AT Chemistry2010

POGIL on TITRATIONS

In this activity we will explore titration, pH curves and acid-base indicators. We will examine two types of titrations: strong acid-strong base titration (relatively simple) and strong acid-weak base/weak acid-strong base titrations (a lot more involved).

Before we get started, there are some terms and language you will need to know and understand:

titrant - one in the buret with known concentration

buret - calibrated volumetric instrument for dispensing a liquid

indicator - substance which changes color in different [H3O+]

titration curve - a plot of pH vs volume of acid (or base) added

endpoint - point at which the indicator changes color

equivalence point - point where moles H3O+ = moles OH-

Let’s first examine strong acid-strong base titrations.

Write the net ionic equation for any strong acid-strong base reaction below:

You will notice the reverse reaction represents the autoionization of water. What is Kw for the autoionization of water?

What will be the K for the reaction between a strong acid and base?

Does the reaction between a strong acid and strong base go pretty much to completion?

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Strong acid-strong base titration calculations are relatively simple because one of three possibilities result after the reaction is complete:

Excess strong acid leftExcess strong base leftEquivalence point

Let’s try the following problem:

Calculate the pH after the following total volumes of 0.2500M hydrochloric acid have been added to 50.00 mL of 0.1500M sodium hydroxide. Again it would be useful to calculate the initial moles of acid and base reacting (from the molarity and volume), do the math to see which runs out, and then calculate the molarity of the excess and the resulting pH.

Note we are adding certain volumes of 0.2500M acid from a buret to an Erlenmeyer flask containing 50.00 mL of 0.1500M sodium hydroxide.

a) 0.00 mL of acid added (this means just plain strong base):

b) 4.00 mL of acid added:

Reaction:

Math part I

C

Do we have excess?

Calculate molar concentration of excess:

Calculate pH:

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c) 30.00 mL of acid added:

Reaction:

Math part I

C

Do we have excess?

What do we call the point in a titration where moles acid = moles base?

In this titration of strong acid and strong base, what species is left at this point?

What is the pH?

d) 40.0 mL of acid added:

Reaction:

Math part I

C

Do we have excess?

Calculate molar concentration of excess:

Calculate pH:

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In the titration of a strong acid and a strong base the pH at equivalence = 7.00 because the only major species that remains is water (and the pH of water = 7.00 at 25.0oC)!

If we graphed the pH vs volume of acid added for the previous titration (starting with strong base), we would get the following curve shown below on the left:

Make special note that when strong acid is added the pH does not dramatically change until almost at the equivalence, and we observe an almost vertical line change in pH. Very near the equivalence point, 2-3 drops of acid results in a dramatic change in pH. We get the same type of curve if we started with strong acid and added strong base, as shown in the graph on the right.

Now let’s consider a little more challenging titration, that of a weak acid with a strong base (weak base-strong acid titrations involve the same ideas).

When doing calculations for this type of titration it is useful to relate our thinking back to our buffered solution problems.

HA(aq) + OH-(aq) A-(aq) + H2O(l)

You will be generating a buffer (weak acid and conjugate base) as a result of the addition of OH -

ions to the solution. You will have a buffered solution until you exceed the buffer capacity of the solution, and then the pH will rise dramatically.

Let's use the titration of formic acid, HCOOH (Ka = 1.8 X 10-4) with NaOH as an illustrative example. The equilibrium constant for the titration can be calculated.

HCOOH(aq) + OH-(aq) = HCOO-(aq) + H2O(l)

can be expressed as a combination of

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1. HCOOH(aq) + H2O(l) = H3O+(aq) + HCOO-(aq) Ka = 1.8 X 10-4 (k1)

2. H3O+(aq) + OH-(aq) = 2H2O(l) 1/Kw = 1.0 X 1014 (k2)

What is the net reaction?

Calculate k for the net reaction:

What does the k value tell you about the extent of the reaction?

We fill a buret with 0.400M sodium hydroxide and we will slowly add (drop by drop) the following volumes of this base to an Erlenmeyer flask containing 50.00 mL of 0.200M formic acid. Calculate the pH after the following total volumes of the NaOH is added to the acid in the flask:

a) 0.00 mL of strong base added (this is just a weak acid problem):

b) 5.00 mL of strong base added:

(note: you can assume the initial concentration of formate ion (HCOO-) is negligible

Neutralization reaction:

Math Ipart

C

moles left

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note that in this reaction you are changing the moles of HCOOH and HCOO-. Let’s now calculate the new molar concentrations of both by dividing the moles by the total volume in liters:

[HCOOH] = [HCOO-] =

Now let’s go back to the original equation for the ionization of formic acid and put in the new concentrations of acid and conjugate base:

Ionization equation:

Use Ka to calculate new [H3O+] and pH:

c) 12.50 mL of strong base added:

Neutralization reaction:

Math Ipart

C

moles left

Let’s now calculate the new molar concentrations of both by dividing the moles by the total volume in liters:

[HCOOH] = [HCOO-] =

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What do you notice about the values for the concentrations of HCOOH and HCOO- at this point in the titration?

When this condition is true, can you see we are at the halfway point in the titration? Notice the ratio of the moles of acid to the moles of base! This is known as the midpoint of the titration.

