Calculations of pH
Definition: pH is abbreviation for
french pouvoir hydrogene. It is
expressed as the negative log of the
concentration of hydronium ion
pH = -log [H3O+]
pH < 7 = acid
pOH
pOH is the concentration of hydroxide
ion
pOH = -log [OH-]
pH > 7 is a base
pOH < 7 is a base
pH + pOH of a solution = 14
 If you know pH, you can
calculation pOH and vice versa
Calculating pH from [H3O+]
pH of an acid with 1 x 10–6 M [H3O+] is 6
 pOH is 8 (6 + 8 = 14)
pH of an acid with 1x 10-5 M [H3O+] is 5
 pOH is 9 (5 + 9 = 14)
Practice: What is the pH of the following?
 1 x 10-3 M HCl
 1 x 10-5 M HNO3
 1 x 10-4 M NaOH
 1 x 10–2 M KOH
Using a Calculator
[H3O+] = 3.4 x 10-5 M, find pH
pH = -log [3.4 x 10-5]
pH = 4.5
Steps: 3.4 EE 5 +/-log +/-
Practice: find pH
 [H3O+] = 6.7 x 10-4
 [H3O+] = 2.5 x 10-2
 [H3O+] = 2.5 x 10-6
Calculating [H3O+] from pH
If pH = 7.52, find [H3O+] and [OH-]
 pH = -log [H3O+]
 [H3O+] = antilog(-pH)
 [H3O+] = antilog(-7.52)
 Steps 7.52 +/- 2nd 10x
 [H3O+] = 3.02 x 10-8
 [H3O+] [OH-] = 1 x 10-14
 [OH-] = 3.3 x 10-7
 Practice, find [H3O+] and [OH-] from pH
 pH = 5.0
 pH = 12.0
 pH = 1.5

Calculations Of Ph

  • 1.
    Calculations of pH Definition:pH is abbreviation for french pouvoir hydrogene. It is expressed as the negative log of the concentration of hydronium ion pH = -log [H3O+] pH < 7 = acid
  • 2.
    pOH pOH is theconcentration of hydroxide ion pOH = -log [OH-] pH > 7 is a base pOH < 7 is a base pH + pOH of a solution = 14  If you know pH, you can calculation pOH and vice versa
  • 3.
    Calculating pH from[H3O+] pH of an acid with 1 x 10–6 M [H3O+] is 6  pOH is 8 (6 + 8 = 14) pH of an acid with 1x 10-5 M [H3O+] is 5  pOH is 9 (5 + 9 = 14) Practice: What is the pH of the following?  1 x 10-3 M HCl  1 x 10-5 M HNO3  1 x 10-4 M NaOH  1 x 10–2 M KOH
  • 4.
    Using a Calculator [H3O+]= 3.4 x 10-5 M, find pH pH = -log [3.4 x 10-5] pH = 4.5 Steps: 3.4 EE 5 +/-log +/- Practice: find pH  [H3O+] = 6.7 x 10-4  [H3O+] = 2.5 x 10-2  [H3O+] = 2.5 x 10-6
  • 5.
    Calculating [H3O+] frompH If pH = 7.52, find [H3O+] and [OH-]  pH = -log [H3O+]  [H3O+] = antilog(-pH)  [H3O+] = antilog(-7.52)  Steps 7.52 +/- 2nd 10x  [H3O+] = 3.02 x 10-8  [H3O+] [OH-] = 1 x 10-14  [OH-] = 3.3 x 10-7  Practice, find [H3O+] and [OH-] from pH  pH = 5.0  pH = 12.0  pH = 1.5