Problems 1, 2, 5, 7, 13, 16, 18, 21, 25
Up to now we have focused on calculations of pH or concentration at a few distinct points. In this chapter we will talk about titration curves, or determining the pH after any arbitrary addition of base to acids or vice versa.. The ability to do this will help to further our knowledge of acid-base reactions
10-1 Titrations of strong acids and strong base
In the reaction of strong acids with strong bases we have something like:
Molecular
HCl(aq) + NaOH(aq) 6 NaCl(aq) + H O(l)2
Complete ionic
H (aq) + Cl (aq) + Na (aq) + OH (aq) 6H O(l) + Na (aq) + Cl (aq)+ - + - 2 + -
Net Ionic
H (aq) + OH (aq) 6H O(l)+ - 2
The NaCl is just along for the ride, so what is the real reaction?
H + OH 6 H O+ - 2
What is the K of this reaction?? (1/K = 1x10w +14 a very strong reaction)
There are only four equations you need to know to describe this titration curve: The initial point, 1 before the equivalence point, one for the equivalent point itself and one for after the equivalence point.
1. Notice that the book concentrates on the problem of a Base being titrated with an acid. You should be able to turn these equations around and to the same analysis for an acid being titrated with a base.
2. While I will follow the approach used in the book, I will plug in slightly different numbers just to show you how thing would change for a slightly different problem.
Let’s say we are titrating 49 mL. of .021 M NaOH with 0.097 M HCL
Step 1 determine the equivalence point
Since we use different equations before, during, and after the equivalence point, your first job is to find the equivalence point.
What is the definition of the equivalence point? (Where you have an equivalent amount of the reactant and titrant.)
Reactant is NaOH, and we have Moles=Molarity x volume
49(.021) or 1.029 mMoles of OH initially-
To be at the equivalence point we need an equivalent number of H+ So if
1.029mM=0.097 X X=10.61 mL.
Inital pH
.021M NaOH
= .021 M Na and .021M OH+ - pOH = -log(.021)
= 1.68 pH=14-1.68
=12.32
Region 1 Before the equivalence point.
Pick a point, any point, say 1 mL
In this region the pH is determined by the excess of OH that has not reacted- with HCl
Total OH = .049L × .021 M = 1.029 mMoles- Total H delivered = .001×.097M = .097 mMoles+ RXN table
H + OH 6+ - H O2
Inital 0 1.029
Added .097 0
RXN -.097 -.097 +.097
Final 0 .932 .097
And [OH ] = .932/Total volume-
= .932mMole/ 49+1 mL
= .01864 M
pOH =-log(.01864) =1.73 pH = 14-pOH = 12.27
Region 2 the equivalence point
This is easy. IF all we have is water and a little salt, what it the pH? 7
NOTE this is only true for strong acid-strong base reactions. It will not be true for anything involving weak acids or bases
Region 3 After the equivalent point
Here the pH is determined by the excess H . Again a long and a short way. + Pick a point, any point, say 20 mL total volume of acid
Initial base = .049 l×.021M = 1.029 mMole Total acid = .020 l ×.097M = 1.94 mMole RXN table
H + OH 6+ - H O2
Inital 0 1.029
Added 1.94 0
RXN -1.029 -1.029 +1.029
Final .911 0 +1.029
And [OH ] = .911/Total volume-
= .911mMole/ 49+20 mL
= .0132M
pH = -log(.0132) = 1.88
10-2 Titration of a weak acid with a strong base
For this let’s use an example from the lab, KHP, Potassium acid phthalate Our example will be a KHP, as the weak acid its structure is:
Pure Phthallic acid has 2 K ’s. 1.12x10 and 3.90x10 . Since wea -3 -6
are working with KHP, the first acid has already been neutralized with KOH, so only the second K is left. We will assume we have .8 g in 50 ml, and we are titrating it with 0.1 M NaOH. Since KHP has a MW of 204.23 this means our acid is
.8×1/204.23×1/.05 = .0783M solution
First what is the reaction that goes on when we dissolve KHP ? It goes to HP- Now what happens when we titrate this with base?
