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Polyprotic Acids: Understanding the Behavior of Diprotic and Triprotic Acids and Bases, Study notes of Stoichiometry

An in-depth analysis of polyprotic acids and bases, which have multiple reactive sites and exhibit complex acid-base equilibria. the stoichiometry, acid-base strengths, and calculations for diprotic and triprotic acids, including examples of well-known acids like carbonic acid. Students will gain a solid understanding of the concepts and be able to apply them to various acid-base problems.

What you will learn

  • What are the steps to calculate the acid-base equilibria for a triprotic acid?
  • How does the presence of multiple reactive sites in polyprotic acids affect their acid-base equilibria?
  • What is the difference between monoprotic and polyprotic acids?

Typology: Study notes

2021/2022

Uploaded on 09/27/2022

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Polyprotic Acids
By now we should be pretty comfortable dealing with monoprotic acids like:
HCl HNO3
HClO4 CH
3COOH
+HNH3
Each of these acids has a single proton to offer to the system. Consequently there is a nice one-to-one stoichiometry to deal
with (one mole of acid yields one mole of protons). The chemical equilibria are also pretty easy.
You either have the strong acid case in which we assume everything goes to completion and what we start with is what we end
with:
HCl Æ H
+ + Cl-
start 100% 0 0
equil. 0 100% 100%
or we have the the weak acid case where very little dissociation occurs, the equilibrium constant, Ka, is very small, and
basically, nothing happens (though that doesn’t stop us from calculating the pH of the weak acid):
HA H+ + A-
initial CHA 0 0
change -x +x +x
equil. CHA-x x x
And from the RICE expression we can easily derive the simplified equations for acids (and their base complements) that pretty
much estimate the pH of any monoprotic acid or basic solution:
[H+] = Ca for strong acids [H+] = (KaCa)0.5 for weak acids [H+] = KaCa/Cb for buffers
If only that was all there was to it.
pf3
pf4
pf5

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Polyprotic Acids

By now we should be pretty comfortable dealing with monoprotic acids like:

H

Cl

H

NO

3

H

ClO

4

CH

3

COO

H

H

NH

3

Each of these acids has a single proton to offer to the system.

Consequently there is a nice one-to-one stoichiometry to deal

with (one mole of acid yields one mole of protons). The chemical equilibria are also pretty easy.You either have the strong acid case in which we assume everything goes to completion and what we start with is what we endwith:

HCl

Æ

H

Cl

start

equil.

or we have the the weak acid case where very little dissociation occurs, the equilibrium constant, Ka, is very small, andbasically, nothing happens (though that doesn’t stop us from calculating the pH of the weak acid):

HA

H

A

initial

C

HA

change

-x

+x

+x

equil.

C

HA

-x

x

x

And from the RICE expression we can easily derive the simplified equations for acids (and their base complements) that prettymuch estimate the pH of any monoprotic acid or basic solution:

[H

] = Ca

for strong acids

[H

] = (K

a

C

a

for weak acids

[H+] = K

a

C

a

/C

b

for buffers

If only that was all there was to it.

To complicate matters, some acids and bases are more fortunate than others in that they are given multiple reactive sites at birth.These molecules are known as polyprotic acids and bases and include:diprotic acids and bases

Ba(

OH

2

H

2SO

4

H

3NCHCO

2H

H

2CO

3

H

OOCCH

2COO

H

triprotic acids

H

3PO

4

tetraprotic acids

H

OOCCH

2-NCH

2CH

2

N-(

H

OOCCH

2

Obviously

polyprotic

systems

are

more

problems

to

deal

with.

First

the

stoichiometry

is

not

a

simple

one-to-one

correspondence and we have to start being careful to account for multiple protons or hydroxides when they are thrown insolution (for example, the two hydroxides on

Ba(

OH

Also, from an equilibrium stand point, we have to recognize that each proton or hydroxide in the molecule had its own acid/basestrength which can be defined in terms of its equilibrium constant. Thus if we have a molecule with three protons, there will bethree acid base equilibria with their own equilibrium constants, K

a

, K

a

, K

a

Understand that because each of these protons has its own affinity for a base, we do not consider the loss of protons to occurindiscriminately or all at once over the molecule.

