From http://wps.prenhall.com/wps/media/objects/3312/3392202/blb1705.html
<< The solubility of almost any ionic compound is affected if the solution is made sufficiently acidic or basic. The effects are very noticeable, however, only when one or both ions involved are at least moderately acidic or basic. The metal hydroxides, such as Mg(OH)2, are examples of compounds containing a strongly basic ion, the hydroxide ion. As we have seen, the solubility of Mg(OH)2 greatly increases as the acidity of the solution increases. As an additional example, the solubility of CaF2 increases as the solution becomes more acidic, because the F– ion is a weak base; it is the conjugate base of the weak acid HF. As a result, the solubility equilibrium of CaF2 is shifted to the right as the concentration of F– ions is reduced by protonation to form HF. Thus, the solution process can be understood in terms of two consecutive reactions:
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[17.18]
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[17.19]
The equation for the overall process is
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[17.20]
Figure 17.17 shows how the solubility of CaF2 changes with pH.
Figure 17.17 The effect of pH on the solubility of CaF2. The pH scale is given with acidity increasing to the right. Notice that the vertical scale has been multiplied by 103.
Other salts that contain basic anions, such as CO32–, PO43–, CN–, or S2–, behave similarly. These examples illustrate a general rule: The solubility of slightly soluble salts containing basic anions increases as [H+] increases (as pH is lowered). The more basic the anion, the more the solubility is influenced by pH. Salts with anions of negligible basicity (the anions of strong acids) are unaffected by pH changes.>>
So your compound NX solubility cheanges depends on the nature of X, it seems to me.
What should be the pH value of the solution to double the solubility of the ionic compound NX?
Ka of HX=2*(10)^-5 M
My thoughts on the question:
Let the solubility of NX=x M
1)
NX(s)<====> N+(aq) + X-(aq)
eq:con: x x
(moldm-3)
X^-<===> HX + OH^- -->(2)
HX<===>H^+ + X^- --->(3)
Kh(hydrolysis constant)=kW/ka
I understand that for the solubility of NX to be twice the initial value,H^+ concentration should be half of the initial vale,from (3) (temperature hasn't change so ka remains the same)
How to proceed on?
2 answers
Since your prof hasn't given a Ksp or a solubility, do you think the pH needed is a fixed number or does the prof want an equation that will give the pH after the solubility is known?