A simple question on solutions is driving me crazy. At this point, I think the textbook's wrong. Here's the question:

A saturated solution of cobalt(III) hydroxide $(K_\mathrm{sp} = \pu{1.6e-44})$ is added to a saturated solution of thallium(III) hydroxide $(K_\mathrm{sp} = \pu{6.3e-46}).$ What is likely to occur?

The answer given is that $[\ce{OH-}]$ is above saturation levels for both the cobalt and thallium in the solution, so that both will precipitate (thallium(III) hydroxide first, since it has a smaller $K_\mathrm{sp}).$

What I don't understand is how they can make such a statement. We don't know the amounts of each solution originally. Furthermore, wouldn't $[\ce{OH-}]$ be somewhere between the two original $[\ce{OH-}]$ depending on amount of each solution? And if so, wouldn't it not trigger precipitation of the cobalt (III) hydroxide (larger $K_\mathrm{sp},$ larger $[\ce{OH-}]$ at full saturation than $[\ce{OH-}]$ of the mixed solutions)?

Because of this, my answer would be that thallium(III) hydroxide precipitates, but cobalt(III) hydroxide remains dissolved.

  • 2
    $\begingroup$ Your reasoning is fine, and the book says nonsense. $\endgroup$ – Ivan Neretin Jun 13 '19 at 5:28
  • $\begingroup$ The reasoning is a little suspect to me. My calculations, paralleling those of @KentdelosReyes lead me to conclude that there is no concentration of $\text{Tl}(\text{OH})_3$ that leads to precipitation because when $[\text{OH}^-]$ is high, $[\text{Tl}^{3+}]$ is low. Quantitatively you get a fourth-degree equation with $3$ negative roots and one positive root corresponding to pure $\text{Tl}(\text{OH})_3$ saturated solution. $\endgroup$ – user5713492 Jun 13 '19 at 7:19
  • $\begingroup$ Oops, I take that back. With a corrected fourth-degree equation I get a precipitate when the fraction of $\text{Tl}(\text{OH})_3$ is between $10.6$% and $100$%. $\endgroup$ – user5713492 Jun 13 '19 at 7:47
  • $\begingroup$ The gist here is that you must assume that there is an excess of both cobalt(III) hydroxide and thallium(III) hydroxide after the solutions are mixed. Thus since both ppt's are in the final solution, it doesn't matter what the initial relative proportions are. $\endgroup$ – MaxW Jun 13 '19 at 8:36

You don't need to know the original amount of the substance because the $K_\mathrm{sp}$ will give you the saturated concentrations of the ions present.

For $\ce{Co(OH)3},$ the $K_\mathrm{sp}$ is $\pu{1.6e-44}$ which means the saturated concentration of the solution is $\pu{4.93e-12 M}$ for $\ce{Co^3+}$ and $\pu{9.86e-12 M}$ for $\ce{OH-}.$ You can calculate this using the equation

$$K_\mathrm{sp} = x\cdot (3x)^3$$

The same applies for $\ce{Tl(OH)3},$ the $K_\mathrm{sp}$ is $\pu{6.3e-46}.$ The saturated solution contains $\pu{2.20e-12 M}$ of $\ce{Tl^3+}$ and $\pu{6.6e-12 M}$ of $\ce{OH-}.$

Now that we clarified the initial concentration dilemma, now we can address the real issue at hand which is the common ion effect. The two slightly soluble solids has a common ion of $\ce{OH-}$ that decreases the solubility of both cobalt(III) hydroxide and thallium(III) hydroxide. This is a complex topic so I recommend you to dig deeper on it.

  • 1
    $\begingroup$ True, if we saturate the solution with both solids together, they both will reach the equilibrium concentrations slightly less than their corresponding solubilities when measured one by one. But that's not what the problem asks for. $\endgroup$ – Ivan Neretin Jun 13 '19 at 7:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.