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## Solubility equilibria

Current time:0:00Total duration:9:28

# Solubility from the solubility product constant

## Video transcript

- [Voiceover] The goal is
to calculate the solubility of copper II hydroxide. We're given the solubility
product constant KSP, which is equal to 2.2
times 10 to the negative 20 at 25 degrees Celsius. So let's conceptualize this problem first. Let's say we have some copper II hydroxide which is blue, so let's say we
put some copper II hydroxide, some solid copper II
hydroxide into a beaker containing water. This is a slightly soluble ionic compound. So not everything we put into the beaker is going to dissolve. Let's say only a small
portion of this dissolves, I'm gonna take my eraser here and I'm gonna take off a
small bit of our solid there at the top and let's say that
small amount turns into ions. So what ions would we have in solution? Copper II tells us we have
copper II plus ions in solutions, so CU two plus and then we
also have hydroxide ions in solution OH minus. So we'd have some hydroxide
ions in solution two. Eventually we reach equilibrium right? So we have a solubility equilibrium where the rate of dissolution
is equal to the rate of precipitation. So let's go ahead and write that out. What's the chemical formula
for copper II hydroxide? We could use this simple little trick here of crossing over your
charges to figure out the chemical formula is
CU with parenthesis OH and a two here. That's our solids and we also have our ions. So CU two plus or ions in solution, we also have hydroxide
ions in solution, OH minus. We need to balance this so we need a two in front of the hydroxide
and everything else here would get a one. Alright so let's set up an ice table. So we have our initial concentration, our change, and then
finally our concentration at equilibrium. Well before that small
amount of copper II hydroxide dissolves, right so that small
amount that I erased earlier, we didn't have anything
for the concentration of our ions in solution. So that's our initial concentration
of our ions, it's zero. Now let's think about the small
amount of copper II hydroxide, the solid that dissolved. Alright so let's say that x
is equal to the concentration of copper II hydroxide that dissolves. So we're going to lose a concentration of copper II hydroxide
which we'll say is x. Look at your mole ratios
for every one mole of copper II hydroxide that dissolves we get one mole of copper
II plus ions in solution. So for losing, if we're
losing x for the concentration of copper II hydroxide
we're going to gain x for the concentration of copper
II plus ions in solution. And for hydroxide ions, for every one mole of copper II hydroxide that dissolves we get two moles of hydroxide ions. Alright so instead of x it'd be 2x. Alright so we're going to
gain 2x for the concentration of hydroxide ions. So at equilibrium, right our
equilibrium concentrations of our ions would be x for copper II plus and 2x for hydroxide. Alright let's write our
equilibrium expression, right? So KSP is equal to, we
look at our products, right so we have CU two plus,
we put the concentration of CU two plus and we
raise the concentration to the power of the coefficient, and here our coefficient is a one. So raise this to the first power. Next our other product here
would be the hydroxide ions, so OH minus and we
raise that concentration to the power of the coefficient
which in this case is a two. So we need to put a two
here and once again we leave this pure solid out of our
equilibrium expression. Alright let's plug in for KSP, the solubility product
constant was given to us it's 2.2 times 10 to the negative 20. So let's plug that in, so this is equal to 2.2 times 10 to the negative 20 and this is equal to the
concentration of copper II plus ions at equilibrium which is x. So we put that in, this
is x to the first power times the concentration of
hydroxide ions at equilibrium raised to the second power. So this would be 2x and then
we have to square it here. And this is where some students
get a little bit confused because if they say well you're
doubling the concentration here, right, and then you're squaring it aren't you kinda like
doing the same thing twice? But remember, these are
two different things. This 2x is because of
the mole ratios, right and we raise it to the
power of the coefficient because that's what you do
in an equilibrium expression. Alright so those are two different things, we're not doing the same thing twice. Alright, when we do our algebra on the right side we would
have x times 4x squared. So that's equal to 4x
cubed and this is equal to 2.2 times 10 to the negative 20. So we need to divide that by four, so we need to divide 2.2
times 10 to the negative 20 by a four, so you could
do that in your head or on the calculator, 2.2 times 10 to the negative 20. Alright we divide that by four and we get 5.5 times 10
to the negative 21st. So we have 5.5 times
10 to the negative 21st is equal to x cubed. Alright so to solve for x we
need to take the cube root of 5.5 times 10 to the negative 21st and unfortunately on this calculator it's a little bit trickier
than on most calculators. In most calculators it's
pretty straight forward and it's pretty easy to do. So let me show you one
way to take the cube root of something on this TI-85 here. So we would put in,
we'll take the cube root, so we put a three in here
and then one way to find this is to go to 2nd catalog and then just move upwards here until
you see the symbol. Alright so I don't see it yet and -- There it is, so right there
is the symbol we want. So we're trying to take the cube root of, we want 5.5 times 10 to the negative 21st and that should give us the cube root, which is equal to, let's
go ahead and round that to 1.8 times 10 to the negative seven. So this is equal to, x
is equal to 1.8 times 10 to the negative seven, and this would be the concentration, right,
this would be molar, this would be the
concentration of copper II plus at equilibrium, right,
let's go back up here. So x is equal to the
concentration of copper II plus at equilibrium and notice
it's also equal to the molar solubility of copper II hydroxide, right? That's how much copper II
hydroxide dissolved, x. So we found the molar solubility
of copper II hydroxide. Our question asked us for solubility, so maybe they meant molar
solubility in which case we're done or maybe they meant
solubility in grams per liter. So let's go ahead and do that now. So this is equal to, this is
equal to the molar solubility. This is the molar solubility, which is moles over liters. What if they wanted grams over liters? Alright, you would need
to have the molar mass of copper II hydroxide. Alright so you could look that
up on your periodic table. So copper II hydroxide has a molar mass of 97.57 grams per mole. So our answer here for molar solubility, this would be moles per liter. So if we wanna go to
solubility in grams per liter, let's look at our units and
see what we'd have to do. We have 1.8 times 10 to the negative seven moles over liters. Alright if you look at the molar mass, alright if you wanna
get to grams over liters all we have to do is
multiply the molar solubility by the molar mass because
the units for the molar mass are grams over moles. And if we multiply, the
moles cancel, right, and we'll end up with grams over liters. So let's go ahead and do
that calculation here. So we have, we rounded
this to 1.8 times 10 to the negative seven,
with it's molar solubility to get to the solubility
in grams per liter we multiply that by 97.57
which is the molar mass of copper II hydroxide, and we get, if we round that to 1.8 times 10 to the negative five. Alright so this is equal to 1.8 times 10 to the negative five, and this would be grams over liters. So this is the solubility, in one liter of solution
you could only dissolve 1.8 times 10 to the negative five grams. So copper II hydroxide is
not very soluble at all. Alright so that's how to
figure out the solubility if you're given the solubility
product constant KSP.