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In this lecture, we're going to continue our

study of these neutralization reactions, in which we

mix together an acid and a base and

examine what the properties of the solution are.

Remember up to this point, this is what we've discussed.

We'll take the base B and the acid HA, and we'll mix them together.

And we've always assumed up to this point, equal moles of acid and base.

Under those circumstances what we've observed is if we

meant to get, mix together, equal members of moles of

strong acid and strong base, then the solution is in fact neutralized.

But if we mix together equal moles of

strong base with weak acid, the solution is basic.

Or equal moles of strong acid with weak base the solution is acidic.

What we'd now like to do is actually go back and test the

consequences of not assuming that they were equal moles of acid and base.

So here is an example, based upon on what we were looking

at before.

We're going to mix together nitrous acid and

sodium hydroxide, just like we did before.

We're going to be, mix unequal numbers of moles of these two substances.

Let's count the number of moles of the nitrous acid that we begin with.

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It is equal to, let's see we have taken 0.5 litres and multiplied

by 0.10 molar, so we

have 0.05 moles, of the nitrous acid. If

we take the number of moles of the ammonia, I'm sorry, sodium hydroxide,

it we've taken 0.25 liters, multiplied by

0.10 moles, gives us a total number of

moles of 0.025 moles. If we compare these two, then, we put in

more moles of a weaker acid and fewer moles of a stronger base.

So it's reasonable to ask the question, simply, which one of those wins?

Is it more important that the base is strong or is

it more important that we put in more moles of the acid?

So is this solution going to be acidic, basic, or neutral when

we carry out this particular reaction

between the hydroxide and the nitrous acid?

Remember when we discussed much earlier the idea of the limiting reactant,

you know, which one of these are we going to run out of first?

Clearly, in this particular case when the reaction takes place, we're going to run

out of the sodium hydroxide before we run out of the

nitrous acid.

Consequently, it is the limiting reactant, and looking at

the reaction above, the number of moles of nitrous

and anion that we produced is clearly limited by

the amount of sodium hydroxide we put into the solution.

And therefore, is equal to 0.025 moles. And

correspondingly, the number of moles of nitrous acid still in the solution

is the amount of nitrous acid which reacted, which is

equal to the amount of sodium hydroxide in the solution.

Subtracted from the amount that we started out with.

So it would be equal to 0.05 moles minus 0.025 moles

which turns out to be equal to 0.025 moles as well.

Now, we need to be careful in calculating the concentrations of these materials

in solution.

So when we do this, we'll calculate the nitrous anion concentration

as being 0.025 moles divided by the volume of the solution.

Let's check the volume of the solution, it's going to

be 750 mL because we mixed together two solutions.

So, that's 0.750 litres and if we take that and just plug

it into the calculator we get a concentration of 0.033 molar.

And, likewise the concentration of the nitrous acid in solution

is equal to 0.33 molar. So I have a

particularly interesting solution here, in which we have in the solution a

combination of the acid and its

conjugate base mixed into the solution at the same time.

So, what are the properties of such a solution?

We try to be very interesting.

Let's try to determine now what we think the pH of this particular solution

will be so we can answer the question whether it's acidic basic or neutral.

And then test that hypothesis.

What we'll then do is take our usual reaction,

nitrous acid plus water. Gives us the hydroxide ion concentration

plus the hydroxide plus the nitrous anion. That's our reaction in our rice table.

Our initial conditions now before we become to equilibrium

are that the concentration of the nitrous acid is 0.033.

The concentration of the nitrous anion

is 0.033 and we'll approximate that the concentration

of the hydronium ion for now is zero.

As we change towards equilibrium we'll assume that

we're going to shift in and out of the

nitrous acid towards the nitrous anion by minus x

that will produce x moles per litre of hydronium.

And x moles per litre of the nitrous anion.

And will reach equilibrium, with concentrations

0.033 minus x, x, and .033 plus x.

We now put those in to our acid ionization equilibrium constant.

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To give us an equation which we can solve, to

find the value of x, which is the hydronium ion concentration.

We'll remember from a previous slide that the value of this equilibrium

constant is 5 times 10 to the minus 4. If that's the

case, then we now have 0.033 plus

x multiplied by x divided by 0.033

minus x is equal to 5 times 10 to the minus 4.

And now we have an equation that we need to solve for x,

to determine what the value of the

hydronium ion concentration is going to be.

We could go through the trouble to write this out as a quadratic equation, but we

could actually take an assumption here, in which x is assumed to be a small number.

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For example, If x is quite a bit

smaller than 0.033, and that's something we can test

in a moment when we solve the equation, then 0.033 plus x is really just the same

as 0.033 to a good approximation. And 0.033 minus x is also equal to x.

To that same approximation. Correspondingly, we ought to be able to

then write, as a good approximation, that Ka is now equal to 0.033

times x divided by 0.033. And that should be equal

to 5 times 10 to the minus 4. And if that's the case, then x

is itself equal to 5 times 10 to the minus 4.

