Question

What is the cobalt (II) ion concentration in a solution prepared by mixing 453 mL of 0.384 M cobalt (II) nitrate with 411 mL of 0.356 M sodium hydroxide? The Ksp of cobalt (II) hydroxide is 1.3

Answer 1

First calculate how many mmoles of each reactant there are.

mmoles Co(NO3)2 = M Co(NO3)2 x mL Co(NO3)2 = (0.384)(453) = 173.95 mmoles Co(NO3)2
mmoles NaOH = M NaOH x mL NaOH = (0.356)(411) = 146.32 mmoles NaOH

Calculate how many mmoles of each reactant there are after the reaction.



The NaOH is completely used up and you have made 59 mmoles of Co(OH)2. But you also have 63 mmoles of Co2+ ion left over. [Co2+] = mmoles Co2+ / total mL solution = 63 / (453 + 411) = 0.000338 M.

Set up an ICE chart for the Ksp reaction.

Molarity . . . . Co(OH)2(s) ==> Co2+(aq) + 2OH-(aq)
Initial . . . . . . . . . . . . . . . . . . ..0.000338 . . . . . . .0
Change . . . . . . . . . . . . . . . . . . . .+x . . . . . .+2x

Ksp = [Co2+][OH-]^2 = (0.000338+x)(2x)^2 = 1.3 x 10^-15

Because Ksp is so small, we can neglect the -x term to simplify the math.

(0.000338)(2x)^2 = (0.000338)(4x^2) = 0.33x^2 = 1.3 x 10^-15
x = 6.3 x 10^-8 M so [Co2+] = 0.000338 + x = 0.000338 M.
The amount of Co2+ present is what is left over from the initial reaction.