00:00:00
in this lesson we're going to discuss
00:00:02
acidbase titrations and this is a very
00:00:04
nice application of all the work that
00:00:06
we've been doing to calculate pH of
00:00:08
various Solutions we've calculated pH of
00:00:12
strong acid strong base Solutions we've
00:00:14
calculated pH of weak acid we weak based
00:00:18
Solutions buffers and of solutions after
00:00:22
some concentration of strong or acid or
00:00:26
base has been
00:00:27
added now you'll recall from your
00:00:29
studies in grade 11 that a titration is
00:00:32
a lab procedure where we are analyzing
00:00:35
an unknown solution we refer to this as
00:00:38
a volumetric analysis because we are
00:00:41
adding a certain volume of known
00:00:43
solution to a certain volume of unknown
00:00:47
solution and the goal is to determine
00:00:49
the concentration of an unknown solution
00:00:52
now what's important to note is that
00:00:54
titrations don't always have to be acid
00:00:56
based titrations we're going to see
00:00:58
later on in the year that we can do
00:01:00
redox titrations we can do precipitation
00:01:03
titrations so it's important to think of
00:01:05
this as a volumetric analyses procedure
00:01:08
rather than something that we use to
00:01:12
analyze acids and bases the bual rec
00:01:16
call is the glassware that we use to add
00:01:20
controlled volumes of the standard
00:01:23
solution or the titrant into our unknown
00:01:27
solution so the standard solution is the
00:01:29
one that is of known
00:01:32
concentration and you learned in grade
00:01:34
11 that there is something that we refer
00:01:37
to as a primary standard and the classic
00:01:40
primary standard in acidbase chemistry
00:01:43
is potassium hydrogen phate so it's
00:01:46
important that you understand that a
00:01:49
primary standard is one where we can
00:01:51
basically be really confident in its
00:01:54
concentration you'll recall that sodium
00:01:56
hydroxide is deliquescent or it absorbs
00:01:59
water so it's really difficult to
00:02:01
prepare a solution of known
00:02:04
concentration of sodium hydroxide
00:02:07
although it's a very useful tirant or
00:02:10
solution that we put in the buet in
00:02:12
order to study acids and bases the
00:02:15
acidbase indicator is the substance that
00:02:17
we put in order to help us to identify
00:02:20
the end point we're going to discuss
00:02:22
indicators in a little bit of detail
00:02:24
later on in the
00:02:25
lesson as you know the classic acidbase
00:02:29
indicator you've used so far is phenol
00:02:31
theine it's colorless in acid and it's
00:02:34
pink or Magenta in base now the end
00:02:37
point of the titration is the point
00:02:39
where we stop titrating so we no longer
00:02:42
add any of the titrant from the buet and
00:02:45
this occurs when we observe a color
00:02:48
change now the goal of completing a
00:02:50
titration is actually to determine the
00:02:53
equivalence point we're trying to find
00:02:56
the point where the two reactants are
00:02:59
sto metrically equal so the moles of
00:03:02
acid for example is equal to the moles
00:03:05
of base or H+ and O minus that is the
00:03:10
equivalence
00:03:12
point now we want to choose an indicator
00:03:15
that's going to change color as close as
00:03:18
possible to the end point because that's
00:03:20
what we're using to identify the
00:03:22
equivalence
00:03:23
point we're going to examine three
00:03:26
different types of acidbase titrations
00:03:28
we're going to look at the the titration
00:03:31
of a strong acid by a strong base we'll
00:03:34
look at the titration of a weak acid by
00:03:37
a strong base and we'll look at the
00:03:39
titration of a weak base by a strong
00:03:43
acid last year the focus was when you
00:03:47
created titration curves the focus was a
00:03:49
strong acid and strong base and you're
00:03:52
going to see from your Advanced
00:03:53
knowledge of acid based chemistry that
00:03:55
these ones the last two have a bit of a
00:03:58
difference and you're going to notice
00:04:00
that we don't do a titration with a weak
00:04:03
base and a weak acid the systems are far
00:04:05
too
00:04:07
complex and so they're not
00:04:11
ideal so this should be familiar to you
00:04:14
this is the picture of the titration
00:04:18
curve for a strong acid with a strong
