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## Miareczkowanie

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# Miareczkowanie słabego kwasu mocną zasadą

## Transkrypcja filmu video

- Here we have a titration
curve for the titration of 50 milliliters of 0.200
molar of acetic acid, and to our acetic
solution we're adding some 0.0500 molar sodium hydroxide. So once again we're
putting pH in the Y axis, and down here in the X axis
is the milliliters of base that we are adding. So in Part A, what is the pH
before you've added any base? Right, so the addition of
0.0 mLs of sodium hydroxide. So we only have to think about a weak acid equilibrium problem here. So our weak acid is acetic acid, right? And it's in water, so acetic
acid donates a proton to H2O, and so H2O turns into H3O plus. And if you take away a
proton from acetic acid, you're left with CH3COO
minus, the acetate anion. Our initial concentration of
acetic acid is 0.0200 molar. Alright, so this is 0.200 molar here, and we're pretending like we don't have any of our products yet, right? So this is what we did in the video on weak acid equilibrium. So the change, alright, whatever we lose from our concentration of acetic acid, since acetic acid turns
into acetate, right, we would gain for the
concentration of acetate. And we would also therefore gain for the concentration
of hydronium, H3O plus. So at equilibrium, right,
our equilibrium concentration would be 0.200 minus X, and then over here we'd have plus X, and over
here we would also have plus X. So when we write our Ka expression, our equilibrium expression, Ka is equal to concentration of products over reactants. And so I'll just go a little bit faster since we've already done
a video on all of this, and so make sure to watch the
video on weak acid equilibrium before you even look at these. Alright, so this would
be X, right, times X. So that's these two X's over here. So our products over our reactants. And that would be 0.200 minus X, leaving water out of our
equilibrium expression. So this would be 0.200 minus X. If we assume that X is a
really small number, right, a very small concentration
compared to 0.200, we can approximate it and
say that 0.200 minus X is about the same as 0.200. So we could rewrite this. Let's rewrite it over here. So we have X squared now,
over 0.200 is equal to Ka. And Ka for acetic acid is 1.8
times 10 to the negative 5. So we can go ahead and solve for X. So let's get out the calculator here. So we have 1.8 times 10 to the negative 5, and we're gonna multiply
that by 0.200, right? So we get 3.6 times 10 to the negative 6. And then we need to take
the square root of that. So the square root of
our answer gives us X, and we can see X is equal to .0019. So let's write that down. X is equal to 0.0019. And remember X represents
the concentration of hydronium ions, so this X right here, right, this X represents our concentration of hydronium ions at equilibrium. And so if this is the
concentration of hydronium ions, right, so this is equal
to the concentration of hydronium ions, find the
pH is very straightforward because we just need to
plug in the concentration of hydronium into our pH calculation. PH is equal to negative log of the concentration of hydronium. So we take this number, 0.0019,
and we plug it into here, and we can solve for the pH. And so let's go ahead and do that. So let's take out the calculator. So the negative log of
.0019 gives us the pH. And so we get 2.72. So the pH is equal to 2.72. So before we've added any base, alright, so 0.0 mLs of sodium hydroxide,
the pH should be 2.72. And so we go over here to
our titration curve, right, and that 0.0 mLs of
base, we go over to here, and so this is our first point, right? So this is A right here, and that should represent a pH of 2.72. Alright next, let's add
some more sodium hydroxide. Now in Part B, our goal is to find the pH after we've added 100.0 mLs
of our 0.0500 molar solution of sodium hydroxide. So let's see how many moles
of base we are adding. So if the concentration of
sodium hydroxide is 0.500 molar, that's the same concentration
of hydroxide ions in solution. So the concentration of
hydroxide ions in solution is 0.0500 molar, and molarity
is moles over liters, right? Molarity is equal to moles over liters. So this is equal moles
over, how many liters is 100 milliliters? We move our decimal place one, two, three, so that's point 1. So that's 0.1000 liters. Alright, so to find moles,
all we do is multiply. Alright, so 0.0500 times
0.1000 give us 0.005, so we're working with 0.00500
moles of hydroxide ions. Alright, so that's how many
moles of hydroxide ions we're adding. How many moles of acid did
we originally have present? Let's go back up here
to our titration curve so we can see those numbers. So we were starting with 50.0 millileters of 0.200 molar acetic acid. Alright, so let's figure
out how many moles of acetic acid we started with here. So the concentration was
0.200 molar, so for our concentration of acetic
acid, our concentration was equal to 0.200 molar,
and that's equal to moles over liters. We started with 50 milliliters,
