Polyprotic Acids

And Acid and Base Salts

Polyprotic Acids So far, we have only dealt with acids that

can give up one proton.

Most acids encountered in biological systems have multiple protons, depending on the pH of the solution.

We call these ‘polyprotic’ acids or bases.

Diprotic acids and bases

How do we calculate the pH of a solution of:

H2A, HA-, or A2- ?

a22(aq)(aq)

-(aq)

a1(aq)(aq)(aq)2

K ;AHHA

K ;HAHAH

Diprotic Acids Treat this like a weak monoprotic acid:

3a1(aq)

-(aq)(aq)2 103.6K ;HAHAH x

3

(aq)2

(aq)-

(aq) 106.3]A[H

]HA][[H

x

H2A HA- H+

I 0.0750 0 0

C -x +x +x

E 0.0750-x x x

Diprotic Acids

0.056MM019.0M0750.0]A[H

M019.0]HA[]H[

case in this ignorecannot

10725.4103.6

103.6-0.0750

))((

2

-

432

3

x

x

xxxx

xx

xx

What about [A2-]?

Diprotic Acids

10a2

2(aq)(aq)(aq)

-

3a1(aq)

-(aq)(aq)2

109.4K ;AHHA

103.6K ;HAHAH

x

x

10

(aq)-

-2(aq)(aq) 104.9][HA

]A][[H

x

The amount of H+ from dissociation of HA- is insignificant relative to H+ from the dissociation of H2A, so total [H+] = 0.019M, and HA- also equals 0.019 M

M109.40.019)(

0.019)()10(4.9

][H

][HA104.9]A[ 10

10

(aq)

(aq)-10

-2(aq)

xxx

Polyprotic Acids Here are three successive ionizations of phosphoric

acid:

The first dissociation constants for phosphoric acid is much greater than the second, about 100,000 times greater

This means nearly all the H+ ions in the solution comes from the first step of dissociation.

H3PO4(s)   + H2O(l)    H3O+

(aq) + H2PO4−

(aq)       Ka1= 7.25×10−3

H2PO4−

(aq)+ H2O(l)    H3O+

(aq) + HPO42−

(aq)       Ka2= 6.31×10−8

HPO42−

(aq)+ H2O(l)    H3O+

(aq) +  PO43−

(aq)        Ka3= 3.98×10−13

Example

Calculate the H+, H3PO4, H2PO4-, HPO4

2-, and PO4

3- concentrations at equilibrium in a 0.10 M H3PO4 solution, for which Ka1 = 7.1 x 10-3, Ka2 = 6.3 x 10-8, and Ka3 = 4.2 x 10-13.

H3PO4(s)   + H2O(l)    H3O+

(aq) + H2PO4−

(aq)       Ka1= 7.25×10−3

H2PO4−

(aq)+ H2O(l)    H3O+

(aq) + HPO42−

(aq)       Ka2= 6.31×10−8

HPO42−

(aq)+ H2O(l)    H3O+

(aq) +  PO43−

(aq)        Ka3= 3.98×10−13

Example

Ka = x2 X = .023

0.10 - x

[H3PO4] = 0.10 - .023 = 0.077 M

[H3O+] = [H2PO4-] = 0.023 M

Example

Substituting what we know about the H3O+ and H2PO4

- ion concentrations into the second equilibrium expression gives:

[HPO42-] = 6.3 x 10-8

Example

Substituting what we know about the concentrations of the H3O+ and HPO4

2- ions into this expression gives

PO43- = 1.2 x 10-18

Salts

In general, salts are ionic compounds composed of metallic ions and nonmetallic ions

Salts dissociate in water. Salt solutions are generally electrolytes.

An electrolyte is a substance that ionizes or dissociates into ions when it dissolves in water (conducts electricity)

Salt + Water

The reaction of a salt and water to form an acid and base is called hydrolysis.

NaCl + H2O NaOH + HCl This is the reverse of a neutralization reaction in

which acid and bases react to form a salt and water.

When acids and bases react, the relative strength of the conjugated acid-base pair in the salt determines the pH of its solution.

Adding a “salt” to water

If the salt is from a strong acid or base, then nothing will happen (like adding table salt to water – no change in pH).

If it is a conjugate of a weak acid or base, then the “salt” is itself also a weak base or acid. So it hydrolyzes and makes some H+ or OH-, which changes the pH.

Adding a “salt” to water

If the salt is from a result of a strong acid and base then the pH is 7, for example KNO3.

A salt formed between a strong acid and a conjugate of a weak base is an acid salt, for example NH4Cl, and the pH will be acidic.

A salt formed between a conjugate of a weak acid and a strong base is a basic salt, for example NaCH3COO, and the pH will be basic.

Acid-base properties of salt solutions: hydrolysis

NaCl (aq) NH4Cl (aq) NaClO (aq)

Hydrolysis example

Which of the following salts, when added to water, would produce the most acidic solution?

a) KBr

b) NH4NO3

c) AlCl3d) Na2HPO4

The answer is B

Example #1 Will an aqueous solution that is 0.20 M NH4F

be acidic, basic or neutral?

NH4+ is the conjugate of a weak base and F- is

the conjugate of a weak acid, so how the two ions compare in their ability to affect the pH must be determined.

NH4+ } ka = 5.6 x 10-10 F- } kb = 1.5 x 10-11

The acid is stronger than the base so the solution will be slightly acidic.

Example #2

What is the pH of a 0.10 M solution of NaOCl? HOCl, ka = 3.0x10-8

1. What type of salt is this? Na+ (from a strong base so does not effect the pH) and OCl- (from the weak acid HOCl)

2. Write the equilibrium expression for the dissolved salt.

OCl- + H2O < HOCl- + OH-

Example #2 cont.

Kb = kw/ka = 3.3 x 10-7

3. From ICE chart kb = x2/0.10

x= 1.8 x 10-4 M,

pOH = 3.74, pH =10.26

Example #3

What is the pH of a 0.20 M solution of hydrazinium chloride, N2H5Cl? Hydrazine, N2H4, is a weak base with kb = 1.7 x 10-6.

Answer: conj. Acid in solution write the acid dissociation and use the ICE chart to determine the H+ and the pH.

x = 3.4 x 10-5, pH = 4.47