00:04:21
base okay a few features of the curve
00:04:25
you can tell by looking at the pH that
00:04:28
we're starting with an a a so in this
00:04:30
case the titrant or the solution in the
00:04:33
buet is the base so we are adding base
00:04:37
in increments and we're measuring the pH
00:04:40
at each of those increments and graphing
00:04:43
the
00:04:46
results now when you have a titration
00:04:48
curve there is always a very steep part
00:04:52
of the titration curve and that always
00:04:53
occurs around the equivalence point the
00:04:57
point where the moles of acid is equal
00:05:00
to the moles of Base in this case we're
00:05:02
titrating A.1 M hydrochloric acid
00:05:06
solution with A.1 molar sodium hydroxide
00:05:09
solution we're starting with 25
00:05:12
milliliters of the acid and so the
00:05:15
equivalence point here occurs at 25
00:05:18
milliliters of base and the pH of this
00:05:21
solution is going to be
00:05:24
seven the reason it's seven is because
00:05:26
the products here are water and Salt N
00:05:30
ACL and as you know from your study of
00:05:32
salts that if a salt is produced from a
00:05:36
strong acid and strong base like NAC is
00:05:39
it will not affect the pH and therefore
00:05:41
the pH is
00:05:43
seven this point of the curve the
00:05:46
equivalence point is actually an an
00:05:48
inflection point so it's the point where
00:05:51
the curve changes from being a curve
00:05:53
that's increasing to a curve that is
00:05:57
decreasing and when we're studying in
00:05:59
these graphs in order to find the
00:06:01
equivalence point we actually need to
00:06:03
take the halfway point of the Steep
00:06:06
point of the curve okay so we take look
00:06:09
at the Steep part of the curve and the
00:06:11
halfway point is where the equivalence
00:06:13
point is which in this case is pH of s
00:06:18
now what we're going to do is look at
00:06:20
how we can put everything together that
00:06:22
we've used already to determine the pH
00:06:26
calculated at different points
00:06:28
throughout the titration
00:06:32
so let's look at the pH before the
00:06:34
titration begins we started with a
00:06:36
solution that is 25 milliliters in .1
00:06:40
molar hydrochloric acid hydrochloric
00:06:43
acid is the only solution present and
00:06:46
it's a strong acid so we know from this
00:06:50
that the concentration of the H+ is
00:06:52
going to be equal to the concentration
00:06:54
of the
00:06:55
HCL so for this particular solution the
00:06:59
pH H is simply the negative log of the
00:07:03
concentration of the H+ which is .1 okay
00:07:08
so initially the pH of the solution is
00:07:14
one which if we look at the table
00:07:17
associated with the Curve it
00:07:21
matches now let's look at the the
00:07:25
concentration of the H+ and subsequently
00:07:28
the pH
00:07:31
after a certain amount of Base has been
00:07:33
added we're talking about the region
00:07:35
before the equivalence point
00:07:38
here so we're dealing with a strong acid
00:07:41
and base and if we're before the
00:07:43
equivalence point we have a scenario
00:07:45
where we have a limiting reactant so
00:07:49
what we need to do is look at the
00:07:52
reaction that we have and remember
00:07:54
anytime you have a reaction with a
00:07:57
strong acid or a strong base so in this
00:07:59
case we've got both it's going to be a
00:08:02
complete reaction so we're going to have
00:08:04
a full Arrow here in order to organize
00:08:07
the data to see what we have left over
00:08:10
we're going to set up an ICF
00:08:13
chart now what we have is we've got a
00:08:16
volume of 35 milliliters for the total
00:08:21
because remember we started with 25
00:08:24
milliliters of the acid and we've added
00:08:27
10 milliliters of the base so a very
00:08:30
very important initial step in all of
00:08:33
the calculations that you're going to do
00:08:34
in this section will involve you
00:08:37
diluting the given Solutions because
00:08:40
during a titration we're not adding
00:08:42
solid quantities we're adding volumes of
00:08:47
solution so if you do the C1 V1 equals
00:08:50
c2v2 with the given 10mil and 25ml
00:08:55
solutions what you see is that you get a
00:08:58
concentration of sodium hydroxide which
00:09:00
is
00:09:02
0.285 and a concentration of HCL which
00:09:06
is
00:09:10
