>> Welcome to a screencast on acid-base titrations.
One of the most important methods of chemical analysis
that chemists use quite frequently is called a titration.
In the titration, we have a titrant that is gradually added
to an analyte, until we reach an end point.
And typically, the end point is something we will be able to see a color change occur,
so we have a clear, colorless solution, for example.
And the solution turns pink.
So this is actually an acid-base titration using an indicator called phenolphthalein,
that is colorless in acidic solution and sort of a pinkish-purple in basic solution.
And what happens is at the end point, more properly referred to as an equivalence point,
we've added enough base to fully neutralize the acid, and then we actually have a little bit
of excess because that's what's needed to see the color change.
And when properly done, titrations can be very, very accurate.
Now, when we do a titration, we typically know one concentration and measure two volumes.
So an example would be we know our volume of analyte, we don't know its concentration.
We know our concentration of titrant, and then we add a known volume, or measured volume,
to get to our end point where the color has changed, and then that allows us
to determine the concentration of the unknown substance, the analyte.
And we have a little bit of math to do.
A little bit of stoichiometric calculations, and there's different variations
of this depending upon what you're trying to find, and we'll do an example or two of this.
Our first example is of a sample of acid mined drainage from the Iron Mountain Mine,
just outside of Redding, California.
And what happens when mining is done, if the rocks that are cut into contain sulfur,
and that is exposed, then to the atmosphere and to water, sulfuric acid is produced
and the iron mountain mine was infamous for having the worst water in the world
and the reason it was designated the worst water in the world was because it was
so concentrated in sulfuric acid.
So here we have a situation where, let's say an environmental chemist is testing the acid mine
drainage to determine how concentrated the mine water is.
And we have a 25 ml sample of mine water, and it required 36.97 milliliters
of a .1834 molar sodium hydroxide solution for it to be neutralized.
Here are some researchers collecting samples at the Iron Mountain Mine,
and then here is a picture of the apparatus.
We have a 25 milliliter sample of sulfuric acid, 25.00 milliliters,
because we've used a volumetric pipette to measure it very accurately.
We titrate using the sodium hydroxide solution of known molarity, in this case, .1834 molar,
the sodium hydroxide solution has been standardized.
We know its concentration very accurately.
And then it took 36.97 milliliters of this solution to neutralize the acid and when
that happens, the solution goes from clear and colorless to a light pink.
And the goal is to get as light a permanent pink as possible.
We would like to not over-titrate by very much.
We always end up over-titrating a little bit.
We'd like to be by as little as possible.
And by doing this, we can find in this case, the concentration or the molarity of this solution.
Alright. So here is our problem.
And first of all, what is the reaction equation for the reaction
that takes place during this titration?
And sulfuric acid, H2SO4, reacts with sodium hydroxide to make sodium sulfate and water.
And when you write out and then balance the equation, you realize that there is 2 moles
of NaOH needed to neutralize 1 mole of sulfuric acid.
Note that this is an acid-base reaction, of course,
and it's also referred to as a neutralization reaction.
Okay, then we move on to the calculation.
And we have 36.97 milliliters of sodium hydroxide solution,
or .03697 liters of sodium hydroxide solution, and we've converted to liters
because we use the concentration of the sodium hydroxide solution, .1834 moles per 1 liter
of the solution to convert from liters into moles.
And then, since there are 2 moles of NaOH required to titrate or react with 1 mole
of sulfuric acid, a second conversion allows us to calculate the number
of moles of sulfuric acid in our sample.
And that turns out to be .003390 moles of H2SO4.
Now, we're not done yet.
The problem asked for the concentration or molarity of the sulfuric acid solution.
So what we do next is remember that molarity is moles per liter, moles of H2SO4 solution,
or moles of H2SO4 in, in this case, 25 ml, or .02500 liters of H2SO4 solution.
And when we do that calculation, we get .1356 moles per liter,
or we often write it .1356 molar, capital M, H2SO4.
So that problem has been solved.
All right.
Well, let's do one more example.
You might end up doing maybe a lab that is similar to this.
We have a sample of a monoprotic acid, that means, it has 1 hydrogen, or one proton,
because hydrogen ion is just simply a proton.
So it's 1 hydrogen in the acid, rather than the 2 hydrogens like in H2SO4,
and then it required 28.32 milliliters
of a .1956 molar sodium hydroxide solution to reach the end point.
And in this case, we're trying to find the molar mass of the acid, rather than the concentration,
because it's just a sample of solid.
Now, what we would do is we would actually dissolve the sample of the monoprotic acid
in some water, put that in a flask, do the apparatus as shown before,
and titrate with the sodium hydroxide, and again,
try to get to our light pink end point using phenolphthalein as our indicator.
And if we do this, and we calculate the molar mass, and we do a careful job,
then we might be able to identify the acid, especially if we have some choices of acid,
and we knew that our unknown monoprotic acid was one of a series of choices,
then if we get its molar mass, we should be able to determine
which substance, or which acid it probably is.
So, first of all, the reaction, monoprotic acid means there's just one hydrogen,
and it's an unknown acid, so it's H-something.
We don't know what the something is, yet.
But it's H-something.
It reacts with NaOH, and we make a salt and a something
and water, and it's a 1 to 1 stoichiometry.
So similar to what we did before, we take the volume of sodium hydroxide, convert to liters,
so .02832 Liters of sodium hydroxide.
We convert from liters to moles for the sodium hydroxide using the molarity.
And then this one is 1 to 1 stoichiometry, so 1 mole of NaOH reacts with 1 mole of HX,
and that will then give us our moles of HX in this sample of our acid.
And now, remember, we're not trying to find moles of HX, we are trying to find Molar mass.
But if you remember what Molar mass is, it's how many grams there are per mole of that substance.
And we now know both the grams of our sample, and the moles of our sample,
and grams to moles is what gives us molar mass, providing both numbers refer to the same amount
of substance, which they do in this case.
So 1.147 grams of HX, divided by .005539 moles of HX, gives us 207.1 grams per mole,
for the Molar mass of our unknown acid.
And now, if we had a list of acids and we knew our unknown acid was one of these,
then which ever one ends up having-- or whatever this Molar mass is closest to for,
with reference to the Molar masses of the other acids,
or the known acids, that's the one we would pick.
And that's it for acid-base titrations.