Video Transcript
In this video, we will learn about
the solubility product constant. We will learn how to calculate it
from ion concentrations in solution. And we will learn how the value of
this constant indicates the degree to which a solute dissolves.
What is the solubility product
constant? The solubility product constant, or
Ksp, is a special equilibrium constant used when a solid ionic substance dissolves
or dissociates in solution. The ionic substance must be a
sparingly soluble substance. Sparingly soluble means it
dissolves to a very small degree or hardly at all, and the solution must be water
based. So for an ionic compound, let’s
call it A, in equilibrium in aqueous solution with its ions C+ and D−, the
equilibrium equation would be aA solid in water giving cC+ aqueous plus dD−
aqueous. Of course, the charges on these
ions may vary. They may be one plus or two plus or
three minus, depending on the ionic compound. So for simplicity, let’s just
remove the plus and minus signs for now.
We can calculate the Ksp in much
the same way as a normal equilibrium constant. Ksp is the solubility product
constant, and this is equal to the concentration of the products multiplied with
each other and each raised to a power equal to its stoichiometric coefficient. We can see how the concentration of
product C is raised to a power of c, and the concentration of product D is raised to
a power of d because D has a coefficient of value small d. Now, because the starting substance
is a solid and because these solids are sparingly soluble, we can eliminate the
denominator. Solids do not affect the
equation. Their concentrations remain
essentially constant, so we can remove them from the Ksp expression.
The concentration of the products
are used in molarity units, moles per liter, which is the same as moles per dm
cubed. Just as with other equilibrium
constants, Ksp is dependent on temperature. This is because temperature affects
how much solute dissolves. So we know the formula equation for
Ksp and the units for the concentration of each product. But how is Ksp useful to us? Ksp is a measure of how much solute
dissolves or the molar solubility of a compound. The more soluble a substance is,
the higher its Ksp value. So if there is a relatively high
concentration of dissolved particles in solution, Ksp will be higher. But if there’s a relatively lower
concentration of these dissolved particles, Ksp will be lower.
Before we do some calculations, I
want to point out something very important. Sometimes we talk about calculating
the Ksp for an insoluble salt. In reality, all salts will dissolve
to some degree. No compound is completely
insoluble. The Ksp values of these so-called
insoluble salts is very small. Now let’s practice writing some Ksp
expressions and calculating some Ksp values.
In this example, the question is,
“At 298 kelvin, the solubility of barium sulfate is 1.05 times 10 to the negative
five moles per dm cubed. Calculate the solubility product
Ksp. The equilibrium equation is BaSO4
solid dissolved in water to give Ba2+ aqueous plus SO42− aqueous.”
The equation is balanced. Ksp is equal to the concentration
of the aqueous products multiplied by each other. Notice that the barium ion is
raised to a power of one because its coefficient is one, and the sulfate ion is also
raised to a power of one. We are not given the concentrations
of the ions, but we do know that for every one barium sulfate, there is one barium
ion and one sulfate ion. And because they are all together
in the same volume of solution, we know that the concentration of the barium and
sulfate ions will be the same as each other and will be equal to the solubility of
the barium sulfate.
This is a small solubility value
because barium sulfate is insoluble in water or, more technically correct, very
poorly soluble. So substituting the solubility
value into the equation, we’ll get an answer of 1.10 times 10 to the negative
10. Note that the units also multiply
with each other, giving us the strange unit in the answer, mole squared per dm to
the sixth. The unit for Ksp is different for
different substances, and sometimes the unit is left out altogether. In this example, because the ratio
of barium to sulfate ions was one as to one, we got this particular unit.
Let’s look at another example.
For the equation given, determine
Ksp in terms of 𝑥 where 𝑥 is equal to the concentration of the lead ion or Pb2+
and the equation is PbCl2 solid dissolved in water to give Pb2+ aqueous plus 2Cl−
aqueous.
Lead chloride is poorly soluble in
water. The Ksp expression for this
equilibrium is the concentration of the lead two plus product raised to the power of
one multiplied by the concentration of the chloride product raised to the power of
two. We are not given the concentrations
of these ions, but we are given the concentration of the lead ion in terms of
𝑥. So if lead’s concentration is 𝑥
and for every one lead ion, there are two chloride ions, then the concentration of
the chloride ions must be two 𝑥. And we mustn’t forget to square the
concentration of the chloride ions. We can solve this different ways,
but one way is to multiply everything out and we get an answer of four 𝑥 cubed.
Now we do not know the unit of
𝑥. But imagine the unit was mole per
dm cubed. We would have mole per dm cubed
multiplied by two mole per dm cubed multiplied by two mole per dm cubed. This would give a final answer with
a unit of moles cube per dm to the nine. So remember in Ksp calculations to
always write out the units so that you get the correct unit at the end.
