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Name: Mohammad
Status: student
Grade: 9-12
Country: Canada
Date: Spring 2009


Question:
How the change in the solution pH effect the solubility of a salt?



Replies:
Mohammad,

Unfortunately, the answer to this will depend on the type of salt. Here is one example (and maybe you can use this example for reasoning out other types of salt):

Salt: CaF2

When CaF2 is dissolved in water, the dissolved salt will fully dissociate to Ca(2+) and 2 F(-).

Since only a small portion of CaF2(s) will dissolve in water, we can imagine a small equilibrium constant for the chemical equilibrium equation: CaF2(s) = Ca(2+) (aq) + 2 F(-) (aq) and the solubility product equation: Ksp = [Ca(2+)][F(-)]^2

However, the fluoride ion can react with water in the equilibrium reaction:

F(-) + H2O = HF + OH(-)

We therefore see that the solution should be basic because the second equilibrium produces OH(-).

If we add more base to the solution (the pH increases), this will drive the second reaction backwards (according to the principle established by Le Chatelier). This will mean more F(-) ions, which means the first equilibrium is also driven backwards - resulting in less dissolved Ca(2+) or more precipitate.

Use the same pattern of reasoning (a combination of knowing what the salt will do in water, and the Le Chatelier Principle), for salts that do not react with water, salts that produce H(+) in water, etc.

Greg (Roberto Gregorius)


Yours is one of those questions that, at first, appears simple but upon further reflection becomes increasingly more complicated. Until finally you (i.e. I) come to the response, "Good Question!". There are several competing factors that determine the solubility of any substance:

(1) Each solute atom/molecule is competing for water molecules, because the hydration of the various solute species is an important, sometimes the most important, driving force that ultimately determines the solubility of the particular solute. This competition can work both ways -- increasing or decreasing the solubility of a particular solute. This is the mechanism behind the so-called "salting out effect".

(2) If the acid/base has an ion common to the salt -- for example HCl added to NaCl, the common ion effect comes into play so I would guess the addition of aqueous HCl to a saturated solution of NaCl would reduce the concentration of Na(+1) because it will combine with the additional Cl(-1) coming from the acid added.

(3) In cases where there is a "real" chemical reaction. For example, adding Ag(NO3) forming insoluble AgCl, the "apparent" solubility of NaCl increases because the excess solid NaCl will disappear.

(4) For any or all of the above reasons, it depends upon what acid or base is used to change the pH. If the acid or base has greater "affinity" for water than the salt, the acid or base will "steal" water, making the water less "available" to the solute salt; if the acid or base has less "affinity" for water than the salt the reverse will occur. I used the terms above in "quotation marks" because I am not sure of a general, measurable, operationally defined method for measuring those terms.

However, I am pretty sure that there is no general method for predicting the change insolubility of a salt with changes in pH. In fact there is no good quantitative model for the solubility salts, or even most neutral solutes especially in the very complex solvent, water.

Vince Calder



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