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Current time:0:00Total duration:8:15

- [Voiceover] Our goal is to
calculate the molar solubility of lead two chloride, which is a slightly soluble compound and we're given the solubility
product constant, Ksp, which is equal to 1.6
times 10 to negative five at 25 degrees. This time now, we're starting out with a .1 molar solution
of potassium chloride. If we have a beaker here and we have a .1 molar
solution of potassium chloride that means the concentration
of potassium ions in solution is .1 molar and the concentration of
chloride in ions in solution is also .1 molar. We have some solid lead two chloride, right, so this is suppose to represent our solid lead two chloride and what our problem wants us to do is to figure out how much
of our lead two chloride will dissolve, if we add it to our solution
of potassium chloride? So some of the lead two
chloride is going to dissolve and eventually reach equilibrium. So let's go ahead and write that out here, we have solid lead two chloride and some of it's gonna
dissolve and give us lead two plus ions in solution and we would also get
chloride and ions in solution so we would get Cl minus. When you balance this, I need a two in front of our chloride anions and everything else would get a one. So we first think about
the initial concentration of our products. So what's the initial
concentration of Pb two plus? Before we've added our lead two chloride, nothing has dissolved yet so we don't have anything
for our concentration of lead two plus. So nothing's dissolved. Before the lead two
chloride has dissolved, we don't have anything
for the concentration of chloride anions from lead two chloride but we do have a concentration
of chloride anions from our .1 molar solution of KCl. We're starting with an
initial concentration of .1 molar for our chloride anions. Next, we think about the change so some of our lead two
chloride is going to dissolve and let's make that concentration x so we're going to lose a
concentration of lead two chloride. For every one mole of lead
two chloride that dissolves we get one mole of lead
two plus ions in solution. So if we lose x for lead two chloride, we gain x for lead two plus and for every one mole of lead
two chloride that dissolves we get two moles of chloride anions. So we're going to gain two x
here for our chloride anions. Our common ion for this
problem is the chloride anion because we have two sources. One source was from our potassium chloride and one source was from
our lead two chlorides. So at equilibrium, our concentration of our
products would be zero plus x for lead two plus or x and .1 plus two x for chloride anions so this is equal to .1 plus two x. Next, we write our equilibrium expression, so our solubility product constant Ksp is equal to concentration of our products so concentration of Pb two plus raised to the power of the coefficient so our coefficient here is one so this is to the first power and we multiply that by the concentration of chloride anions in solutions, Cl minus. We raise that to the
power of the coefficient, our coefficient is a two so we raise that to the second power. Next, we can go ahead and plug in our solubility product constant, so Ksp is equal to 1.6 times
10 to the negative five so 1.6 times 10 to the negative five is equal to the concentration
of lead two plus ions at equilibrium, which we
already figured out was x so this is x to the first power, times the concentration of
chloride anions at equilibrium which is .1 plus two x squared. So we have .1 plus two x squared. Next, we have to solve for x. All right, so on the left side we have 1.6 times 10 to the negative five is equal to x to the first. Over here .1 plus two
x, we could do that math but to make our lives easier we usually make the assumption
that x is really small compared to .1. If x is really small compared to .1, .1 plus two x is
approximate the same as .1. Let's go ahead and put in .1 so we have .1 instead of .1 plus two x. .1 squared is .01. So in the right side we have x times .01 and this is equal to 1.6 times 10 to the negative five. When you solve for x, x is equal to .0016 molar and so this would the concentration of lead two plus ions in solution. If we go back up here, we can see that lead, the concentration of
lead two plus ions is x so that's our concentration and that's also the molar
solubility of lead two chloride, that's how much dissolved. Let's go ahead and write that in here. Our goal is to calculate
the molar solubility and it was equal to .0016 molar. Now, keep in mind this
is the molar solubility for lead two chloride in our solution of potassium chloride. This is in potassium chloride. In the video on solubility
product constant we found the molar solubility
when it was just pure water. If you look at that video, we
calculate the molar solubility to be .016 molar so this was in pure water so we didn't have any
potassium chloride here. Notice, what the addition of
the potassium chloride did to the molar solubility, it decreased the solubility
by a factor of 10. If you compare these values here so the solubility in pure water was .016 and now the solubility has decreased, it's .0016. So 1/10 the original solubility and this is due to the
presence of a common ion and our common ion was the chloride anion. According to Le Chatelier's principle, a system that's disturbed from equilibrium will shift its equilibrium
position to relieve the applied stress, so we can use that idea in this situation. If we think about our reaction
of being an equilibrium, you increase the concentration
of your product here, if you increase the concentration
of your chloride anions that's our stress. Our stress is increase
concentration of a product. The equilibrium shifts
to decrease the stress so the equilibrium shifts to the left and the excess chloride ions that we have, combine with lead two plus ions to form more of our solid. Therefore, we decrease the solubility of lead two chloride due to
the presence of our common ion. Notice that Ksp doesn't change, Ksp is still 1.6 times
10 to the negative five but the molar solubility has been affected by the presence of our common ion. So a common ion decreases the solubility of our slightly soluble compounds. In this case, by a factor of 10. We can use this concept
in a laboratory separation so let's say that we had, let's say we had some
solid lead two chloride and our goal was to
isolate all of the solid. So before we filtered it, we could add a common ion, we could add a source of chloride anions. Due to the common ion effect that decreases the solubility
of lead two chloride which means we are gonna
get more of our solid because our goal is to isolate as much of our solid as possible. So that's one use for
the common ion effect in the laboratory separation.