Question

After obtaining 0.546g of CaCl2, 10mL of 0.3M NaOH the pH of the filtrate was 11.61 and the pH of the saturated solution was 11.79. How can I fill an ICE table determining the equilibrium concentrations of Ca2+ and OH for the Ksp of filtrate and saturated solution?
(Also, which method should be most accurate)

Answer 1

mol of CaCl2 = mass/MW = 0.546/110.98 = 0.004919 mol of CaCl2

mol ofNaOH = MV = 0.3*(10*10^-3) = 0.003 mol of NaOH

pH of filtrate = 11.61

pH salutration = 11.79

Ca(OH)2 <-> Ca+2 + 2OH-

Ksp = [Ca+2][OH-]^2

Total volume = 10 mL

we need [OH-] in solution (saturated)

so..

[Ca+2] = mol/V = 0.004919 /(10^-3) = 4.919 M

[OH-] = 10^-pOH = 10^-(14-pH) =  10^-(14-11.79) = 0.006165

so..

initially:

[Ca+2] = 4.919

[OH-] = 0.3

after equilibirum

[Ca+2] = 4.919 - x

[OH-] = 0.3 - x = 0.006165

so

Ksp = [Ca+2][OH-]^2

Ksp = (4.919)(0.006165^2)

Ksp = 1.869575*10^-4