Chem 30: Acid-Base Equilibria
Welter Class Notes
3.0 pH Scale and Calculations
pH stands for the potential of hydrogen, that is, the concentration of hydrogen
ions in solution.
pH = - log [H+]
Example: Find the pH, given (a) [H+] = 1.0 x 10-3 and (b) [H+] = 2.5 x 10-11.
Solution:
(a) pH = - log [1.0 x 10-3] = 3
(b) pH = 10.6
3.1 The pH Scale

ranges from 0 to 14 to represent how acidic or basic a solution is
o acids have a pH <7
o the lower the pH, the stronger the acid
o a neutral solution has a pH of 7
o bases have pH >7
o the higher the pH, the stronger the base

the scale is logarithmic, meaning that every change of one unit on the scale
represents a ten-fold change in concentration (101)
o a solution with pH 3 is 10 times more acidic than a solution with pH 4
o a solution with pH 13 is 100 times (10 x 10) more basic than a solution
with pH 11
3.2 Calculations involving pH
Example 1: Calculate the pH of a 0.01M HNO3 solution.
Solution:
HNO3 (aq) → H+(aq) + NO3-(aq)
HNO3 is a strong acid with a 1:1 ratio between HNO3 and H+. So… [HNO3] = [H+] =
0.01 M
pH = - log [H+]
= - log [0.01]
= 2.0
Chem 30: Acid-Base Equilibria
Welter Class Notes
Example 2: Find the pH of a 0.01 M solution of ammonia.
NH3 (g) + H2O(l) ↔ NH4+(aq) + OH-(aq)
Solution:
NH3 is a weak base with Kb= 1.8 × 10-5 (always look up with weak acids/bases).
Since ammonia is a base, first calculate [OH-]. Then use Kw to determine [H+].
Next we calculate [H+]:
Kb
=
1.8 ×10-5
=
x2
=
(1.8 × 10-5) (0.01)
x2
=
1.8 × 10-7
x
=
4.2 × 10-4
[OH-]
=
4.2 × 10-4 M
Kw
[H+ ]
Solve for pH:
[NH4+] [OH- ]
[NH3]
(x) (x)
0.01
=
=
[H+] [OH-]
Kw
[OH-]
1.0 ×10-14
[H+]
=
[H+]
=
4.2 × 10-4
2.4 × 10-11
pH = - log [H+]
= - log [2.4 × 10-11]
= 10.6
Example 3: We have a solution with a pH = 8.3. What is [H+]?
Solution: [H+] = antilog (–pH)
= antilog (-8.3) = 5.0 × 10-9 M
Chem 30: Acid-Base Equilibria
Welter Class Notes
Example 4: A 0.24M solution of the weak acid, H2CO3, has a pH of 3.49. Determine
Ka for H2CO3 (carbonic acid).
Solution:
H2CO3 (aq) ↔ H+(aq) + HCO3-(aq)
Ka =
[H+][HCO3-]
[H2CO3]
[H+] = antilog (–pH)
= antilog (-3.49) = 3.2 × 10-4 M
1:1 ratio between [H+] and [H2CO3]
Ka =
Ka =
[H+][HCO3-]
[H2CO3]
[3.2 x 10-4][ 3.2 x 10-4]
[0.24]
Ka = 4.3 x 10-7
3.3 Calculations involving pOH
pOH = - log [OH-]
For bases, once we find [OH-] for a base, we can quickly determine pOH.
Example: Find the pOH, given [OH-] = 4.2 x 10-4.
Solution:
pOH = - log [4.2 x 10-4] = 3.4
pH + pOH = 14
Once we find pOH, it is a simple matter to find pH: pH + pOH = 14
pH = 14 – pOH
pH = 10.6
Chem 30: Acid-Base Equilibria
Welter Class Notes
3.0 pH Scale and Calculations
pH stands for the potential of______________, that is, the concentration of
hydrogen ions in solution.
Example: Find the pH, given (a) [H+] = 1.0 x 10-3 and (b) [H+] = 2.5 x 10-11.
Solution:
(a) pH = - log [1.0 x 10-3] = 3
(b) pH = 10.6
3.1 The pH Scale

ranges from 0 to 14 to represent how _________ or ________ a solution is
o acids have a pH <7
o the lower the pH, the stronger the acid
o a neutral solution has a pH of 7
o bases have pH >7
o the higher the pH, the stronger the base

the scale is _____________, meaning that every change of one unit on the
scale represents a _____ -fold change in concentration (101)
o a solution with pH 3 is ___ times more acidic than a solution with pH 4
o a solution with pH 13 is ___ times (10 x 10) more basic than a solution
with pH 11
3.2 Calculations involving pH
Example 1: Calculate the pH of a 0.01M HNO3 solution.
Solution:
HNO3 (aq) → H+(aq) + NO3-(aq)
HNO3 is a strong acid with a 1:1 ratio between HNO3 and H+. So… [HNO3] = [H+] =
0.01 M
pH = - log [H+]
= - log [0.01]
= 2.0
Chem 30: Acid-Base Equilibria
Welter Class Notes
Example 2: Find the pH of a 0.01 M solution of ammonia.
Solution:
NH3 is a weak base with Kb= 1.8 × 10-5 (always look up with weak acids/bases).
Since ammonia is a base, first calculate [OH-]. Then use Kw to determine [H+].
Next we calculate [H+]:
Solve for pH:
Example 3: We have a solution with a pH = 8.3. What is [H+]?
Solution: [H+] = antilog (–pH)
= antilog (-8.3) = 5.0 × 10-9 M
Chem 30: Acid-Base Equilibria
Welter Class Notes
Example 4: A 0.24M solution of the weak acid, H2CO3, has a pH of 3.49. Determine
Ka for H2CO3 (carbonic acid).
Solution:
H2CO3 (aq) ↔ H+(aq) + HCO3-(aq)
Ka =
[H+][HCO3-]
[H2CO3]
3.3 Calculations involving pOH
For bases, once we find [OH-] for a base, we can quickly determine pOH.
Example: Find the pOH, given [OH-] = 4.2 x 10-4.
Solution:
pOH = - log [4.2 x 10-4] = 3.4
Once we find pOH, it is a simple matter to find pH: pH + pOH = 14
pH = 14 – pOH
pH = 10.6