Instead, there is a nice systematic series of reactions with the first reaction

involving the most acidic proton, the resulting conjugate base reacting to yield the next most acidic proton, and so on.This is actually good news because it means we can talk about the equilibria separately.So let’s write the equilibria down in sequence for the general case of a triprotic acid, H

3A, including equilibrium equations:

Polyprotic acid calculations: From a perspective in which we would want to work acid/base problems, the important thing to look for in polyprotic acids iswhether the K

a

values are really far apart in size. This is of interest to you because:

When K values in a polyprotic system are 3 or more orders of magnitude apart, we can treat the chemicalproblem as involving cases of single equilibria.

In other words, we can treat most diprotic and triprotic acid systems as just a collection of monoprotic acids to which we canapply all the stuff we’ve learned the last couple of lectures.

[H

] = Ca

for strong acids

[H

] = (K

a

C

a

for weak acids

[H+] = K

a

C

a

/C

b

for buffers

There is one exception among the calculations and that is for what we call the amphiprotic acid case, and we will see that in amoment. But first, let’s consider the simpler calculations for polyprotic acids by specifically looking at the diprotic acid,carbonic acid: H

2CO

3

Here are the two equilibria

[H

][HCO

3

]

H

2CO

3

H

HCO

3

Ka

1

____________

= 1 x 10

[H

2

CO

3

]

[H

][CO

3

]

HCO

3

H

CO

3

Ka

2

= 1 x 10

[HCO

3

]

(By the way carbonic acid equilibria are

famous because they explain everything from why Diet Coke is acidic to why Tums

are good for an upset stomach to why Round Rock water spots your dishes and your car to why I can write on the board withchalk.)All that is left to do then, is make sure we choose the right equilibrium to solve a problem.

As we look at the word problems, let’s make things easier for ourselves by visualizing all the possible beakers containing formsof a diprotic acid and think about where we are on a titration curve if we were titrating a weak diprotic acid with a strong base.: Case 1. What is the pH of a 0.1 M H

2

CO

3

solution with K

a

-

Where are you on a titration curve? Beginning of weak acid titrationEquation used to determine the pH. H

= (K

a

C

a

Approximate pH.

Which is what you would expect for a weak acid. Case 2. What is the pH of a 0.1 M H

2

CO

3

solution mixed with 0.1 M NaHCO

3

with K

a

-

Where are you on a titration curve? In the buffer region for the first dissociation.Equation used to determine the pH. H

= K

a

C

a

/C

b

Approximate pH.

Which is what you would expect for a weak acid buffer. Case 3. What is the pH of a 0.1 M NaHCO

3

(the most famous antacid)with K

a

-

and K

a

-

Where are you on a titration curve? At the first equivalence point.Equation used to determine the pH. H

= (K

a

K

a

Approximate pH.

H

2

A

H

2

A

H

2

A

HA

HA

HA

Case 5. What is the pH of a 0.1 M 0.1 M Na

2CO

3

with K

a

-

Where are you on a titration curve? At the second end point.Equation used to determine the pH. This is a weak baseSo we use the weak base equation: OH

= (K

bC

b)

Approximate pOH

2 and

pH 12

Which makes sense since our K

b

value is pretty large.

Generalizing our calculation for a polyprotic acid: there are three different kinds of calculations you perform

Weak acid and weak base approximations at the beginning and the end of the titration when you are either fullyprotonated or deprotonated.

Buffers calculations when you have both an acid and its conjugate base in solution—note that diprotic acids have twobuffers, triprotic acids have three buffer regions, etc.

Amphiprotic calculations happen when you have a species capable of both adding and losing a proton. These compoundsneed two equilibrium expressions to explain them which is why they are the only equations so far in our work to requiretwo equilibrium constants.

Now go work the triprotic example in worksheet 7. It is just like this but far more painful and it needs to get done if you aregoing to do these problems on an exam.

A

=

A

=