And at the end of lecture on Wednesday we
talked about the multi-step process that
takes place when you put zinc hydroxide
in solution, and then add ammonia.
So, are you guys gonna talk
the whole class period today?
Okay, thank you.
So, we talked about going through and
thinking about what could happen
with a complex ion formation.
Now the first question everyone asks is,
how do I know that zinc
forms that complex ion?
What do we, how can I tell,
I don't know enough about
the transition metal chemistry yet to even
know if a complex ion is going to form.
And the thing that is going to give us
the evidence that a complex ion will form,
is going to be your KF table,
which is your formation constant.
Just like we look at particular salts,
like you have sodium chloride,
there's no Kf value for that.
The reason for
that is because it's completely soluble.
So any salt that has a Ksp value,
we know it's gonna be insoluble,
at least according to our Chemistry
121 definition, same thing happens for
complex ion formation.
If you see a complex ion on the KF table,
we know that a complex ion's gonna form.
So when I tell you,
you have Zinc two plus in solution and
you put ammonia as a reagent in there,
then you're gonna know that
a complex ion's gonna form.
And the only way you can tell
is if you look at the k f table,
and if you don't look at that table you're
never gonna get this problem right.
So on the k f table is this
complex ion that forms so we know.
That zinc is gonna form a complex ion
when ammonia is interacted with that.
So make sure you're familiar with
the table that is given in the lab manual.
And when you're taking an exam, like I
said, you want to look at that first.
So we went through and
we showed that there was evidence
that a complex ion would form.
And then the next question is, how can
we approximate the concentration, or
the solubility when we add a zinc
solution and ammonia together?
So the question we're gonna start with is
what is the solubility of zinc hydroxide.
In 15 molar NH3.
So, we don't have an aqueous
solution anymore,
the solution is completely different.
This is gonna tell us, and if we look at
our KF table that a complex ion can form.
We showed that overall reaction at
the end of lecture on Wednesday,
and it's written here.
Remember, this is a two step process.
The first thing that happens is the zinc
hydroxide forms in equilibrium.
With zinc two plus ions and Minus ions.
The zinc two plus ions are then gonna
interact with the ammonia to form this
complex ion.
So how can we look at
the various steps of this and
how can we set up a problem to solve?
Again, we're gonna use
an ice table the zinc
hydroxide is a solid,
does not show up in our ice table.
The ammonia we have an initial
concentration of 15,
and before we add any of the ammonia or
initially.
We're gonna have a concentration of 0 for
the complex ion and 0 for
the When we allow this reaction or
this equilibrium to be established,
the changing concentration is gonna
be +x for the complex ion, +2x for
the Ion, and minus 4x for the ammonia.
This gives an equilibrium
concentration of 15 minus 4x for
our ammonia, x for
the complex sign and 2x for the hydroxide.
From the previous notes on Wednesday,
our K expression is equal to the KSP,
times the KF, because that's
the two steps that are occurring.
If we multiply those two together,
we get 1.3, times 10 to the minus 7.
So this is going to be our K for
this overall reaction.
The K expression that we can set up
will be the equilibrium concentration
of the ZN, NH342+ complex ion,
times the concentration of Iron squared.
This is divided by the equilibrium
concentration of NH3 to the fourth power,
and this is going to equal
1.3 times 10 to the minus 7.
So we can now plug in our equilibrium
concentrations in order to solve for x.
So in this particular case our complex
ion concentration is gonna be x.
The concentration of
minus is gonna be two x.
We need to square that and
the concentration of the ammonia
is gonna be 15 minus 4x to the 4th power.
This is gonna give 1.3
times 10 to the minus 7
which is our equilibrium constant for
the overall reaction.
Know a lot of these
problems are tricky and
they involve high numbered exponents.
So when I say to calculate
the concentration of ammonia.
We are going to be using
approximations in every case.
It doesn't matter if they don't follow the
same rules as they did in chemistry 122.
But for these complex ion questions,
I would never give you
an expression with 15 minus 4x,
expect you to raise that to the fourth
power, and try to solve that polynomial.
Okay?
It's just, the math is just too
complicated and I don't want you to
deal with that math in this class.
Okay?
Especially when you're using
a TI30 calculator on an exam.
So, what we're gonna do is we're
gonna approximate that 15 minus 4x
is gonna be roughly equal to 15, so
we can ignore this part of the expression.
That makes the math so much easier.
So, if we go through and
calculate this, our X is gonna equal
0.118 molar, and again,
that's a rough approximation.
So what we can say about this value for
x right here, is that if we assume,
X moles of zinc hydroxide,
which is ZnOH2 dissolve
than x is the molar solubility
of zinc hydroxide.
In 15 molar ammonia.
So, this is gonna be drastically different
than what we see in aqueous solution.
So, If I come back and
we kind of compare,
if we look at our experimental results.
If we went into the lab and
performed an experiment,
we would show that 8.9 times ten to
the minus seven grams of ZN OH2 will
dissolve in four millilitres of water.
Zinc Hydroxide has a very,
very low molar solubility in water,
we're not gonna get many grams
to dissolve in solution.
But if we compare that to
the result that we just calculated,
we can say that 0.0469
grams of zinc hydroxide or
ZnOH2 will dissolve In four millilitres.
Of 15 molar NH3.
What I wanna point out here Is
that these two values right here,
are a dramatic difference.
And this dramatic difference can be
attributed to complex ion formation.
So this shows us that we
can really manipulate,
and we can control the solubility
by using this complex
ion formation, and
using it to our advantage.
So in certain cases when you
looked at your group one analysis,
you need to take that silver complex,
and you have silver chloride,
which really doesn't want to
dissolve in anything, but
if you add concentrated ammonia, it will
form a complex ion and it will dissolve.
Okay.
There's other cases through out
the qualitative analysis scheme, and
you're going to notice that particularly
with the silver complex, we used a lute
AGCL because we want to try to separate
the silver from all the other ions.
If we would add concentrated HCL, the
complex ion would form much more readily.
So that's one of the things to look at,
and
when you're performing the experiments, or
when you're performing the steps in this
particular laboratory, keep it in the back
of your mind which effect is happening?
When is the complex ion forming and
how is that allowing us to
manipulate the solubility?
Another very common question that
will get involving the complex ions,
is that a complex ion is a nice way to
remove a particular ion from solution.
Because in this case,
Zinc2 plus is completely different
than the complex ion that
forms that is ZnNH322 plus.
So we can effectively remove ions from
solution using this complex ion formation.
