Question

If the Ksp for Co(OH)₂ = 5.92 X 10-15, what is the molar solubility of Co(OH)₂ in a buffer solution composed of 1.80M trimethylamine (= (CH₃)₃N) (Kb = 6.5 X 10-5) and 1.00M trimethylaminium hydrochloride (= (CH₃)₃NH⁺Cl^-)?

Answer 1

Let us calculate the pOH of the buffer solution by using Henderson–Hasselbalch equation

pOH = pKb + log [Salt/Base]

         = -log[6.5 x 10^-5) + log [1.0/1.8]

         = 5.187 + (-0.25527)

         = 4.9312

pOH  = -log[OH] = 4.9312

  [OH]   = 10^-4.9312 = 1.171 x 10-5 M

Co(OH)2 = Ca²⁺ + 2 [OH-], that means one mole of Ca2+ = 2 moles of OH-

Hence Ca2+ = 2 x 1.171 x 10-5 M = 2.342 x 10-5 M

Solubility product (Ksp) for Co(OH)2 is 5.92 X 10-15

i.e Ksp = [Ca2+][OH]^2 / [Ca(OH)2] = 5.92 X 10-15

5.92 X 10-15 =   [2.342 x 10-5] [1.171 x 10-5 ]^2 / [Ca(OH)2]

[Ca(OH)2]   = 0.5424 M

Hence the molar solubility of Co(OH)₂ = 0.5424 M