# Mixing cobalt(III) hydroxide and thallium(III) hydroxide

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.

• Your reasoning is fine, and the book says nonsense. – Ivan Neretin Jun 13 '19 at 5:28
• 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. – user5713492 Jun 13 '19 at 7:19
• 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 $100$%. – user5713492 Jun 13 '19 at 7:47
• 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. – MaxW Jun 13 '19 at 8:36

This should be a comment, but I decided to post it as an answer, since from my point of view, the present answers fail to shed light on the absolute simplicity of the truth.

We can state from the present answers that there will be no precipitates, but not because the concentration is smaller than an atom per liter, we don't know the volume used since it was not stated.

The simple fact is that the $$[\ce{OH^-}]$$ concentration stays the same, it will be $$10^{-7}$$ before and after the mixing of the solutions, no matter how much from each solution we will mix together, since the solubility product is so small and the dominating factor for the $$[\ce{OH^-}]$$ concentration comes from the autoprotolysis of water. Accepting this little knowledge, no matter what we will never reach the saturation concentration after mixing. The saturation concentration will stay the same always (in this problem), so we will just dilute the initial concentration.

Imagine we mix an equal amount of both solutions, then the before saturated concentration of the substances will be halved. So in the end in this specific case for an example the solution will be half saturated with both hydroxides. That is why there is no precipitation.

Edit: Also there is no sense talking about common ion effect, since this phenomenon arises from the charge balance equation. $$3[\ce{Tl^{3+}}]+3[\ce{Co^{3+}}]+[\ce{H^+}]=[\ce{OH^-}]$$

As we postulated before $$[\ce{H^+}]=[\ce{OH^-}]$$, since the autoprotolysis is the domination factor, both of the concentration of the ions will be zero (simplified). In truth the $$[\ce{OH^-}]$$ concentration will be just slightly higher than the $$[\ce{H^+}]$$ concentration, but we can neglect this equation since we assumed the only $$[\ce{OH^-}]$$ concentration comes from autoprotolysis, meaning we don't have to solve a system of equations, just plugging in the value in the solubility product.

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.

• 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. – Ivan Neretin Jun 13 '19 at 7:16
• @Kent de los Reyes: Please check your calculations. The $[\ce{OH-}]$ should be much higher than what you have calculated. – Mathew Mahindaratne Aug 7 '20 at 16:32

The calculations made by Kent de los Reyes are based on the hypothesis that the solvent (water) does not produce any $$\ce{OH-}$$ ions. He calculates that the concentrations of $$\ce{OH-}$$ is about $$10^{-12}$$ in saturated solutions of $$\ce{Tl(OH)3}$$ or $$\ce{Co(OH)3}$$. But the real concentration of $$\ce{OH-}$$ is much higher, because of the water dissociation. If the saturated solution is prepared in pure water, $$[\ce{OH-}]$$ should be equal to $$10^{-7}$$ M. And the concentrations of the ions $$\ce{Co^{3+}}$$ and $$\ce{Tl^{3+}}$$ are of the order of $$\dfrac{10^{-45}}{({10^{-7}})^3} = 10^{-24}$$.

Well! Such a concentration is smaller than one atom per litre. So it is of no use discussing any precipitate whatsoever.