Acids and bases when added to water disturb an equilibrium that already exists.The changes in the concentration of H3O + and OH - can be determined using the equilibrium law expression stated in the previous section. The calculation is different for strong vs weak acids and bases. We will consider strong acids and base calculations first.
Strong Acids / Bases
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Important
Idea
When strong acids are added to water the [H30+] in the solution will become equal to the concentration of [H30 +] produced from the dissolved acid (unless the concentration of the acid is very dilute). The same is true of [OH-] and bases.
This means that if the concentration of the [H3O +] produced by an acid is 2.0 mol/l the concentration of [H3O +] in the water solution will also be the same.
Explanation:
Consider adding 2 mol/l HCl to water. The equation below describes it's dissolving.
HCl (aq)
+
H2O(l)
H3O +
+
Cl-
  • The acid dissolves to form H30 + ions. Notice that the mole ratio of HCl: H30 + is 1:1.
  • therefore the [HCl] = [H3O +].
  • The concentration of [H3O +] added to this solution is therefore 2.0 mol/l.
  • This concentration is added to pure water whose [H3O +] is 1.0 x 10 - 7 mol/l (see above). If we add the two concentrations we get 2.0 mol/l + 1.0 x 10 - 7 mol/l = 2.0000007 mol/l.
    Considering significant figures then the concentration of the solution = 2.0 = concentration of HCl


Calculating concentrations of [H3O +] for strong acids and [OH- ] for strong bases
To calculate the concentrations of strong acids and bases we need to perform the following procedure
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important
procedure
Calculating [H3O + ] or [OH-] for strong acids and bases
Write the dissolving of the strong acid or base in water.
Calculate the concentration of the strong acid or base in mol/l.
Using the mole ratio from the equation calculate the concentration of [H30 +] or [OH-]
Using the Kw formula calculate the concentration of the other (H3O or OH-)
Examples and Assignments
 Question:
Calculate the [H3O +] and [OH-] concentrations of a solution created when 4.5 grams of Ca(OH)2 is dissolved in 300 ml of water.

Answer:
Dissolving reaction: Ca(OH)2 ------> Ca 2+ + 2 OH-
Concentration of Ca(OH)2 in mol/l = 4.5 g ÷ 74.10g/mol ÷ 0.300 L = 0.020 mol/l
Mole ratio of Ca(OH)2: OH is 1:2 . Therefore the [OH - ] = 2 x [Ca(OH)2]
[OH-] = 0.040 mol/l
 
[H3O+] = Kw/ [OH-] : Therefore [H3O+] = 1.0 x 10 -14 / 0.040 mol/l
[H3O+] = 2.5 x 10-13
 

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Check your understanding of calculating concentrations of [H3O+] and [OH-]
 click on the check mark and attempt the question

Assignment #1:Calculating [H3O+] and [OH-] for strong acids and bases
The following assignment gives you some drill and practice using the procedure above to calculate [H3O+] and [OH-]