# Homework Help: I need to understand this derivation on solubility

1. Mar 28, 2013

### Sunil Simha

1. The problem statement, all variables and given/known data
This is a derivation of the relation between solubility of a salt and pH of the solution in which it is dissolved from our textbook.
Consider a salt MX whose Ksp=[M+][K-]

If this were dissolved in a solution of the acid HX whose Ka=[H+][X-]/[HX]
then what would be its solubility?

Now Ka/[H+]=[X-]/[HX]

or Ka/([H+]+ Ka)=[X-]/([HX]+[X-])=f say
It can be seen that f decreases as pH decreases.

If S were the solubility,

Then Ksp=(S)(f*S)
= S2(Ka+[H+])/Ka

Could anyone please explain how Ksp=(S)(f*S) because I have no idea.

2. Mar 28, 2013

### epenguin

You would help yourself and us if you told us both what your book is defining solubility as in this case. It could be the total concentration of M or the total concentrtion of X in solution at saturation. I'll assume it is X.

Your equation Ksp=[M+][X-] remains valid all the time.

But in the simple case of the salt alone you have that [M+] = [X-] and it reduces to a simple square, [M+]2 or [X-]2 - same thing - which you also call S.

In your more complicated case I guess they are defining a K'sp as [M][Xtotal]. The Xtotal is ([X-]+[HX]). Which is ([X-] + [X-][H+]/Ka) from which your rìequation follows.

If you have any more questions and come back, please quote your source more completely.

3. Mar 28, 2013

### Sunil Simha

Hello epenguin,

I'm sorry about my lack of clarity in the question but I myself did not understand what the textbook defined as solubility in this section. It doesn't say a word about it in the parts before and after this derivation and the derivation itself is vague about it. Attached to this post is a picture of that section.

Thanks a lot for taking the time to help me.

Sunil

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4. Apr 10, 2013

### Sunil Simha

I'm really sorry about my clumsiness in typing the question.

Please see the attachment in the previous post. It reads Ksp= S2(Ka)/(Ka + [H+]).

Now can anyone please clarify why this is correct?

Thanks a lot and I apologize again for my goof-up.

5. Apr 13, 2013

### epenguin

There seem to be two things to understand, this phenomenon and the theory of it, and what this author says about it, of which the first is more important. Possibly someone who needs to invert an equation twice in two lines is not the easiest expositor to follow.* Instead of following, work it out yourself - that is often easier, advantageous and necessesary in this field. Give yourself a permission to do that! As a habit. Use the book as check of your conclusions.

In this case I'll give my way. We define solubility, S, as the molarity of salt MX in solution when saturated. It isn't really MX molecules in solution but we call it that. This molarity we can write [MX] but it's also equal to [M+] and to [X-] in the simple case without protonations: in which case

[MX] = [M+] = [X-] and at saturation = S.
So then Ksp = [M+].[X-] ...(1)
= S2

Now if X- can be protonated while M not be deprotonated, you have a weak acid and a strong base and a solution of MX will be alkaline.

Say you now add acid and protonate some of the X-

Ksp = [M+].[X-] still. We should suppose the added acid is, say HNO3 with a soluble M-salt.** Then [M+] is still our conventional [MX] and, at saturation, S, so Ksp = S.[X-]. …(2)
But now [X-] ≠ [M+] (in fact we would have [M+] = [X-] + [NO3-]). And [X-] ≠ [MX], for some of the X has been sequestered as soluble HX.

The fraction of total MX that is still free is
$$\frac{[X^-]}{[X^-]+[HX]} = \frac{1}{1+[H^+]/K_{a}}$$

So the [X-] in eq.(2) is now related to [MX] and, at saturation to S by
$$[X^-] = [MX] . \frac{1}{1+[H^+]/K_{a}}$$
so substituting in (2) will give

$$K_{sp} = S^2 . \frac{1}{1+[H^+]/K_{a}}$$

So if we defined an apparent Ksp at any [H+] as KspH+ = [MX]2 at saturation at that [H+], i.e. the S2 in the last equation above, we'll get
KspH+ = Ksp(1 +[H+]/Ka)

- a typical type of expression for a pH-dependent phenomenon. I suppose he writes this f to save writing out the same function of H+ every time, and also it would be more general, e.g. if something had a soluble ammonia complex you might find expressions invoving (1 + [NH3]6/K') or something instead of
(1 + [H+]/Ka)

* Mine could be boiled down a bit too.:rofl:

**This is not mentioned on the quoted page, and you could get confused if you are led to think of being able to change [H+] independently without this other change.
Slightly different to think about is the case where you acidify by adding the acid HX. Need to think about whether that is really comprised in the formula – you can no longer call this a solution of MX – but I think you can work out that this addition will cause no extra precipitation of MX.

Last edited: Apr 13, 2013
6. Apr 13, 2013

### Sunil Simha

Thanks a lot epenguin and I shall follow your advice next time I have such a problem. This finally makes sense to me now.