0.714 now in this case you can see that
00:09:12
the sodium hydroxide is limiting which
00:09:15
makes sense because we're before
00:09:19
the equivalence point and so the
00:09:22
limiting reactant is going to be
00:09:24
completely
00:09:26
consumed HCL is in a one one ratio with
00:09:30
the NaOH so the concentration of the HCL
00:09:34
that's left over here is
00:09:37
0.04
00:09:40
28 now when we were dealing with buffers
00:09:43
we would have needed to look at this
00:09:45
side of the
00:09:47
solution and look at the ions that are
00:09:50
put into the solution but because sodium
00:09:53
chloride does not affect the pH and H2O
00:09:56
is a pure liquid it doesn't affect our
00:09:58
calculations
00:10:00
there's no need for us to look at how
00:10:02
much nhcl is in this
00:10:05
solution now if we look at this what we
00:10:07
have left over in the solution at this
00:10:09
point is strong acid this is where the
00:10:13
key decision making comes in you always
00:10:15
need to be conscious that you're making
00:10:18
a decision that you've got a strong acid
00:10:21
the strong acid will dissociate
00:10:23
completely so that will put into our
00:10:26
solution a concentration of H+ at0
00:10:32
0.428 okay and as expected it's
00:10:35
decreased originally it was1 so it's
00:10:38
decreased a little bit so the pH
00:10:41
here is
00:10:46
1.36 okay so so far the pH has risen
00:10:49
from 1 to
00:10:53
1.36 and it's not too large of an
00:10:55
increase the beginning part of a
00:10:58
titration the pH changes
00:11:03
gradually let's look at just before the
00:11:06
equivalence point so when we've added
00:11:12
24.90 milliliters of the sodium
00:11:15
hydroxide solution which if you look at
00:11:17
the beginning it's 0.1 in its
00:11:21
concentration okay so right now our
00:11:24
total
00:11:26
volume is 49
00:11:33
9 so again we've got the reaction of
00:11:37
NaOH with
00:11:45
HCL okay I'm going to do my ICF
00:11:49
chart the concentration of sodium
00:11:52
hydroxide in the solution is 0.499
00:12:02
and the concentration of HCL is
00:12:11
0.0
00:12:16
501 and so the sodium hydroxide again is
00:12:19
our limiting reactant so it's going to
00:12:21
go to zero and the concentration of HCL
00:12:24
that I'm left with is 2 * 10 the4
00:12:29
so the pH still hasn't risen too too
00:12:33
much but it is making its way upwards
00:12:37
and the pH at that point is
00:12:41
3.9 oh sorry it's
00:12:45
3.7 put it
00:12:47
over okay so it's
00:12:55
3.7 now if we looked at the point
00:12:59
24.99 we would see that the pH would
00:13:02
rise to
00:13:05
4.70 at the equivalence
00:13:08
point with the addition of only 0.01
00:13:11
milliliters of Base the pH jumps to
00:13:15
seven and we've got a solution of just
00:13:21
NAC okay so small additions right around
00:13:24
the equivalence point do cause rapid
00:13:27
changes in PH
00:13:30
and it's really important to know always
00:13:32
that when you have the titration or the
00:13:34
reaction of a strong acid and strong
00:13:36
base at the equivalence point the pH is
00:13:39
going to be seven okay there's no
00:13:41
calculation necessary for that now if we
00:13:44
look at beyond the equivalence point so
00:13:47
when we've added
00:13:50
one very small quantity of sodium
00:13:54
hydroxide in excess so I'm talking about
00:13:58
25 5.01 milliliters of Base being added
00:14:03
at this point so my total volume is
00:14:11
50.01 at this point now the sodium
00:14:14
hydroxide is going to
00:14:17
be our limiting reactant okay and it's
00:14:20
important that as you can see when we're
00:14:22
dealing with such
00:14:25
small differences here you need to keep
00:14:28
lots of sign ific diges in your
00:14:30
preliminary calculations so that you can
00:14:32
actually see the difference so now at
00:14:35
this point beyond the equivalence point
00:14:37
the sodium hydroxide is actually in
00:14:40
excess in a concentration of 2 *
00:14:45
105 so this being a strong base
00:14:47
dissociates completely so I know the
00:14:50
hydroxide concentration is 2 * 10 to
00:14:53
the5 so I can calculate the P of 4.70
00:15:00
and the pH is therefore
00:15:05
9.30 okay so as you can see between
00:15:10
24.99 and 25.01.2013
00:15:29
and that's what really drives the pH
00:15:31
upwards when you're doing a titration it
00:15:34