HP +OH 6 P- - -2 +H O 2
What is the K for this? Well it looks sort of like the acid reaction
HP + H O6P + H- 2 2- +
But it has an OH instead of H and it has the water on the wrong side- +
What if we combine the above reaction with the reverse of the water dissociation:
H + OH W H O+ - 2
We get the correct equation. This that in mine, what is the K of the reaction? K
a w
= K ×1/K
=3.90x10 ×1/1x10 = 3.9x10 -6 -14 8
A very favorable reaction. I think you can safely say this is also a strong reaction.
IN this titration there are 4 regions to worry about 1. Initial pH before you have added NaOH
2. From the initial addition of NaOH up to the equivalence point 3. The equivalence point
4. After the equivalence point
With this buildup its actually quite easy, you have seen all these calculations before!
Where do we start? Locate the Equivalence point
Moles of Acid = .8 g/204.23 = 3.917 mMoles Volume of base = to 3.917 mMoles
3.917=.1(X) X=39.17 ml
1. Initial pH
This is just the pH of a dilute weak acid [H ] = X; X /(F-X) = K+ 2 a
In this example
X /(.0783-X)=3.90x102 -6
We will assume that X is small compared to .0783 so we get X /.0783 = 3.90x10 ; X=5.526x10 ,2 -6 -4
the assumption looks good; pH=3.25 If we do it the long way
X /(.0783-X)=3.90x10 ; using the solver to get an exact solution2 -6 X=5.506x10 (original answer off by .02 or .3%)-4 pH= 3.26
2. After addition of NaOH, but before equivalence point.
Now you should recognize that in the region we have HA and A in solution , what does- this mean?? BUFFER REGION - USE HH
All we need to do is figure out HA and A- So let’s try for say 20 mL. of base added
For each mole of OH we add we change 1 mole of HA to A- So if we add 2 mM of OH how much A have we made? 2mM- -
For each mM of A we make we remove a mole of HA, so how much HA is there?- 3.917-2=1.917mM
Now all we have to do is to plug in to HH pH=pK + log (A /HA)a -
Notice something sneaky here if we are going to use the ratio of A /HA, we don’t need- to convert to molarity because all that would do is divide both the numerator and the denominator by the same volume.
Thus
pH=5.41 + log (2/1.917); pH= 5.43 3. At the equivalence point
This should be another easy one. You were already shown how to determine the pH of A earlier-
it goes by the form of X=[OH ]-
b w a
X /(F-X) = K =K /K2
The only trick part is that F is no longer .0783 M!
Why not?? Because we have added titrant and diluted it ! What is F now??
F=F X (Initial volume of HA/ Total volume at EQ point)org
Here we have
F=0.07834 x (50/(50+39.17) = .05663M
b w a
K = K /K = 2.564x10 ;-9 2.564x10 =X /(.05663-X)-9 2
This K is pretty small, so let’s assume that .05663-X..05663 2.564x10 (.05663) =X-9 2
X=1.20x10 ;pOH=4.920, pH=9.080-5
(You get the same answer if you use the quadratic and solve exactly) Region 4 After the equivalence point
This is just like after the equivalence point for a strong acid and weak base
Let’s just do the quick and dirty
Say we are at 45 mL., 45-39.17 or 5.83 mL. past the equivalence point [OH ]=.1(5.83/50+45)-
=6.14x10 ; pOH=2.21, pH=11.79-3
Now that you have done a titration curve, what do you think would happen to that titration curve is you had a weaker acid? (pK goes up, middle of curve rises) Note that as the acid gets weaker and weaker, the center rises until the point where you don’t have a clear endpoint (figure 10-10)
Another problem is what happens with concentration (also this figure) as the acid gets more dilute, the curve gets shallower and shallower
10-3 Titration of weak base with strong Acid
Essentially the same of weak acid with strong base but just reverse acids and bases. Go over yourself. May be tested on it.
10-4 Finding the End Point
Remember what we actually want to find is the equivalence point, but using physical methods what we actually find is the End Point, where we end the titration.
