# Solubility Product Calculation

• Max364
In summary: If you are interested in finding amount of precipitate, you need to use other tools. Scaling index is useless for that. All it says is "this solution will produce a precipitate". There can be tons of the precipitate, there can be micrograms.
Max364
I have a question about calculating solubilities of sparingly soluble salts.

Eg Ksp CaF2 = 4 x 10-11
So, Saturation Index of CaF2 is:

SICaF2 = IP / Ksp
Where IP = Ionic Product = [Ca2+] x [F-]2
[Ca2+] and [F-] are molar concentrations of each ion.

Example:
We have 400 ppm Ca and 12 ppm F in a water

Ion product IP = [400/40000] x [12/19000)]2 = [0.01]. [3.99 x10-7] = 3.99 x 10-9

Saturation Index SICaF2 = 3.99 x 10-9 / (4 x 10-11 ) = 100

All is fine so far, but let's say we now have a massive amount of extra Ca, say 5,000ppm

Now, IP = [5000/40000]. [3.99 x10-7] = 4.99 x 10-8
So, Saturation Index now SICaF2 = 4.99 x 10-7 / (4 x 10-1 ) = 12.475

So massively increased due to increased Ca.
But my question is: How accurate is this new SICaF2 ?

Because we have a limited amount of F (only 12ppm) the it should not matter how much extra Ca we have? As it can only react with a fixed amount of F, ie, 1 ppm Ca reacts with 2ppm of F and once all the F is used up then excess Ca distorts the SI calculation?

What is the meaning of the saturation index?

I must admit I have never seen it before, but I have a feeling values over 1 can only be used to predict whether to expect precipitation. From the way you have defined it if SI is below 1, nothing happens, if it is above 1, CaF2 precipitates till SI equals exactly 1.

Borek said:
What is the meaning of the saturation index?

I must admit I have never seen it before, but I have a feeling values over 1 can only be used to predict whether to expect precipitation. From the way you have defined it if SI is below 1, nothing happens, if it is above 1, CaF2 precipitates till SI equals exactly 1.

Yes that's correct, any value over 1 means ppt...SI is just a way to predict scaling.
The point is does having excessive one species over another making the actual calc incorrect for prediction?

No, it still predicts the precipitation correctly.

OK let's put it another way: In the above eg i had 400ppm Ca and 12ppm F to give me SI of 100 - so that water will scale.

But if we had 6% Ca (60,000ppm) and only 1ppm F this will give me the same SI for CaF2, ie, 100, yet this water will not have any noticeable scale in terms of grams deposited as we only have a very small amount of F - so you will not see the deposit?
But the calc suggests both waters are equally scaling? this is why i say its misleading?

Scaling index doesn't tell anything about amount of scale. For that you would need some other index. If you are trying to use SI to predict amount of scaling you are using a wrong tool - no wonder you are being mislead by the result.

So what do you understand by the two examples above - both giving SI of 100?
Are you saying that if you had to predict scaling of these two waters based on SI you would conclude that they are both equally scaling?

No, I would conclude that in both cases some scaling can be expected, but there is no information about AMOUNT of scaling to be expected.

If you are interested in finding amount of precipitate, you need to use other tools. Scaling index is useless for that. All it says is "this solution will produce a precipitate". There can be tons of the precipitate, there can be micrograms.

I disagree, using only the SI information, both examples being 100 in this case, looking at the actual concentrations used in the calculation one can conclude that one water is more scaling than the other, ie, 400ppm Ca and 12ppm F is more scaling than 6% Ca and 1ppm F, simply because irrespective of the excess Ca, you only have a limited amount of counter ion.
Example: if you are treating a boiler you would be right to suspect CaF2 scaling with the higher F levels and would require treatment, whereas the water with lower F but same SI of 100 would pose a negligible risk.

Max364 said:
using only the SI information, both examples being 100 in this case, looking at the actual concentrations

You are contradicting yourself - you either use only the SI information, or the SI information with the actual concentrations. Don't you see it is not the same?

I have a solution A and a solution B. Solution A has SI of 100, solution B has SI of 100. Using only SI information, which will produce more scaling?

Yes but SI values are not plucked out of the air, they are calculated using concentrations, i am saying that if you were doing the calcs for SI using concentrations then you WILL be able to tell which is more scaling.
But you are right that if one only had SI values that someone quoted then you have no idea which is more scaling...or you may even say, wrongly in the case, that both waters are equally scaling?

Max364 said:
Yes but SI values are not plucked out of the air, they are calculated using concentrations, i am saying that if you were doing the calcs for SI using concentrations then you WILL be able to tell which is more scaling.

If I am given concentrations I am not going to calculate SI. I am going just to calculate the reaction quotient (this product with correct powers) and compare with the Ksp. I don't need additional division to see if what I calculated is lower or higher than Ksp

But you are right that if one only had SI values that someone quoted then you have no idea which is more scaling...or you may even say, wrongly in the case, that both waters are equally scaling?

Define "scaling". I guess you mean "amount of solid that can precipitate", but this is not obvious. Especially taking into account some of the precipitated solid doesn't end as a scale (attached to the surface) but just as a suspension that can be easily removed (by filtration, or even just by replacing the water with a new one).

We compare ionic product concentrations with Ksp so that we know whether a given solution is near saturation or not etc.
If you read my original question again, i mention that excess Ca distorts the SI calc in this example etc...so i referred to concentrations throughout - no contradiction there.
In industry, the way to predict scaling is using SI, even if its only a suspension etc, in a closed system you can not or may not want to just replace solution - water recycling, zero discharge etc all means that its not that simple.

So going back to the industry standard of using SI - my question still stands:

Massively increasing one species relative to the other distorts the predicted SI?

Then you have to find someone using SI in your form in their daily practice. I have never heard about SI before and from a quick search it doesn't look like something widely popular. Not to mention the fact what I was able to find was different from what you have listed -- I found several indices (PSI, LSI, RSI), but they are defined differently and they all deal with just carbonate hardness, not with any substance that can crash out of the solution.

Scaling tendency (ST) seems to be identical with your definition.

All these indices are just proxies for the real equilibrium present in the solution, and as is the case with every proxy, they have limited range of applicability.

## 1. What is the solubility product (Ksp) and how is it calculated?

The solubility product (Ksp) is a measure of the maximum amount of a solute that can dissolve in a solvent at a given temperature. It is calculated by multiplying the concentrations of the ions in a saturated solution, each raised to the power of their respective coefficients in the dissociation equation.

## 2. What is the significance of the solubility product in determining the solubility of a compound?

The solubility product is a measure of the equilibrium between a solid and its dissolved ions in a solution. It indicates the maximum amount of solute that can be dissolved in a particular solvent and is used to predict the solubility of a compound under different conditions.

## 3. How does temperature affect the solubility product?

Generally, the solubility product increases with increasing temperature. This is because as temperature increases, the kinetic energy of the molecules increases, leading to more collisions and a greater chance for solute particles to break apart and dissolve.

## 4. Can the solubility product be used to determine the concentration of individual ions in a solution?

Yes, the solubility product can be used to determine the concentration of individual ions in a solution. By knowing the Ksp and the concentration of one ion, the concentration of the other ion can be calculated using the dissociation equation and the equilibrium expression.

## 5. Are there any limitations to using the solubility product calculation?

Yes, there are some limitations to using the solubility product calculation. It assumes that the solution is at equilibrium, which may not always be the case. Additionally, it does not take into account factors such as pH, complex formation, or the presence of other ions in solution, which can affect the solubility of a compound.

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