doesn't necessarily mean that you're
00:15:37
going to have an incorrect calculation
00:15:40
if you've added you know half a drop in
00:15:42
excess okay because the amount of volume
00:15:46
that causes the pH to jump significantly
00:15:49
is really really
00:15:57
small now when we're hydrating a weak
00:16:00
acid by a strong base we're going to be
00:16:02
dealing with a variety of situations
00:16:05
we're going to deal with weak acid
00:16:07
calculations we're going to deal with
00:16:08
buffer calculations we're going to deal
00:16:10
with the addition of a strong base to a
00:16:13
buffer the scenario that we're looking
00:16:15
at here is with the titration of
00:16:17
vinegar okay and B sodium
00:16:22
hydroxide as you can see here with a
00:16:25
weak acid the pH initially is a little a
00:16:28
little bit higher you're also going to
00:16:31
see that this here is the buffer region
00:16:34
so there's a very uh small changes in PH
00:16:38
over this range as we move towards the
00:16:41
equivalence point the steepness around
00:16:45
the equivalence point the height of
00:16:48
it steepness is the same but the length
00:16:51
of the section where it's steep is not
00:16:53
uh quite as high because of the lower
00:16:57
initial pH here and this is where it's
00:17:00
going to be really important for us to
00:17:02
use the technique of taking the middle
00:17:06
the midpoint of the Steep section in
00:17:09
order to determine the equivalence point
00:17:11
pH one of the major differences of
00:17:14
titrating with weak Solutions is that
00:17:17
the equivalence point doesn't actually
00:17:18
lie at seven in this case the
00:17:21
equivalence point is at
00:17:23
8.72 and we'll see in our calculations
00:17:25
why that makes
00:17:27
sense so we're going to go through a
00:17:29
series of calculations here looking at
00:17:32
the pH before the titration begins okay
00:17:36
uh during the titration halfway to the
00:17:38
equivalence point and at the equivalence
00:17:41
point so before the titration begins we
00:17:44
have a solution of 25 milliliters of
00:17:47
acetic acid it's important to make a
00:17:51
conscious recognition that this is a
00:17:53
weak acid so if I want to know the pH of
00:17:57
a weak acid solution I need
00:18:01
to do a hydrolysis
00:18:08
calculation so we've got a 0.1 molar
00:18:12
solution got none of this at the
00:18:14
beginning plus X Plus
00:18:18
X this is - X 0.1 minus X and it's
00:18:23
approximately equal to 0.1 and then I
00:18:27
can use my Ka so I've got x^2 is equal
00:18:31
over .1 is equal to 1.8 *
00:18:35
105 and we get a pH here of
00:18:40
2.87 okay so initially before anything's
00:18:43
been added it's simply a weak acid
00:18:48
calculation next if we want to calculate
00:18:50
the pH during the titration so we're
00:18:53
before the equivalence
00:18:55
point we are adding strong base
00:18:59
to a solution so we need to write the
00:19:02
reaction for the acid that's going to
00:19:04
react with the base which is the vinegar
00:19:07
plus the
00:19:11
hydroxide this will produce
00:19:15
acetate and
00:19:18
water because we're adding a strong base
00:19:20
this is a complete
00:19:23
reaction and we're going to use an ICF
00:19:26
chart to organize our data
00:19:30
the concentration of the sodium
00:19:32
hydroxide that we're using is
00:19:34
0.1 and the volume that we have at this
00:19:37
point is 30 milliliters because we've
00:19:39
added 5 milliliters to our initial 25mil
00:19:47
solution okay so we see that we've got a
00:19:50
0.833 and
00:19:54
0.167 got none of this initially the
00:19:57
water I'm not worried about okay so the
00:20:00
hydroxide is going to be our limiting
00:20:05
reactant and so it goes down to zero and
00:20:08
it helps us know
00:20:10
what our x value is which is
00:20:21
0.0167 so we can figure out what the
00:20:23
concentration of the acetic acid and the
00:20:25
acetate ion are in the solution once the
00:20:28
hydroxide has been consumed now at this
00:20:31
point whenever ever you've completed
00:20:33
your ICF chart you need to look at it
00:20:35
and decide what you have so right now
00:20:38
we've got a solution that contains 0.0
00:20:42
666 mole per liter in the vinegar and
00:20:46
0.1 0167 mole per liter in the acetate
00:20:50
so this you must must must recognize is
00:20:54
