A. Calculate the solubulities of the hydroxides. B. What would be the order of p

Question

A. Calculate the solubulities of the hydroxides.

B. What would be the order of precipitation if the OH- as added gradually to the solution containing the metal ions? Are you going to be able to separate Cr(OH)3, Cr(OH)2, Co(OH)2 and Co(OH)3? 2. Qualitative Analysis of cations and Anions Given a solution containing chromium and chromium (III and cobalt (ll) and (I cobalt (III), (a) Calculate the solubilities ofthe hydroxides.The Ksp are as follows Cr(OH)3 6.3 x 1031 Cr(OH)2 2 10-16 Co(OH)2 1.6 x 10-15 Co(OH)3 1.6 x 10-44 (b) What would be the order of precipitation if the OH- as added gradually to the solution containing the metal ions? Are you going to be able to separate Cr(oH)3, Cr(OH)2, Co(OH)2 and Co(OH)

Explanation / Answer

Cr(OH)3 <==> Cr^3+ + 3OH-

Ksp = s * (3s)^3 = 27s^4

or, 6.3*10^-31 = 27s^4

or, s =1.24*10^-8 M

Solubility of Cr(OH)3 is 1.24*10^-8 M

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Cr(OH)2 <==> Cr^2+ + 2OH-

Ksp = s* (2s)^2

or, 2*10^-16 = 4s^3

or, s = 3.68*10^-6 M

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Co(OH)2 <==> Co^2+ + 2OH-

Ksp = s*(2s)^2

or, 1.6*10^-15 = 4s^3

or, s = 7.37*10^-6 M

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Co(OH)3 <==> Co^3+ + 3OH-

Ksp = s*(3s)^3

or,1.6*10^-44 = 27s^4

or, s= 8.39*10^-16

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the hydroxide will precipitate from the solution when the ionic product of the hydroxide exceeds the Ksp values. The hydroxide with a smaller Ksp value will precipiate faster.The order of precipitation is : Co(OH)3> Cr(OH)3>Co(OH)2>Cr(Oh)2.

It will be difficult to separate Co(OH)2 from Cr(OH)2 as the solubility product values are very close. Other two hydroxides can be easily separated due to large difference in their Ksp value.

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