Hey there!
Hi there, a quick textbook question:
"Explain why we cannot compare ksp values to determine the relative solubility between two compounds."
Is this because ksp is dimensionless? I always thought that we'd use ksp to judge if something was good at dissolving or not...
Thanks!
You can actually use K
sp to compare the relative solubility between compounds, but
if and only if they dissociate to produce the same ratio of ions. For example, you can compare the relative solubility of potassium chloride and sodium hydroxide, or even calcium carbonate, because they dissociate to produce the same ratio of ions as can be seen here:
KCl
(aq)⇌K
+(aq)+Cl
-(aq)NaOH
(aq)⇌Na
+(aq)+OH
-(aq)CaCO
3(aq)⇌Ca
2+(aq)+CO
32-(aq)Another example of this would be say magnesium hydroxide and calcium chloride:
Mg(OH)
2(aq)⇌Mg
2+(aq)+2OH
-(aq)CaCl
2(aq)⇌Ca
2+(aq)+2Cl
-(aq)What you can't do is compare two compounds that when ionised, form different ratios of ions. This is fundamentally because of the mathematical definition of K
sp, because for example in the compound AB⇌A
++B
-If we now consider the compound AB
2⇌A
++2B
-, our K
sp becomes
One of the reasons we can't is that when considering the units of the former K
sp, it is mol L
-1, while for the latter it is mol
2 L
-2, but also because you have an extra value when considering the solubility product. However, in addition to this, in the former case, when you're looking at the actual solubility, you get that K
sp=s
2, while in the latter case you get that K
sp=4s
3 simply because of ion ratios. Thus, even if they had really similar K
sp values, the actual solubility is heavily skewed. What you need to compare is not the K
sp, but the actual solubility s. Note that this doesn't occur with the compounds with the same ion ratios, since it'll just be s
2 for both, or 4s
3 or something or other. Hence, this is why you (usually) can't compare K
sp to compare relative solubility between two compounds.
Hope this helps
EDIT:
helloo,
i was wondering why insoluable and sparingly soluable salts reach equilibrium and not soluable salts??
Hey
Soluble salts will not reach equilibrium generally because they will just dissolve completely and dissociate into their ions, hence leaving none of the compound. The system will just go to completion instead of equilibrium. The opposite is true for insoluble and sparingly soluble salts; they don't have as much compound dissociating, so they will be 'stuck' in equilibrium. To put it in terms of K
sp, we know that if K
sp is small, the substance is not as soluble, and if it is large, it is more soluble. eg. For a soluble salt, more ions will be produced at equilibrium, producing a more 'top-heavy' solubility product which is larger --> more soluble. In your case, this is when the substance is completely soluble, and there are no reactants left, leading to a K
sp value that tends to infinity. For an insoluble or sparingly soluble salt, the K
sp value will be zero or close to zero respectively, indicating that at equilibrium, the solubility product is 'bottom-heavy' with a lot more reactants, and thus that the salt is not as soluble, thus forming an equilibrium.
Hope this explanation makes sense