pH & pOH of Strong Acids & Bases
Strong Acids
- Most strong acids are monoprotic and fully ionize in aqueous solutions
- Examples include HCl (hydrochloric acid) and HNO3 (nitric acid)
- Due to the acid fully ionizing, there is a high concentration of H+/H3O+ ions resulting in the solution having a low pH
- The H+/H3O+ concentration is equal to the molar concentration of the acid:
- [H+ (aq)] = Mstrong acid
- This can be used to calculate the pH using:
- pH = -log10[H+ (aq)]
- The position of the equilibrium is so far over to the right that you can represent the reaction as an irreversible reaction
Diagram to show the ionization of a strong acid
The diagram shows the complete dissociation of a strong acid in aqueous solution
Strong Bases
- Strong bases, such as Group 1 metal hydroxides, completely dissociate in aqueous solutions
- A large concentration of OH– ions are present in the solution causing it to have a high pH
- Group 1 hydroxides produce one hydroxide ion per formula unit but others, such as Ca(OH)2 will produce more than one
- Therefore the concentration of hydroxide ions can be calculated using:
- [OH-] = Mstrong base x number of OH- ions per mole
- This can then be used to calculate pOH:
- pOH = -log [OH–]
- pOH can then be used to calculate pH:
- pH = 14 - pOH
- The dissociation of a strong base can be represented using the diagram below
- The position of the equilibrium is on the right and so the reaction is irreversible
Diagram to show the dissociation of a strong base
The diagram shows the complete dissociation of a strong base in aqueous solution
Worked example
1. Calculate the pH for a solution with a hydrogen ion concentration of 1.6 x 10-4 M.
2. Calculate the pH of a solution containing 0.00052 M Ba(OH)2.
Answers:
Answer 1:
- The pH of the solution is:
- pH = -log [H+]
- pH = -log 1.6 x 10-4
- pH = 3.80
Answer 2:
- The pH of the solution is:
- [OH-] = 0.00052 x 2 = 0.00104
- pOH = -log 0.00104 = 2.98
- pH = 14 - 2.98 = 11.02