a buffer and so you can
00:20:57
use the h hassle block
00:21:01
equation to very quickly and easily
00:21:05
calculate the pH so the pH at this point
00:21:09
is
00:21:17
4.14 okay now next we're going to look
00:21:21
at halfway to equivalence point
00:21:38
halfway to equivalence point is a
00:21:40
special point in the
00:21:42
titration so the volume at this point
00:21:48
is
00:21:50
37.5 milliliters
00:22:02
so this is the point when half of the
00:22:04
vinegar has been neutralized so we
00:22:07
started with 25 milliliters of the acid
00:22:10
at 0.1 mole per liter and so that is
00:22:13
half titrated or half neutralized when
00:22:17
we've added 12.5 milliliters of sodium
00:22:21
hydroxide of the same
00:22:23
concentration so at this point when you
00:22:27
look at the halfway to equivalence point
00:22:31
you're going to notice that this is a
00:22:33
really special point in the
00:22:40
titration and it will eventually become
00:22:42
a very quick
00:22:46
calculation okay so when we've added
00:22:49
37.5 milliliters what we have is our
00:22:52
initial concentration of the vinegar is
00:22:56
0.667
00:22:58
we've
00:22:59
added half that amount of
00:23:07
Base our base is still
00:23:10
our limiting reactant so it goes to zero
00:23:14
and it helps us to determine what our x
00:23:18
value
00:23:19
is and now what we see here is that
00:23:22
we've got a buffer so we still have some
00:23:24
of the vinegar and the acetate ions
00:23:27
present
00:23:29
this is a particularly important point
00:23:31
of the buffer though because at this
00:23:33
point in the titration the concentration
00:23:37
of the acetate is equal to the
00:23:39
concentration of the
00:23:40
vinegar and when we use the Henderson
00:23:43
hul
00:23:48
equation we've got our point where
00:23:54
the log portion of the Henderson Hassel
00:23:57
block equation isal equal to zero
00:23:59
because if our base concentration and
00:24:02
our acid concentration are equal then
00:24:05
we'll have log of one which is equal to
00:24:07
zero so halfway to equivalence the pH is
00:24:11
always equal to the pka of the
00:24:19
acid so you'll be able to if you
00:24:22
recognize this special scenario you'll
00:24:24
be able to save yourself some
00:24:26
significant time in making calculations
00:24:29
because you can simply report that if
00:24:32
you're halfway to equivalence the pH is
00:24:34
equal to the
00:24:40
pka at the equivalence point is another
00:24:43
special calculation this one is the most
00:24:46
challenging of the calculations that you
00:24:52
do so the equivalence point is the point
00:24:56
whereby the amount of acid is
00:24:59
stochiometrically equal to the amount of
00:25:05
Base so we're dealing with a one:1 ratio
00:25:07
here and our concentrations are the same
00:25:11
so this is the point where 25
00:25:13
milliliters of Base has been added so
00:25:16
our total
00:25:18
volume is 50
00:25:21
milliliters and so our
00:25:24
concentrations are 0.05 and 0.5
00:25:27
respectively
00:25:29
so in this case there is no limiting
00:25:30
reactant they both go to zero they're
00:25:33
both consumed but the acetate ion is
00:25:36
still being
00:25:39
produced so when we looked at the
00:25:41
equivalence point of a strong acid
00:25:43
strong base hydration we automatically
00:25:45
knew that we had a pH of seven and that
00:25:49
was because of the nature of the salt
00:25:51
that was produced in that reaction this
00:25:55
salt coming from a we acid affects the
00:26:00
pH okay this is the conjugate base of
00:26:07
the acetic acid solution so what we need
00:26:10
to do is now look at the hydrolysis
00:26:14
reaction that
00:26:17
occurs with the acetic
00:26:25
acid okay the concentration of
00:26:30
the acetate ion is
00:26:38
0.05
00:26:41
this
00:26:51
hydroxide and now the other major Point
00:26:54
here is that we've got the KA for the
00:26:57
acetic acid but we need to calculate the
00:27:00
KB using Ka is equal
00:27:05
to sorry Ka time KB is equal to KW so
00:27:09
the KB in this case is
00:27:12
5.56 * 10 the
00:27:15
-10 it's small enough to make the
00:27:17
Assumption so we've got that is equal to
00:27:19
x^2 over
00:27:21
0
00:27:23
05 so then you can calculate the p
00:27:28
at
00:27:30
5.28 and the
00:27:32
pH at
00:27:36
8.72 so calculating the pH at the
00:27:39
equivalence point of a weak acid