Notice that 5 times 10 to the minus 4 is in fact dramatically smaller than 0.033.

Very substantially smaller.

Consequently, this is a good approximation.

And that means that this is a

good approximation to the answer to the equation.

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We can use this value of x because it's equal to the hydronium ion concentration,

according to our table up above, to calculate

what the pH of this particular solution is.

And the pH of the solution then turns out to be equal to 3.3.

So now we have an answer to our

question, the solution is in fact, acidic.

But it is more important that we have excess acid

in solution, because actually the base has been essentially fully neutralized.

There is no hydroxide still floating around from the sodium hydroxide,

it's a little hydroxide left in there from the auto-ionization of water.

But, overall, what we primarily have are the acid and its conjugate base.

Nitrous acid is a stronger acid than nitrous anion is a base.

And so the solution, turns out to be acidic.

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Those turn out

to have the very interesting property.

That they have relatively constant pH even when we add a

strong ace, a strong acid or a strong base to it.

For example if I were to modify this particular problem by instead of having it

the outset taking 0.50 litres of the nitrous

acid, and 0.025 litres of the sodium hydroxide.

Imagine we dumped in another 50 millilitres of this sodium hydroxide.

That's quite a bit of additional base. That would change this volume here

to 0.30 litres. Change the number of moles of sodium

hydroxide to 0.30. That would also change the amount of

nitrous, nitrous acid produced to 0.030. It would change the amount of nitrous acid

still available to 0.020.

We could go down here and actually modify the entries in the table quite quickly.

But in the end what we would discover, is that in this modified solution, if we were

to add in instead 50 millilitres of 0.1 molar solution, it

turns out in that particular case the pH becomes 3.4.

Whereas just

a little while ago, sorry, it's actually 3.5.

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Whereas a little while ago, we determined that the solution actually has pH 3.3.

The addition of 50 millilitres of concentrate, of fairly concentrated

sodium hydroxide solution, has only slightly changed the pH of the solution.

That's a property of a buffer solution, and

it's in fact what gives them their name.

That apparently changes in pH are buffered in these solutions, by the

presence of both the acid and the base at the same time.

Why would this be true?

Well, one way to understand that is actually

to consider it in terms of Le Chatelier's principle.

Let's re-write our reaction here again. H, N, O, 2.

Plus H2O goes to NO2 minus,

plus H3O plus, oops, not on the screen.

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If I were to add strong acid to this particular reaction.

That strong acid would react with the nitrous

anion and be consumed by it, neutralized by it.

Removing some of this material from the solution will cause

the HNO2 to shift back over creating some additional hydronium ion.

But buffered against it, to a certain extent, by just the fact

that, as we have added, say, the extra acid,

the solution responds to try to consume that acid.

Or, if I were to add base to this solution it will neutralize a certain amount of the

HNO2, causing the reaction to shift from right to

left, consuming some of the hydronium, but balancing those consumptions.

In the way that Le Chatelier's principle says

that we will always respond to shift the

stresses away from the react, the change that we have made as we go to equilibrium.

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That we made the approximation that the concentration of the acid is

about the same at equilibrium as it was when we started off.

Because we assumed that x was tiny compared to the initial

concentration of the acid, and it turned out that it was.

X was quite small as we looked down below.

And likewise the concentration of the anion we

assumed was the same at equilibrium as it was initially because x is quite small.

If that's the case, then we might take

each of the concentrations here of the anion

and the acid and approximate that they are

the same, at equilibrium as they were initially.

The expression we've written right now is exact.

We're now going to make an approximation to that equation

by replacing the concentration of the anion with its initial concentration.

And the concentration of the acid with its initial concentration.

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Now let's just move this term to the other side of the equation.

The pH of a solution, of a buffer solution is

equal to the P Ka of the con, of the weak acid, which is in that solution plus

the logarithm base 10 of the concentration of the

anion that we began with. Divided by the concentration of the

acid that we begin with. This particular equation goes by a name.

It's called the Henderson-Hasselbach Equation.

I've prepared it up here on the screen

so that hopefully, it's a little easier to read.

But notice one of the consequences of this equation

is that first, if the an-ion concentration and the

acid concentration are the same, then this ratio is one.

The logarithm base ten of one is zero, and the pH will become equal to the P Ka.

In other words, the pH of, a buffer solution in

which we have equal amounts of the acids and base.

Can be predicted just by looking a the P Ka.

Secondly, any changes

that we make in either A minus or HA, result not in

changes in this ratio in the pH, but rather only in the logarithm.

And as a consequence, the pH will vary only slightly as

we say add some acid, causing the base to shift to acid.

Or add some base causing the base the acid to shift to base.

Then in either of those circumstances the pH will remain relatively constant.

Buffer solutions have wide applicability in chemistry.

Most notably in bio chemistry and bio molecular systems.

Where balancing the pH and maintaining a relatively constant

pH can be important for biophysical processes, for example.

And this lecture helps explain why buffers have the property that they do.