1. pH

2. Hydrogen Ions and Acidity
Water molecules go through self ionization to form hydrogen ion and hydroxide ions.
H2O(l)
H+(aq) + OH-(aq)
In aqueous solution, hydrogen ions H+ are always joined to a water molecule as hydronium ions
In pure (neutral) water, the self-ionization of water results in 1 x 10-7 M of H+ ions and 1 x 10-7 M of OH- ions
H2O (l) + H2O (l)
H3O+(aq) + OH-(aq)

3. Keq = [H3O+]*[OH-] Called the Kw or the ion product constant
H2O (l) + H2O (l)
H3O+(aq) + OH-(aq)
You can write an equation for an equilibrium constant for this!!
Called the Kw or the ion product constant
Kw ALWAYS equals 1x10-14 at 25ºC, regardless of what else is in the solution, like an acid or base
In a basic/alkaline solution, the hydroxide ion (OH-) is greater than 1x10-7 M and the hydrogen ion (H+) is less 1x10-7 M
In a acidic solution, the hydrogen ion (H+) is greater than 1x10-7 M and the hydroxide ion (OH-) is less 1x10-7 M
Keq = [H3O+]*[OH-]
You can use [H+] as an alternative here as well.

4. Kw = [H+]*[OH-] 1x10-14 = (2.6*10-3 M)* [OH-] 3.85* 10-12 = [OH-]
A can of Coke has [H+]= 2.6*10-3 M. What is the concentration of OH-?
Kw = [H+]*[OH-]
1x10-14 = (2.6*10-3 M)* [OH-]
3.85* = [OH-]

5. KW = [H+]*[OH-] 1x10-14 = (5.99*10-8 M)* [H+] 1.67* 10-7 = [H+]
Calculate the [H+] concentration in each of the following:
[OH-] = 5.99*10-8M
[OH-] = 1.43 * M
KW = [H+]*[OH-]
1x10-14 = (5.99*10-8 M)* [H+]
1.67* 10-7 = [H+]
1x10-14 = (1.43 * M)* [H+]
6.99* 10-3 = [H+]

6. The pH Concept pH + pOH = 14 pH = -log[H+] pOH = -log[OH-]
pH is the negative logarithm of the hydrogen ion concentration
A neutral solution H+ = 1x10-7 has a pH = -log[1x10-7]= 7
A solution in which the [H+] is greater than 1x10-7 M and has a pH less than 7.0 is acidic
A solution with a pH greater than 7 is basic and has a [H+] concentration of less than 1x10-7 M
You can also calculate pOH which is the negative logarithm of hydroxide ion concentration
pOH = -log[OH-]
pH + pOH = 14

7. pH and significant figures
Hydrogen ion concentrations should always be reported to two significant figures
pH and pOH calculations should always be reported to two decimal places
Rules are due to the sensitivity of pH meters
How to solve pH/pOH problems: pH
[H+]
[OH-]
pOH

8. pH practice Calculate the pH of a solution with [H+] = 7.42*10-5M
Calculate the pOH of a solution with [OH-] = 4.21*10-9M
pH = -log[H+]
pH = -log[7.42*10-5]
pH = 4.13
pOH = -log[OH-]
pH + pOH = 14
pOH = -log[4.21*10-9]
pOH = 8.38
pH = ?
pH = 5.62

9. pH practice The pH of a solution is 3.85. What is the [H+]?
What is the [OH-] concentration of a solution with pOH = 10.52?
pH = -log[H+]
3.85 = -log[H+]
-3.85 = log[H+]
Find the inverse log (inv log) or 10x of this number on calculator
1.41*10-4 = [H+]
pOH = -log[OH-]
10.52 = -log[OH-]
= log[OH-]
3.02*10-11 = [OH-]

10. Calculate the pOH of a solution with [H+] = 6.32*10-5M.
Kw = [H+]*[OH-]
pH = -log[H+]
pH = -log[6.32*10-5]
1x10-14 = (6.32*10-5M)* [OH-]
1.58* = [OH-]
pH = 4.20
pOH = -log[OH-]
pH + pOH = 14
pOH = -log[1.58* ]
pOH = 14
pOH = 9.80
pOH = 9.80