Now let’s go back to the original equation for the ionization of formic acid and put in the new concentrations of acid and conjugate base:

Ionization equation

Calculate new [H3O+] and pH:

From the Ka expression you noticed that [HCOOH] = [HCOO-] and the Ka = [H3O+]. If

Ka = [H3O+]

Then pKa = pH (this is always true at the midpoint!)

d) 25.00 mL of strong base added:

Neutralization reaction:

Math Ipart

C

moles left

Compare the moles of acid titrated with the moles of base added at this point in the titration?

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What do we call this point in the titration?

How many moles of acid and moles of base are left?

What species is left in solution at this point?

What does the remaining species do in water (donate or accept a proton)?

Let’s write the reaction of this species (should be a weak base) in water:

We now have a weak base problem. To calculate the pH of this solution, we will need to calculate the molar concentration of the base and get the Kb for the base.

To calculate the molar concentration of the base, we take the moles of the base (which we calculated previously) and divide it by the total volume in liters.

What is the total volume (in liters)?

Calculate [HCOO-] =

How do we calculate the value for Kb?

Kb =

Now go back to the reaction of formate ion in water and do an ICE problem for a weak base.

Reaction:

I

C

E

Calculation for pH:

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Notice that the pH at the equivalence point in the titration of a weak and with a strong base is basic. This is to be expected because at the equivalence point of a titration between a weak acid and strong base the only species left in solution will be a weak base. The same idea applies for a strong acid-weak base titration. In this case the only species left at equivalence will be a weak acid and the pH will be acidic. The only case where the pH at equivalence will be neutral is that involving a strong acid-strong base, because at equivalence we only have water.

e) 30.00 mL of strong base added: (this is post equivalence)

Neutralization reaction:

Math Ipart

C

moles left

NOTE: we have passed the equivalence point! WE now have strong base in excess and the excess strong base dictates the pH. So all we need do is calculate the molar concentration of the excess hydroxide ion and calculate the pH.

Excess [OH-] =

A pH titration curve for a weak acid-strong base is shown below:

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Label the following parts of the curve:

From the start of the titration to the equivalence point we the buffer region Equivalence point (already labeled) Post equivalence

You should also be aware that the slope of the line near equivalence is less vertical compared to a strong acid-strong base titration and the pH at equivalence is basic.

Now let’s try a weak base strong acid titration problem. These kinds of titrations are handled using the same strategies as those involving weak acids and strong bases. We have the buffer region, equivalence point, and post-equivalence.

Calculate the pH at each of the following points in the titration of 50.00 mL of a 0.100 M methyl amine (CH3NH2) solution with 0.100 M HCl solution (Kb for CH3NH2 = 4.38 X 10-4).

a) Initially: (before any acid is added, this is just a weak base problem)

Ionization of methyl amine in water:

Reaction:I

C

E

Calculation for pH:

b) midpoint:

neutralization reaction:

Math Ipart

C

moles left

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Let’s now calculate the new molar concentrations of both by dividing the moles by the total volume in liters. We know the volume of base (50.00 mL). How can we calculate the volume of acid added at this point?

Volume HCl:

[CH3NH2] = [CH3NH3+] =

Now let’s go back to the original equation for the ionization of methyl amine and put in the new concentrations of base and conjugate acid:

Ionization equation

Calculate new [OH-] and pH:

Was there an easier way to determine the pH?

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c) equivalence point:

neutralization reaction:

Math Ipart

C

moles left

once again we ask what species is left and does it accept or donate a proton. To calculate the molar concentration of this species you will need to calculate the total volume, get the corresponding K value and do an ICE problem calculation. Good luck!

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pH Titration Curve for a Strong Acid-Weak Base

Titration Curves for Weak Acids of Different Strengths (i.e. Ka’s) with Strong Base:

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Polyprotic Acid Titrations

Above is a titration curve for the titration of a polyprotic acid (phosphoric acid) by a strong base. The curve shown is for 100.0 mL of a 0.1000M H3PO4 titrated with NaOH. No clear third equivalence point is seen at 300 mL because Ka for HPO4

2- is not much larger than Ka for water in aqueous solution.

Acid-Base Indicators

The purpose of an indicator is to allow you to know when you have reached the equivalence point in a titration. Acid-base indicators are normally organic acids. Thus, when you add an indicator to a solution and perform a titration, you are titrating the indicator along with your substance of interest. This requires some small extra amount of titrant.

The volume of titrant at which you visually detect the equivalence point is called the endpoint. The volume at the equivalence point is never exactly equal to the endpoint, but it is usually quite close.

There are two criteria that are used to choose an indicator for a given titration:

the pKa of the indicator should be within + 1 unit of the pH of the solution at the equivalence point.

the color change at the endpoint should be clearly distinguishable.

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Problem: Given three acid-base indicators:

methyl orange - endpoint at pH 4

bromthymol blue - endpoint at pH 7

phenolpthalein - endpoint at pH 9

Select the best indicator for each of the following titrations:

a) sulfuric acid with potassium hydroxide

b) ammonia and hydrobromic acid

c) sodium nitrite with hydroiodic acid

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