Professor Dave again, let’s make some predictions.
We just learned all about the solubility product and how to calculate it.
But what are the applications of the solubility product?
For one thing, we can use a Ksp to look at a solution and predict if a solid precipitate
will form as a result of a chemical reaction.
Let’s learn how to do this now.
Calcium carbonate, when in aqueous solution, will establish an equilibrium between the
solid form and dissociated form.
When equilibrium is reached, the reaction quotient, Q, for this equilibrium must be
equal to Ksp, which is 4.8 x 10^-9 for this substance.
Let’s say that instead of dissolving calcium carbonate, we mix together aqueous solutions
of calcium nitrate and sodium carbonate, both of which will completely dissolve to produce
the respective ions.
With calcium ions and carbonate ions in solution but no solid calcium carbonate, this equilibrium
will shift left, causing some precipitation of calcium carbonate to occur, as long as
Q is greater than Ksp.
So a comparison of Q and Ksp is the way to predict whether precipitation will occur in
solution.
For example, to prepare magnesium metal, we allow magnesium hydroxide to precipitate from
seawater by adding calcium hydroxide to it.
In a typical sample of seawater, the magnesium ion concentration is 0.0537 moles per liter,
and let’s say we want to find out if enough calcium hydroxide is added to give a hydroxide
concentration of 0.001 molar, will magnesium hydroxide precipitate form, given that it
has a Ksp of 2.1 x 10^-13?
As we said, we simply need to compare Q and Ksp, so let’s calculate the reaction quotient, Q.
If Q is equal to magnesium ion concentration times hydroxide ion concentration squared,
then plugging in those concentrations will give us a reaction quotient of 5.4 x 10^-8.
This is significantly larger than Ksp, and we know that if Q is greater than K for any
type of equilibrium, the equilibrium will shift left, in this case producing solid,
thus precipitate will form until Q becomes equal to Ksp.
If we consider a similar situation, where the only difference is that Q is less than
Ksp, a precipitation will not occur, and all ions will remain in solution.
This also means that given a Ksp and one ion concentration, we can calculate the concentration
the other ion we will need to reach in order for precipitation to begin to happen.
We just need to find the concentration at which Q will become equal to Ksp.
For example, look at calcium oxalate.
This substance can be in equilibrium with the calcium ion and oxalate ion, and it has
a Ksp of 2.27 x 10^-9.
If we have a solution of this substance that is completely dissociated, and the concentration
of calcium is 2.2 x 10^-3 moles per liter, what must the oxalate ion concentration climb
to in order for precipitation to occur?
Once again, precipitation occurs when Q equals Ksp, so we can simply write out the Ksp expression,
and plug in the two known values, that would be the Ksp itself and the calcium ion concentration.
Solving for oxalate concentration, we get 1 x 10^-6 moles per liter, which is the necessary
oxalate concentration to initiate precipitation of calcium oxalate.
This concept can be modified to translate an ion concentration into a pH when acidity
and basicity are concerned.
An application of this is that water with a manganese ion concentration above 1.8 x
10^-6 moles per liter will stain clothing when doing laundry, which is why manganese
concentration is sometimes reduced by adding base.
If the pH is maintained above a certain level, the manganese can precipitate as manganese
hydroxide, and will thus be unable to stain the clothing.
So given that this substance has a Ksp of 4.5 x 10^-14, what pH is required to provide
enough hydroxide to keep the manganese concentration at or below 1.8 x 10^-6?
We use the same process as before.
We are looking at the point where precipitation occurs, so we can just write out the Ksp expression
and plug in the Ksp and the manganese concentration we are interested in.
This will allow us to solve for the hydroxide concentration, which will be 1.6 x 10^-4.
From this, we take the negative log to get the pOH, and then subtract from 14 to get
the pH, which will be 10.2.
Therefore, if there is enough base in the detergent to keep the pH above 10.2, the manganese
ion concentration will remain below a threshold that would stain the clothes.
And that illustrates just a few applications of the solubility product.
Let’s check comprehension.