Why Is Magnesium Precipitated from Sea Water Using Calcium Hydroxide?

[jaumzaum]

For we to obtain metallic magnesium from sea water we can precipitate it in the form of magnesium hydroxide if we add calcium hydroxide:

Mg²⁺ + Ca(OH)₂ → Mg(OH)₂ + Ca²⁺

Why does this reaction happen?

 

[member 392791]

Have you checked the usual suspects? Such as being favored thermodynamically or an increase in entropy?

 

[jaumzaum]

Have you checked the usual suspects? Such as being favored thermodynamically or an increase in entropy?

Hi Woopydalan.
Actually I would like a more qualitative explanation, not a quantitative one. If you had no numbers like the entalpy/entropy/gibbs energy of the substances, how would you explain that a reaction like this would happen?

 

[member 392791]

The qualitative answer is that it is favored by thermodynamics to happen or there is an increase in entropy. Whether or not you had the values at your disposal would not really matter. That is how you would explain it happening. I know you are looking for another answer, but simply that is the reason reactions occur.

It's like asking ''Why is a banana yellow?'' the answer would be ''It doesn't absorb wavelengths between 570-590 nm'' but then you say ''well what if I don't know the wavelength of yellow light, then how do I explain it?'' then my response would follow ''It doesn't absorb yellow light.''

 

[Ygggdrasil]

Try looking up the solubility (specifically, the solubility product constant K_sp) for each of these substances.

 

[jaumzaum]

Try looking up the solubility (specifically, the solubility product constant K_sp) for each of these substances.

I thought the solubility could have something to do with it, but I still cannot explain it. Actually the calcium hydroxide is much more soluble than the magnesium one.

Solubility of magnesium hydroxide is 1.4mg/L while solubility of calcium hydroxide is 1.73g/L at 20 °C
The Ksp are 1.5 10^-11 for magnesium and 4.68 10^-6 for calcium

How can I explain it now, knowing that Mg(OH)2 is much less soluble than Ca(OH)2

 

[Borek]

Calculate concentrations of Ca²⁺ and OH^- in the saturated solution of Ca(OH)₂.

Introduce any magnesium salt in such an amount concentration of Mg²⁺ becomes the same as concentration of Ca²⁺.

What will happen now?

 

[jaumzaum]

Calculate concentrations of Ca²⁺ and OH^- in the saturated solution of Ca(OH)₂.

Introduce any magnesium salt in such an amount concentration of Mg²⁺ becomes the same as concentration of Ca²⁺.

What will happen now?

Actually when I was trying to find the concentration of Ca2+ I've got another doubt.

When I try to find the Ksp from the solubility, I get an absurd result.

solubility of calcium hydroxide is 1.85g/L at 0°C, that is 2.5 10^-2M
Ca(OH)₂ → Ca²⁺ + 2OH^-
2.5 10^-2-----0-------0------
---0-----2.5 10^-2----5. 10^-2

Ksp = 4 (2.5 10^-2)³ = 6.25 10^-5 that is different from the Ksp_0°C = 8 10^-6

Why does this happen?

And answering your question. Mg²⁺ would precipitate as it it less soluble than Ca²⁺. But what does this have to do with the first question?

 

[Borek]

Solubility of Ca(OH)₂ is higher than that predicted by Ksp alone, as Ca²⁺ gets complexed by OH^-.

And answering your question. Mg²⁺ would precipitate as it it less soluble than Ca²⁺. But what does this have to do with the first question?

And you still don't see that it means the reaction proceeds to the right?

 

[jaumzaum]

Solubility of Ca(OH)₂ is higher than that predicted by Ksp alone, as Ca²⁺ gets complexed by OH^-.



And you still don't see that it means the reaction proceeds to the right?

Not yet, haha :(
Could you explain it to me? (I'm not very good in chemistry...)

 

[Borek]

To be honest, I have no idea how to help you further. You have came to the right conclusion that when there are two weakly soluble salts competing for an anion (or cation, doesn't matter), reaction will proceed in such a way that the less soluble salt will be the main product. That's exactly what is happening here.

 

[jaumzaum]

To be honest, I have no idea how to help you further. You have came to the right conclusion that when there are two weakly soluble salts competing for an anion (or cation, doesn't matter), reaction will proceed in such a way that the less soluble salt will be the main product. That's exactly what is happening here.

But sea water isn't saturated of calcium, it is saturated of magnsium. And we are not adding Mg(OH)₂, we are adding CaOH₂

I don't see the equivalence

 

[Borek]

Will writing the reaction as

Mg²⁺ + 2OH^- -> Mg(OH)₂

help?

 

[Ygggdrasil]

Perhaps it is helpful to think of things in terms of energy as Woopydalan suggested. Based on the Ksp values, you require 28 kJ/mol of free energy to break Ca(OH)₂ into Ca²⁺ and two OH^-, but forming Mg(OH)₂ from Mg²⁺ and two OH^- releases 57 kJ/mol of free energy (under standard conditions). Therefore, breaking apart calcium hydroxide and using the hydroxide ions to form magnesium hydroxide gives an overall decrease in the free energy of the system, and therefore, the reaction is thermodynamically favorable.