1. Polyprotic Acids
And Acid and Base Salts

2. 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.

3. Diprotic acids and bases
How do we calculate the pH of a solution of:
H2A, HA-, or A2- ?

4. Diprotic Acids Treat this like a weak monoprotic acid: H2A HA- H+ I
0.0750 C -x +x E x x

5. Diprotic Acids
What about [A2-]?

6. Diprotic Acids
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 M

7. 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

8. Example
Calculate the H+, H3PO4, H2PO4-, HPO42-, and PO43- 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
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

9. Example Ka = x2 X = .023 0.10 - x [H3PO4] = 0.10 - .023 = 0.077 M
[H3O+] = [H2PO4-] = M

10. Example
Substituting what we know about the H3O+ and H2PO4- ion concentrations into the second equilibrium expression gives:
[HPO42-] = 6.3 x 10-8

11. Example
Substituting what we know about the concentrations of the H3O+ and HPO42- ions into this expression gives
PO43- = 1.2 x 10-18

12. 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) 12

13. 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. 13

14. 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.

15. 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.

16. Acid-base properties of salt solutions: hydrolysis
NaCl (aq) NH4Cl (aq) NaClO (aq)

17. Hydrolysis example
Which of the following salts, when added to water, would produce the most acidic solution?
a) KBr
b) NH4NO3
c) AlCl3
d) Na2HPO4
The answer is B

18. 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 F- } kb = 1.5 x 10-11
The acid is stronger than the base so the solution will be slightly acidic.

19. Example #2
What is the pH of a 0.10 M solution of NaOCl? HOCl, ka = 3.0x10-8
What type of salt is this? Na+ (from a strong base so does not effect the pH) and OCl- (from the weak acid HOCl)
Write the equilibrium expression for the dissolved salt.
OCl- + H2O < HOCl- + OH-

20. Example #2 cont. Kb = kw/ka = 3.3 x 10-7 From ICE chart kb = x2/0.10
x= 1.8 x 10-4 M,
pOH = 3.74, pH =10.26

21. 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