So far, this is what we have
learnt. When an ionic solid A dissolves
sparingly in aqueous solution to form positive and negative ions C+ and D−, this is
the corresponding Ksp expression. And I removed the positive and
negative signs because this is just a general equation and a general Ksp
expression. We know Ksp is a constant for
equilibrium conditions and an equilibrium will only form in a saturated
solution. But what if the following
occurred? Imagine you multiplied the molar
concentrations of ions C and D with each other for a particular solution. And the answer you got was less
than the Ksp. This would only happen in the case
of an unsaturated solution. This means more of the ionic
substance A can dissolve.
Unsaturated solutions are not in
equilibrium, and so we cannot use the Ksp expression for an unsaturated
solution. In this last scenario, if we
multiply the ion concentrations with each other in a particular solution and we get
a value greater than the equilibrium solubility constant, this tells us that a
precipitate will form. The concentrations of C and D in
solution are too high for a stable solution, and solid ionic substance A will begin
to form to decrease the concentrations of C and D and bring the system back to
equilibrium, which is more stable. Now It’s time for some
practice.
Taking the solubility product of
aluminum hydroxide to be 1.90 times 10 to the negative 33 at 298 kelvin, what is the
concentration of Al3+ ions in a saturated solution? Give your answer in scientific
notation to two decimal places. And the answer options are (A) 2.90
times 10 to the negative nine moles per dm cubed. (B) 5.02 times 10 to the negative
nine moles per dm cubed. (C) 6.60 times 10 to the negative
nine moles per dm cubed. (D) 5.95 times 10 to the negative
12 moles per dm cubed. Or (E) 2.58 times 10 to the
negative 11 moles per dm cubed.
We are given a solubility product,
or Ksp value, of 1.9 times 10 to the negative 33. This is for aluminum hydroxide or
Al(OH)3. And we are asked to find the
concentration of aluminum three plus ions in a saturated solution. Now aluminum hydroxide is sparingly
soluble. It only partially dissolves in
water to give Al3+ ions and OH− or hydroxide ions according to this balanced
equation. We are told that the solution is
saturated, so we know that it is at equilibrium. We can write the Ksp expression for
this sparingly soluble salt. Ksp or the solubility product
constant is equal to the molar concentration of the products multiplied with each
other. So we can write both ions in square
brackets.
Now the stoichiometric coefficient
of Al3+ is one. And so we raise the concentration
of Al3+ to the power of one. And the stoichiometric coefficient
of the hydroxide ion is three. So we raise the concentration of
the hydroxide ion to a power of three. Now we are looking for the
concentration of the Al3+ ion. So putting in some data that we
know, specifically the Ksp value of 1.90 times 10 to the negative 33 — and by the
way, in this example, we are not given the units of Ksp though Ksp does have units —
and knowing that according to the balanced equation for every one mole of the solid
ionic substance, one mole of aluminum ions is formed and three moles of hydroxide
ions.
So if we make the concentration of
Al3+ ions equal to 𝑥, we could even write let the concentration of Al3+ equal 𝑥,
then the concentration of hydroxide ions must be three 𝑥 because of the
stoichiometric coefficients. You can write this information down
on the side if you wish. We must remember to raise the
concentration of hydroxide ions three 𝑥 to a power of three. You don’t have to expand three 𝑥
to the three, but I have here just to make it clear what numbers we are dealing with
and to do our calculation correctly. Multiplying everything out, we get
27𝑥 to the power of four. Let’s solve for 𝑥. So simplifying, we get 1.90 times
10 to the negative 33 divided by 27 equals 𝑥 to the four. To get rid of the fourth power, we
take the fourth root of both sides, and this gives 𝑥 equals 2.90 times 10 to the
negative nine.
Now 𝑥 is equal to the
concentration of Al3+, which is what the question was asking for. Since the concentrations of the
ions in a Ksp expression are given as molar concentrations, we can write the unit
for 𝑥 as moles per dm cubed. The answer is in two decimal places
and in scientific notation. So the concentration of Al3+ ions
in the saturated solution is 2.9 times 10 to the negative nine moles per dm
cubed.
Let’s summarize everything we have
learnt. Ksp is the solubility product
constant used for sparingly soluble ionic compounds which dissolve in aqueous
solution. Ksp is an equilibrium constant. For the general equilibrium
equation shown here, we get the following Ksp expression. Ksp is equal to the concentration
of ion product C raised to the power c multiplied by the concentration of ion
product D raised to the power d.
Ksp concentrations are molar
concentrations. We learnt that the more soluble an
ionic substance is, the higher its Ksp value and that Ksp can be calculated for an
insoluble salt because nothing is truly insoluble in reality. And finally, we learnt that if the
concentration of the products multiplied by each other and raised to their
respective coefficients is not equal to Ksp, then the system is not in
equilibrium.