procedure. The following data were collected for Trial 1. (See Report Sheet.) Complete determinations. Record the calculated values to the correct number of significant figures. Molar Solubility and Solubility Product of Calcium Hydroxide Calculation Zone

Answer 1

5: Volume of HCl added = final buret reading - initial buret reading = 13.90 - 1.70 = 12.20 mL = 0.0122 L

6: Moles of HCl added = M x V(L) = 0.0480 mol/L x 0.0122 L = 5.856x10^-4 mol

7: HCl + OH- ------ > H2O + Cl-

Hence moles of OH- in saturated solution = moles of HCl reacted = 5.856x10^-4 mol

8: Total volume after neutralization, Vt = 25.00 mL + 12.20 mL = 37.20 mL = 0.0372 L

Hence equilibrium [OH-] = moles of OH- / Vt = 5.856x10^-4 mol / 0.0372 L = 0.015742 mol/L

9: 1 mol of Ca(OH)2 contains 2 mol OH- and 1 mol Ca2+ ion. Hence

[Ca2+] = [OH-] / 2 =  0.015742 M / 2 = 0.007871 mol/L

10: Ca(OH)2 -----> Ca²⁺(aq) + 2OH-(aq)

-------------------------- S ------------- 2S

molar solubility, S = [Ca2+] = 0.007871 mol/L

11: Ksp = [Ca2+] x [OH-]² = 0.007871 x (0.015742)² = 1.95 x 10^-5 (answer)