A solution contains 0.0500 M Co3 (aq), 0.0270 M S2–(aq), and 1.65 M NH3. Cobalt(III) ions in aqueous solutions complex with NH3 to produce Co(NH3)63 . (Kf = 5.0 x 1031) (Part 1) What will be the concentration of Co3 (aq) when Co(NH3)63 forms? (Part 2) Will Co2S3 precipitate? (Ksp = 4.0 x 10–21)
Part 1. Co+3+6NH3[Co(NH3)6]+3 For the hexaamminecobalt complex formation the formation constant is very high indicating it's very much proceeded to right side of the equilibrium. As 1 mole of Co+3 reacts with 6 moles of ammonia, Co+3 is the limiting reagent(0.05 M Co+3 +1.65 M NH3). So no uncomplexed ie free Co+3 will be there in the aqueous solution.
If ionic product of Co2S3 is greater than it's solubility product(KSP) then only it'll precipitate. Now let's first calculate the ioni roduct of Co2S3=[Co+3]^2[S-2]^3=(0.05)^2(0.0270)^3=4.9*10^-10 It's much much greater than the KSP of Co2S3. So definitely Co2S3 will be precipitated.
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