which is 0.500 liters. So we multiply 0.200 by 0.0500, and we get 0.0100 moles of acetic acid. So that's what we're starting with here. So we have an acid and a base. Our acid is acetic acid, and
our base is the hydroxide ions. The hydroxide ions are going to neutralize the acid that's present. So we get a neutralization reaction. Let's go ahead and write what happens. Our acid is acetic acid, and
we're adding some hydroxides. So the hydroxide is going
to take the acetic proton from acetic acid, the
hydroxide is gonna take this proton right here. And OH minus an H plus,
of course, form H2O. So once hydroxide takes a
proton from acetic acid, we're left with the conjugate
base for acetic acid, which is, of course, the acetate ion. So we also make CH3COO minus here. Alright, what were we starting with? For acetic acid, right, we were
starting with 0.0100 moles. So let's go ahead and
write that, 0.0100 moles. And for hydroxide we were adding 0.00500. Alright, so we're adding
0.00500 moles of hydroxide. So the hydroxide ions
are going to neutralize the acetic acid, alright,
so we're going to lose all of our hydroxide. So all of it reacts. So all of this reacts here,
and so we're left with nothing. We use up all of our hydroxides, alright, and that much hydroxide
is going to neutralize the same amount of acetic acid. So we're gonna lose that much acetic acid. So 0.0100 minus 0.00500,
right, this is gonna give us 0.0050 moles of acid left over. So this is how much acid
was not neutralized. Alright, so if we lose
that much acetic acid, we're also going to gain that
much acetate anion, right? So if we start with zero over here, we're going to gain the same amount. So 0.00500 is how much we gain over here. So we're gonna finish,
after neutralization, with 0.00500 moles of acetate anion. Alright next, let's figure
out the concentration of acetic acid in the acetate anion. So we have moles for both,
alright, we have moles for both, but we need to find the volume. Well, let's think about that. So let's go back up here to the problem, and we're adding 100 milliliters, alright? And we started with 50
milliliters of our acid solution. Alright, so 50 plus
100 is 150 milliliters. Alright so now our new
volume is 150.0 milliliters, 50 plus 100, and that's
equal to 0.15 liters, right? We move our decimal place. So let's figure out our concentrations. We have moles, we have
volume, so let's find the concentration of acetic
acid now, after neutralization. Alright, so we have
0.00500 moles, alright, so let's go ahead and put
that in, 0.0050 moles. And our volume is 0.1500,
right, so 0.1500 is our volume. So we can go ahead and
do our calculations. I'll just take out the calculator here. So we have 0.005 divided
by .150, and we get .033. So the concentration is 0.033 molar. Alright, what about the
concentration of acetate? Alright, so the concentration
of acetate anions in solution, alright, would be equal to- Look at the numbers, they're the same. We have 0.00500 here, and once again, our total volume is .1500,
so it's the same calculation. 0.0050 over 0.1500, and we
would therefore also get a concentration of 0.033 molar. So we have equal
concentrations of a weak acid and its conjugate base. And that makes me think
about a buffer solution. So that's what we've formed here. As we drip base, as we drip hydroxide ions into our original acidic solution, we're slowly a buffer solution. And here we have equal concentrations. So our goal was to find
the pH, and since we have a buffer solution now,
it's easiest to just use the Henderson Hasselbalch equation. Alright, so the Henderson
Hasselbalch equation was pH is equal to the pKa,
plus the log of the concentration of A minus
over the concentration of HA. Alright, and the Ka for acetic acid, we talked about earlier, was
1.8 times 10 to the negative 5. So the pKa would just be
the negative log of that. So let's take out the calculator. Let's take the negative log of 1.8 times 10 to the negative 5. And so the pKa is equal to 4.74. So let's write that down here. PKa is 4.74. So we plug that in to our
Henderson Hasselbalch equation right here. So the pH is equal to 4.74, plus log of the concentration of A minus. A minus would be acetate,
right, which is 0.033, over HA. HA would be acetic acid, which is 0.033. So we have the same concentrations. So this is log of 0.033 over 0.033. And that of course is one. Alright, so we have log of one here, and you probably already know
what log of one is equal to. I'll do it on the calculator
so you can see that log of one is equal to zero. And so all of this is equal to zero. So the pH is equal to 4.74 plus, you know this would all be zero over here. So the pH is equal to 4.74. And this is the half equivalence point. We've neutralized half
of the acids, right, and half of the acid remains. And using Henderson Hasselbalch
to approximate the pH, we can see that the pH is
equal to the pKa at this point. So let's go back up here
to our titration curve and find that. Alright, so the pH is 4.74. The pH is 4.74 after we've
added 100 mLs of our base. So we go right up here to 100 mLs. And so this would be our second point. This is what we did in Part B. And the pH is approximately
4.74 at this point.