Question

A solution contains 0.0500 M Co3 (aq), 0.0280 M S2–(aq), and 1.85 M NH3. Cobalt(III) ions in aqueous solutions complex with NH3 to produce Co(NH3)63 . (Kf = 5.0 x 1031) What will be the concentration of Co3 (aq) when Co(NH3)63 forms?

[Co3] = M

Will Co2S3 precipitate? (Ksp = 4.0 x 10–21)

yes/no

Answer 1

Co3+ (aq) + 6NH3 ---> [Co(NH3)6]3+ (aq) ,

0.05 moles Co3+ will react with 3 x 0.05 i.e 0.15 moles NH3 to give 0.05 moles [Co(NH3)6]3+

NH3 left = 1.85 -3(0.05) = 1.7 M

now we consider reverse equilibrium

at equilibrium [Co(NH3)6]3+ =0.05-X , [Co3+] = X . [NH3] = (1.7+X) = 1.7 (since X will be very small)

1/Kf = (1/5x10^31) = [Co3+][NH3]^6/[Co(NH3)6]3+]

2 x 10^-32 = (X)(1.7)^6 /( 0.05-X)

4.143 x 10^-35 = X = [Co3+] is free conc of Co3+

now Co2S3 (s) ---> 2Co3+(aq) + 3S2-(aq)

Ionic product = [Co3+]^2 [S2-]^3   = (4.143 x 10^-35)^2 x ( 0.028)^3 = 3.77 x 10^-74

now ionic product < Ksp , hence we get Co2S3 precipitate