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 K
sp = [M
+].[X
-] ...(1)
= S
2
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
-
K
sp = [M
+].[X
-] still. We should suppose the added acid is, say HNO
3 with a soluble M-salt.** Then [M
+] is still our conventional [MX] and, at saturation, S, so K
sp = S.[X
-]. …
(2)
But now [X
-] ≠ [M
+] (in fact we would have [M
+] = [X
-] + [NO
3-]). 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 K
sp at any [H
+] as K
spH+ = [MX]
2 at saturation at that [H
+], i.e. the S
2 in the last equation above, we'll get
K
spH+ = K
sp(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 + [NH
3]
6/K') or something instead of
(1 + [H
+]/Ka)
* Mine could be boiled down a bit too.
**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.
