- [Instructor] For this problem, our goal is to figure out
whether or not a precipitate will form
if we mix 0.20 liters over 4.0 times 10 to the negative
third Molar solution of lead two nitrate,
with 0.80 liters of an 8.0 times 10 to the negative third
Molar solution of sodium sulfate.
The first step is to figure out the identity
of the precipitate that might form.
So we're mixing an aqueous solution of lead two nitrate
with an aqueous solution of sodium sulfate.
In the lead two nitrate solution they're lead two plus
cations and nitrate anions.
In the sodium sulfate aqueous solution,
there are sodium cations and sulfate anions.
So we take the cation from one and the anion from the other.
So one possible product would be lead sulfate.
So let's write on here, PbSO4, after we cross over
our charges and we take the other cation and the other anion
and so the other product would be sodium nitrate.
So we write it in here and a NaNO3.
To balance the equation, we need a two in front of NANO3.
Since nitrates are soluble, sodium nitrate is an aqueous
solution and lead sulfate would be our possible precipitate.
Now that we know are possible precipitate,
let's go ahead and write a net ionic equation
showing the formation of that precipitate.
So lead two plus ions, would come together with sulfate
anions to form PBSO4.
So PBSO4 is the possible precipitate.
Since lead sulfate is our possible precipitate,
we really only care about the concentration of lead two plus
ions and sulfate anions in solution.
We don't need to worry about sodium cations or nitrate
anions because those are the spectator ions
in our overall reaction.
Running the overall equation, and the net ionic equation
are really optional for a problem like this.
But we really need to do is identify the precipitate
and then write out the disillusion equation.
So PBSO4 would be our possible precipitate.
And if it dissolves in water, we would form lead two plus
cations in aqueous solution, and sulfates anions
in aqueous solutions, right?
aq over here.
The reason why it's important to write out the dissolution
equation is because we can write a KSP expression
from it.
So KSP is equal to, it would be the concentration
of lead two plus raised to the first power
because we have a coefficient of one
in the balanced equation,
times the concentration of sulfate also raised
to the first power and pure solids are left out
of equilibrium, constant expressions.
Therefore we're not going to include lead sulfate.
For lead two sulfate KSP is equal to 6.3 times 10
to the negative seven at 25 degrees Celsius.
The concentrations of lead two plus and sulfate in the KSP
expression, are equilibrium concentrations.
For our problem, we're gonna calculate QSP,
which has the same form as KSP,
the differences the concentrations can be
at any moment in time.
And we're gonna calculate QSP at the moment,
our two solutions are mixed,
and then we're going to compare QSP to KSP.
I've drawn out some diagrams to help us understand
how QSP compares to KSP and what that means
for the solution.
However, these aren't perfect diagrams
they're just to help get the point across.
If QSP is less than KSP, the solution is unsaturated,
which means no precipitate would form.
For an unsaturated solution,
you can dissolve more led two sulfate in it.
So lead two sulfate as a white solid,
so if we were to put a small amount of lead two sulfate
in our unsaturated solution, it would dissolve,
and it would continue to dissolve until QSP is equal to KSP.
And the system is at equilibrium.
At equilibrium, the solid is turning into the ions
at the same rate, the ions are turning back into the solid.
Since the rate of dissolution is equal to the rate
of precipitation when the system is at equilibrium,
the concentrations of lead two plus ions
and sulfate ions are constant,
and this represents a saturated solution.
And since the solution is saturated at equilibrium,
if we tried to add some more lead two sulfate
at the same temperature,
we wouldn't be able to dissolve any more,
we would just increase the pile of lead two sulfate
on the bottom of the beaker.
That concept helps us understand what happens when QSP
is greater than KSP.
When QSP is greater than KSP, the solution is oversaturated.
So it's exceeded the limit of what can dissolve,
and therefore you can imagine some lead two plus ions
combining with some sulfate ions to form a precipitate.
Therefore, when QSP is greater than KSP,
a precipitate will form.
The precipitate will continue to form,
until QSP is equal to KSP,
and the system reaches equilibrium.
Next, we need to go back to what we were given
in our initial problem,
when we mixed our two solutions together.
Remember we only cared about the concentration of lead
two plus ions and sulfate ions.
So we're gonna calculate the concentration of those two ions
at the moment in time, when the two solutions are mixed.
First let's calculate the concentration of lead two
plus ions.
The original solution of lead two nitrate
had a concentration of 4.0 times 10 to the negative
third Molar.
So molarity is equal to moles over liters.
So we can plug in the concentration,
and we can also plug in the volume of that solution,
which was 0.20 liters, and solve for X,
X equal to 8.0 times 10 to the negative fourth moles.
That's how many moles of lead two nitrate
there are and that's also how many moles of lead two
plus ions there are.
Therefore to find the concentration of lead two plus ions
after the solutions are mixed,
we plug in 8.0 times 10 to the negative fourth moles,
and for the volume we're adding these two solutions
together, so the total volume of the solution
is 0.20 plus 0.80.
So the concentration of lead two plus ions
will be equal to 8.0 times 10 to the negative fourth Molar.
We can do the same type of calculation
to find the concentration of sulfate ions
after the two solutions have been mixed.
So we take the concentration of the original solution
of sodium sulfate and plug that into the molarity equation,
plug in the volume solve for X and 6.4 times 10 to negative
third moles is how many moles of sodium sulfate there are.
That's also how many moles of sulfate ions there are,
so we plug in that number into the concentration
for sulfate, and once again, since we're adding
the two solutions together, we divide that by the total
volume to get a concentration of sulfate ions
of 6.4 times 10 to the negative third Molar.
Now that we know the concentrations of lead two plus ions
and sulfate ions, after the two solutions have been mixed,
we can plug those concentrations into our QSP expression
and solve for QSP.
So at this moment in time, QSP is equal to 5.1 times 10
to the negative six.
At 25 degrees Celsius, the KSP value for lead two sulfate
is equal to 6.3 times 10 to the negative seventh.
QSP at this moment in time is 5.1 times 10
to the negative six.
Therefore QSP is greater than KSP.
Since QSP is greater than KSP, we've exceeded the limit
of what can dissolve and therefore the solution
is oversaturated.
So yes, a precipitate will form and the precipitate
of lead two sulfate will continue to form
until QSP is equal to KSP.