11. Strong and Weak Acids and Bases
A strong acid completely ionizes in water – simple Keq
Weak acids ionize only slightly in aqueous solution, which means there is a still an aqueous component to them – need to rewrite equilibrium constant
Remember – concentrated ≠ strong
HCl(aq) + H2O(l)
H3O+(aq) + Cl-(aq) % ionized
CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO-(aq) <1% ionized

12. pH of 0.10 M Solutions of Common Acids and Bases
Compound  pH
HCl (hydrochloric acid) 
H2SO4 (sulfuric acid) 
H2SO3 (sulfurous acid) 
H3PO4 (phosphoric acid) 
HF (hydrofluoric acid) 
CH3CO2H (acetic acid)
H2CO3 (carbonic acid) (saturated solution)
H2S (hydrogen sulfide) 
NH4Cl (ammonium chloride) 
HCN (hydrocyanic acid) 
Na2SO4 (sodium sulfate) 
NaCl (sodium chloride) 
NaHCO3 (sodium bicarbonate) 
Na2SO3 (sodium sulfite) 
NaCN (sodium cyanide) 
NH3 (aqueous ammonia) 
Na2CO3 (sodium carbonate) 
Na3PO4 (sodium phosphate) 
NaOH (sodium hydroxide, lye) 

13. Acid Disassociation Constant
The equilibrium constant for weak acids (HA) can be written as:
Acid
Conjugate base
H3O+(aq) + A-(aq)
HA(aq) + H2O(l)
Keq
[H3O+] X [A-]
[HA] X [H2O] =
For dilute solutions, the concentration of water is a constant, and can be combined with Keq to give the acid dissociation constant (Ka)
Keq X H2O = Ka
[H3O+] X [A-]
[HA] =
Ka is sometimes referred to as the ionization constant
Weak acids have small Ka values, while stronger acids have larger Ka values

14. Base Disassociation Constant
The equilibrium constant for weak Bases (B) can be written as:
base
Conjugate acid
BH+(aq) + HO-(aq)
B(aq) + H2O(l)
Keq
[BH+] X [HO-]
[B] X [H2O] =
Again, for dilute solutions, the base dissociation constant (Kb)
Keq X H2O = Kb
[BH+] X [OH-]
[B] =
The magnitude of Kb indicates the ability of a weak base to compete with the very strong base OH- for hydrogen ions
The smaller the Kb the weaker the base

15. Calculating Dissociation Constants
Problems involving Ka of weak acid or the Kb of a weak base:
Write the expression for Ka or Kb using the actual compounds in the chemical equation. Rearrange for the unknown variable.
Substitute the known numbers of all the variables at equilibrium into the expression for Ka or Kb.
Solve for the unknown.

16. A M solution of acetic acid is only partially ionized and has a pH of What is the acid dissociation constant (Ka) or ethanoic acid?
H3O+(aq) + C2H3O2-(aq)
HC2H3O2(aq) + H2O(l)
pH = -log[H+]
Ka =
2.87 = -log[H+]
-2.87 = log[H+]
1.35*10-3 = [H+]
= [C2H3O2-]
Ka =
Ka = 1.82*10-5

17. The Ka of nitrous acid is 4. 10-4
The Ka of nitrous acid is 4.0 *10-4. What is the concentration of [H+] in a 0.85M solution of nitrous acid? What is the pH?
H3O+(aq) + NO2-(aq)
HNO2(aq) + H2O(l)
Ka =
pH = -log[H+]
pH = -log[1.84*10-2]
pH = 1.73
Ka * [HNO2] = [NO2-]*[H+]
Ka * [HNO2] = x*x = x2
√(Ka∗[HNO2] =x
√(4.0∗ 10 −4 ∗0.85)=x
1.84*10-2= x = [H+]