pH and pOH of Strong Acids and Bases
Quick Notes
- Strong acids and bases fully dissociate and ionize in aqueous solution.
- For strong acids: [H3O+] = initial acid concentration
- can use pH = −log[H3O+]
- For strong bases: [OH−] = initial base concentration
- can use pOH = −log[OH−]
- For group II hydroxides, remember each formula unit contributes 2 OH− ions. (i.e. Ca(OH)2 = 2 × OH− ions).
- We can use the relationship pH + pOH = 14 at 25 °C.
Full Notes
Strong Acids and Ionization
Strong acids are substances that completely ionize in aqueous solution. This means that when a strong acid dissolves in water, every molecule donates a proton (H+) to water, forming hydronium ions (H3O+) and the conjugate base of the acid.
Example: Examples of strong acids include:
- Hydrochloric acid (HCl)
- Hydrobromic acid (HBr)
- Hydroiodic acid (HI)
- Nitric acid (HNO3)
- Perchloric acid (HClO4)
- Sulfuric acid (H2SO4) – for the first proton only
Because ionization is complete, the concentration of hydronium ions in solution is equal to the initial concentration of the acid (assuming no dilution or secondary reactions). This makes it easy to calculate the pH of strong acid solutions:
Example: If [HCl] = 0.010 M, then [H3O+] = 0.010 M
pH = −log(0.010) = 2.00
Note – this works because one mole of HCl reacts with one mole of H2O to form one mole of H3O+ (and one mole of Cl−)
HCl + H2O → H3O+ + Cl−
The direct relationship between acid concentration and [H3O+] is a key characteristic of strong acids and is useful when performing pH calculations.
Note - acids can be further classified as monoprotic, diprotic and triprotic. Monoprotic acids form one H3O+ ion per acid molecule (such as HCl and HNO3), diprotic acids form two H3O+ ions per acid molecule (such as H2SO4) and triprotic acids form three H3O+ ions per acid molecule (such as H3PO4).
What is the pH of a 0.010 M HCl solution?
- Recognize the acid is strong (full dissociation)
[H3O+] = 0.010 M - Apply the pH equation
pH = −log(0.010) = 2.00
Answer: The solution has a pH of 2.00 (acidic).
Strong Bases
Strong bases also dissociate completely in water to produce hydroxide ions, OH−. Group I hydroxides (like NaOH and KOH) release one hydroxide ion per formula unit, while Group II hydroxides (like Ca(OH)2 and Ba(OH)2) release two hydroxide ions per formula unit.
Just as we can use pH = -log10[H3O+] for strong acids, we can use pOH = -log10[OH-] for strong bases.
What is the pH of a 0.0050 M NaOH solution?
- Identify base type
NaOH is Group I → 1 OH− per formula unit, so [OH−] = 0.0050 M. - Find pOH
pOH = −log(0.0050) ≈ 2.30 - Convert to pH
pH = 14.00 − 2.30 = 11.70
Answer: The solution has a pH of 11.70 (basic).
What is the pH of a 0.0050 M Ba(OH)2 solution?
- Identify base type
Ba(OH)2 is Group II → 2 OH− per formula unit, so [OH−] = 2 × 0.0050 = 0.010 M. - Find pOH
pOH = −log(0.010) = 2.00 - Convert to pH
pH = 14.00 − 2.00 = 12.00
Answer: The solution has a pH of 12.00 (basic).
Key Equations
- pH:
- pOH:
and don't forget pH + pOH = 14 (at 25 °C)
Always check if the base provides more than one OH− ion per unit. Remember that temperature affects the value of Kw; pH + pOH = 14 only holds at 25 °C.
Summary
- For strong acids (like HCl, HNO3, HBr, etc.), they ionize completely in solution. This means the concentration of hydronium ions [H3O+] is equal to the initial concentration of the acid. You can calculate the pH directly using the formula: pH = −log[H3O+]
- For group I strong bases (like NaOH or KOH), they also fully dissociate, so the concentration of hydroxide ions [OH−] equals the initial concentration of the base. To find pH, calculate the pOH first: pOH = −log[OH−] Then use: pH = 14 − pOH
- For group II strong bases (like Ca(OH)2 or Ba(OH)2), each mole of base provides two moles of hydroxide ions. So, multiply the base concentration by 2 to get [OH−], then proceed as above.
- Finally, remember the key identity that relates pH and pOH at 25°C: pH + pOH = 14