Indicators
Most of this material was covered in the last chapter. Just as a reminder. You want the pK to match the pH of the equivalence point. And you want to add aa
minimum amount of indicator. If either of these are off you may have indicator error, and you may have to run a blank sample to eliminate this error
Using a pH electrode
While we have just calculated a titration curve, this is something that is usually done experimentally in the lab. IF you are luck and have several thousands of dollars you can by a machine that measures the amount of titrant added and the pH of the solution automatically and makes pretty plots for you. If not, or you are in the lab, you have to manually add some titrant, and read the pH off a pH meter.
When you do this you should get a nice curve like we have shown in our theoretical plots. The first question is how do you locate the endpoint in these plots.
Notice most of the time there is a sharp rise in pH at the beginning. This is not an equivalence point, it is just the buffer getting established
Next we have the sharp rise in pH at the end point. Which point in here is the
true endpoint?? The one that has the largest slope. How do you find that?? Does the term derivative sound familiar?
Derivative is the slope of the line
So if you have two point ,40.1 ml pH=8.71 And 40.2, pH 8.8, what is the derivative or slope?
)y/)x = (8.8-8.71)/(40.2-40.1)
Since this slope is between two point, you would plot it at the average of the two points or at 40.15
When you plot the derivative you should get a plot like this
The peak here is the point with the largest slope, ie your end point.
If you want to get even more exact, what does the second derivative do?
(Will cross from + to - at the equivalence point) This is an even better way to find the eq pt.
See Figure 10-4
Now if you’ve got a machine, this works out nicely, you can even get the machine to calculate the derivative for you
But what about real manual data. Here you usually have errors, you read the pH wrong, you read the volume wrong etc. This makes small errors in your pH vs. Vol plot, but nothing too bad. What happens when you get to your derivative plot? The errors get magnified and it looks worse than it does in the book.
If you get really lousy data (your amino acid in the lab) when you have a very weak endpoint on top of noisy data these plots can be worthless, so we need another approach
The Gran plot
This book does not cover another tool called the Gran plot, but let’s look at it anyway because it can be useful. The Gran plot uses some more involved equations, but I won’t try to derive them for you, let’s just cut to the chase and show you how it works.
One problem with all the above analysis is that it is often pretty hard to nail down a good E.P, especially is you are dealing with either a very dilute sample or a very weak acid or base. Typically you have lots of good data out in the buffer region with only 1 or two scattered points near the E.P. The Gran plot is alternate way of plotting the that uses information from the buffer region to find the E.P in a more robust manner.
Plot {Volume of base × [H ]} on the Y axis and {Volume of base} on the X axis.+ The Y axis
Volume of base x [H ] = volume of base × 10+ -pH
Volume of base X 10-pH vs volume of base you transform the data into a new form. In
this plot the X intercept of the plot is your equivalence point and the slope of the plot
HA A-
=Ka(( / ( )
To get a good linear plot, you should use only the data within the last 10-20% fo the titration, so if you have a titration where you think the endpoint is about 30 mls, you should make this plot for 24-32 mls (20% of 30 is 6, 30-6=24, plus a few mls after the suspected end point)
I think you will see in the lab how useful this plot can be.
10-5 Practical notes on titrations
1. A long time ago we mentioned Primary Standard. Primary standard for acid/base reaction are listed on table 12-5 Note that KOH and NaOH are not primary standards, thus we have to calibrate them in the lab
2. Strong base solution tend to pick up carbonate. In the lab this is avoided by using a 50% NaOH stock solution where the carbonate is precipitated out (If you use this solution, and you see your aliquot is cloudy throw it out and start over)
3. Even your standardized NaOH solution will pick up carbonate. Keep tightly closed, don’t use if over a week old
4. Strong bases attack glass. Don’t store base in class bottles, don’t leave in buret for long times.
10-6 Kjeldahl Nitrogen analysis Will Skip
10-7 Spreadsheets
While I like the idea of making a spreadsheet to calculate an entire titration curve, the way he does it is to use some equations that he doesn’t get to until Chapter 12, so I think we will skip this as well.