00:27:43
titrated by a strong base it's going to
00:27:46
be slightly basic because of the
00:27:48
presence of the acetate ion which is the
00:27:50
conjugate of the
00:27:54
vinegar we can do a similar series of
00:27:57
calculations
00:27:59
with the titration of a weak base by a
00:28:02
strong
00:28:03
acid so there is space here for you to
00:28:08
walk through the
00:28:10
calculations okay it's very similar I'm
00:28:12
going to ask you to change this
00:28:16
to10 mole per
00:28:19
liter so your job is to go through each
00:28:22
of
00:28:23
the scenarios so you're going to look at
00:28:26
before the titration during the
00:28:28
titration halfway to
00:28:31
equivalence and at the equivalence
00:28:35
point you can compare the values that
00:28:37
you get to the
00:28:39
chart that is right here you should be
00:28:42
getting the same pH as that we have at
00:28:44
the particular points and a few features
00:28:47
of this you can see that we've got our
00:28:48
buffer region here where the pH doesn't
00:28:51
change too graph drastically leading up
00:28:53
to our equivalence point you're starting
00:28:56
with an acid because we're sorry we're
00:28:58
starting with a base the ammonia
00:29:00
solution and you're adding acid to it so
00:29:02
the pH is grad gradually changing over
00:29:05
time the equivalence point lies at the
00:29:09
midpoint of the Steep portion of the
00:29:18
curve now if you move ahead in the notes
00:29:21
there's a few little points here so
00:29:23
first of all we already discussed that
00:29:25
we would not titrate a weak acid with a
00:29:27
weak base
00:29:28
there's not really a clear end point
00:29:30
because of the weak base that's put into
00:29:34
the solution by the salt of this weak
00:29:37
acid and the weak acid that's put into
00:29:40
the solution by the salt of this weak
00:29:43
base so it's a competing reaction so
00:29:47
it's easier to avoid this kind of
00:29:49
complex
00:29:52
scenario now when we titrate with a pH
00:29:55
meter we're going to be able to actually
00:29:58
determine the KA of a particular acid
00:30:02
you're going to do a lab where you
00:30:04
determine the dissociation constant of
00:30:07
an acid by using a half
00:30:10
titration because we know that the pH is
00:30:13
equal to the pka halfway to the
00:30:15
equivalence point so it makes for a very
00:30:17
simple method to calculate the
00:30:22
KA now selecting an indicator to do a
00:30:25
titration is a really important lab
00:30:29
technique and what we are looking for is
00:30:32
we are looking for a solution that's
00:30:35
going to change color as close to the
00:30:39
equivalence point as
00:30:42
possible now indicators themselves are
00:30:45
actually weak assets and so the presence
00:30:47
of H+ in the solution is actually
00:30:50
shifting a small equilibrium to change
00:30:53
the color of the particular indicator
00:30:58
the color
00:31:00
change
00:31:01
occurs when at the breaking point
00:31:05
between when the concentration of one
00:31:08
color is actually equal to the
00:31:10
concentration of the other color so it's
00:31:14
a point where the concentration of the
00:31:17
acidic form of the indicator is equal to
00:31:20
the basic form of the
00:31:22
indicator this point actually occurs at
00:31:27
the pka of the particular
00:31:29
indicator so when we're looking for a
00:31:33
indicator we want to look for an
00:31:35
indicator that will have a PKA close to
00:31:39
the pH of our equivalence
00:31:44
point so if we look at these indicators
00:31:47
here okay phenol theine the indicator
00:31:50
that you use most commonly has a pka of
00:31:54
9.3 so this is ideal
00:31:58
for the titration of a weak acid by a
00:32:02
strong base because the pH at
00:32:06
equivalence is close to
00:32:09
9.3 the other reason that phenol phalion
00:32:12
is such a good indicator is because it
00:32:14
has such a drastic color change so those
00:32:17
two features we're looking for okay but
00:32:20
a classic AP question is to identify
00:32:23
which indicator is appropriate for a
00:32:26
particular acidbase hydration by using a
00:32:30
set of PKA values okay so if you have
00:32:33
something that the equivalence point is
00:32:36
around four okay bromoenol blue might be
00:32:40
an appropriate indicator for that
00:32:41
titration
