Question

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when 25.0 ml of 0.100 M Co(NO3)2 is mixed with 45.0 mL 0.100 M KOH(aq), show by calculation whether a precipitate of Co(OH)2(s) will for. Ksp of Co(OH)2 is 1.3 * 10^-15

Answer 1

Co(NO₃)₂ + 2KOH  $\to$ Co(OH)₂ + 2KNO₃

mole ratio of Co(NO₃)₂ and KOH = 1:2

mmoles of Co(NO₃)₂ = volume × molarity

= 25×0.100 = 2.5

mmoles of KOH = 45×0.100 = 4.5

Now, moles of Co(NO₃)₂ reacted = ( mmoles of KOH /2) = (4.5/2) = 2.25 .

mmoles of Co(OH)₂ formed = mmoles of Co(NO₃)₂ reacted = 2.25.

Total volume of solution = (25+45) = 70 mL

Concentration of Co(OH)₂

= (mmoles of Co(OH)₂/ total volume)

= (2.25/70)

= 0.032 M.

Ksp of Co(OH)₂ = 1.3×10^-15

Let solubility of Co(OH)₂ = S ( mol/L or M)

Co(OH)₂ $\to$ Co²⁺ + 2OH^-

then, Ksp = [Co²⁺] [OH^-]²

Or, Ksp = S × (2S)²

So, Ksp = 4S³

Or, 1.3×10^-15 = 4S³

Or, s³ = (1.3×10^-15/4)

Or, s³ = 3.25×10^-16

Or, S = (3.25×10^-16)^1/3

Or, S = 6.875×10^-6 M

As, concentration of Co(OH)₂ in the solution is greater than the solubility of Co(OH)₂ , so Co(OH)₂ will form precipitation.