,n,,mnn

1. pH = - log [H3O+]
[H3O+] = 10- pH mol/L
For pure water at 25oC
pH = - log (1.0 x 10-7) = 7.00
For a change in pH by 1, H3O+ concentration changes by 10
Higher pH, lower H3O+ concentration
pH of pure water is 7
pH of an acidic solution is less than 7
pH of a basic solution is greater than 7

3. pOH = - log [OH-]
pKw = - log Kw
pKw = 14.00 at 298 K
[H3O+ (aq)] [OH- (aq)] = Kw
- log[H3O+ (aq)] - log[OH- (aq)] = - log Kw
pH + pOH = pKw
At 298 K pH + pOH = 14.00

4. Strengths of Acids and Bases
The pH of 0.10 M HCl(aq) will be recorded as close to 1
The pH of a 0.10 M solution of CH3COOH(aq) solution is
recorded as ~ 3.
H3O+(aq) concentration in 0.10 M HCl(aq) is greater than that
in 0.10 M CH3COOH(aq)
HCl(aq) + H2O(l)  H3O+(aq) + Cl- (aq)
CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO- (aq)

5. CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO- (aq)
Ka =
[CH3COOH(aq)]
[H3O+(aq)]
[CH3COO- (aq)]
At 298 K, Ka for CH3COOH(aq) = 1.8 x 10-5
NH3 (aq) + H2O(l)  NH4
+(aq) + OH- (aq)
Kb =
[NH3(aq)]
[NH4
+(aq)]
[OH- (aq)]
At 298 K, Kb for NH3(aq) = 1.8 x 10-5
Ka: acidity constant
or acid dissociation
constant
Kb: basicity constant
or base dissociation
constant

6. The proton donor strength of an acid is measured by the
value of Ka; higher Ka, stronger the acid
The proton acceptor strength of a base is measured by Kb;
higher Kb, stronger the base
pKa = - log Ka
pKb = - log Kb
The larger the pK values, weaker the acid or base

9. Ka Kb = [H3O+(aq)] [OH- (aq)] = Kw
Or pKa + pKb = pKw
NH3 (aq) + H2O(l)  NH4
+(aq) + OH- (aq)
Kb =
[NH3(aq)]
[NH4
+(aq)]
[OH- (aq)]
NH4
+(aq) + H2O(l)  H3O+(aq) + NH3 (aq)
Ka =
[NH4
+( aq)]
[NH3 (aq)]
[H3O+(aq)]
Relationship between conjugate acid/base pairs

10. The stronger the acid/base, the weaker its conjugate
base/acid
pKa - pink
pKb - blue
HClO2(aq)/ ClO2
-(aq)
HOCl(aq)/ OCl-(aq)
CH3COOH(aq)/ CH3COO-(aq)
NH4
+(aq) / NH3(aq)
CH3NH3
+(aq) / CH3NH2(aq)

12. Using tabulated Ka and Kb values determine which species
is stronger as an acid or base
1) as acid HF(aq) or HIO(aq)
2)as base C6H5NH2(aq) or (CH3)3N(aq)
3) as acid C6H5NH3
+(aq) or (CH3)3NH+(aq)

13. Molecular Structure and Acid Strength
The more polar or weaker the H-A bond, the stronger the acid
Effect of bond strength
HF < HCl < HBr < HI
H - I bond is weakest
H2O < H2S < H2Se < H2Te
H-Te bond weakest

14. For an acid HA, greater the electronegativity of A, stronger
the acid
electronegativity difference
N-H 0.8
F-H 1.8
HF is an acid in water, NH3 is a base

15. Solutions of Weak Acids/Bases
For a strong acid and base; assume that
deprotonation/protonation reactions go to completion
HCl(aq) + H2O(l)  H3O+ (aq) + Cl- (aq)
pH = - log [H3O+ (aq)]
Knowing the concentration of HCl, can determine pH
For weak acids/bases, set up equilibrium table to determine
the H3O+ (aq) / OH- (aq) concentration at equilibrium,
knowing the value of Ka/Kb.

16. CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO- (aq)
Calculate the pH and percentage deprotonation of 0.10 M
CH3COOH(aq) given that Ka is 1.8 x 10-5.
Ka =
[CH3COOH(aq)]
[H3O+(aq)]
[CH3COO- (aq)]
CH3COOH(aq) CH3COO-(aq) H3O+(aq)
Initial 0.10 0 0
Change - x x x
Equilibrium 0.10 - x x x
1.8 x 10-5 = x2/(0.10 - x)

17. 1.8 x 10-5 = x2/(0.10 - x)
Since Ka is so small, assume that x << 0.10
1.8 x 10-5 ≈ x2/(0.10)
x = 1.3 x 10-3 M
[H3O+(aq)] = 1.3 x 10-3 M
pH = 2.89
% deprotonation = 100% x ([CH3COO-(aq)]/[CH3COOH]initial)
= 100% x (1.3 x 10-3 M)/(0.10 M) = 1.3 %
Note: x < 5% of 0.10 , OK to make this approximation

18. For a weak base
B(aq) + H2O(l)  HB+ (aq) + OH- (aq)
Use a similar approach to determine pOH knowing Kb, and
then determine pH
Determine the pH and percentage protonation of a 0.20 M
aqueous solution of methylamine, CH3NH2. The Kb for
CH3NH2 is 3.6 x 10-4.
pH = 11.9
% protonation = 4.2%

19. pH of Salt Solutions
CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l)
“neutralization” reaction
If a 0.3M solution of CH3COOH(aq) is added to a 0.3M solution
of NaOH, pH of resulting solution is not 7.0 but ~ 9.0
Solution of a salt is a solution of an acid (usually the cation)
and a base (usually the anion), and the pH depends on
their relative strength.
CH3COO-(aq) determines the pH of the solution

20. Ni(H2O)6
2+(aq) + H2O(l)  H3O+(aq) + Ni(H2O)5(OH)+(aq)

22. Estimate the pH of 0.15 M NH4Cl(aq).
Kb (NH3(aq)) = 1.8 x 10-5
NH4
+ (aq) is an acid and Cl- (aq) is neutral; expect pH < 7
NH4
+ (aq) + H2O(l)  H3O+ (aq) + NH3(aq)
Ka =
[NH4
+ (aq) ]
[H3O+(aq)]
[NH3(aq)]
NH4
+ (aq) NH3 (aq) H3O+(aq)
Initial 0.15 0 0
Change -x x x
Equilibrium 0.15-x x x

23. Ka (NH4
+ (aq)) =
Kw
Kb (NH3 (aq))
5.6 x 10-10 =
x2
0.15 - x
Assume x << 0.15
x ≈ 9.2 x 10-6 (agrees with the assumption)
pH = - log(9.2 x 10-6 ) = 5.04

24. Polyprotic Acids & Bases
A polyprotic acid can donate more than one H+
Carbonic acid: H2CO3(aq); dissolved CO2 in water
Sulfuric acid: H2SO4(aq)
Phosphoric acid: H3PO4(aq)
A polyprotic base: can accept more than one proton
Carbonate ion: CO3
2-(aq)
Sulfate ion: SO4
2-(aq)
Phophate ion: PO4
3-(aq)
Treat each step of protonation or deprotonation sequentially

25. H2CO3 (aq) + H2O(l)  H3O+(aq) + HCO3
-(aq) Ka1 = 4.3 x 10-7
HCO3
-(aq) + H2O(l)  H3O+(aq) + CO3
2-(aq) Ka2 = 4.8 x 10-11
Typically:
Ka1 >> Ka2 >> Ka3 >>…
Harder to loose a positively charged proton from a negatively
charged ion, because of attraction between opposite
charges.