Calculating the pH of a Strong Acid or Base

Video tutorial

The pH Scale

The pH scale

The concentrations of solutions of acids and bases can vary over many orders of magnitude. It is very inconvenient to have to report and deal with these concentrations as numbers so a logarithmic or 'p' scale:

• pH is related to the concentration of H₃O⁺(aq) (often written as just H⁺(aq)):

  pH = -log₁₀[H₃O⁺(aq) = -log₁₀[H⁺(aq)]
  
  where "log₁₀" means log to the base 10 and is written as "log" on calculator.

• pOH is related to the concentration of OH^-(aq):

  pOH = -log₁₀[OH^-(aq)

In water, [H₃O⁺(aq)] × [OH^-(aq)] is a constant equal to K_w at a given temperature. This allows pH and pOH to be interconverted:

• At room temperature (25 ° C or 298 K), K_w = 1.00 × 10^-14 meaning that:

  pH + pOH = 14.00
  pH = 14.00 - pOH
  pOH = 14.00 - pH

Although pOH or pH could be reported, almost always it is the pH that is given even for bases.

Strong acids

Strong acids

Strong acids like HCl, HBr, HI, HNO₃ and HClO₄ are completely dissociated in water:

• HCl+ H₂O(l) → H₃O⁺(aq) + Cl^-(aq)
• HBr + H₂O(l) → H₃O⁺(aq) + Br^-(aq)
• HI + H₂O(l) → H₃O⁺(aq) + I^-(aq)
• HNO₃ + H₂O(l) → H₃O⁺(aq) + NO₃^-(aq)
• HClO₄ + H₂O(l) → H₃O⁺(aq) + ClO₄^-(aq)

So if a bottle says a solution contains 1.0 mol per litre of HCl, it really contains 1.0 mol per litre of H₃O⁺(aq). A 0.10 M solution of HNO₃ has a concentration of H₃O⁺ of 0.10 M.

The pH of a strong acid can therefore be calculated straight from the concentration of the acid:

1. A 1.0 M solution of HCl has [H₃O⁺(aq)] = 1.0 M so:

   pH = -log₁₀[H₃O⁺(aq)] = -log₁₀(1.0) = 0.00

2. A 0.10 M solution of HNO₃ has [H₃O⁺(aq)] = 0.10 M so:

   pH = -log₁₀[H₃O⁺(aq)] = -log₁₀(0.10) = 1.00

Concentration of H3O+		10.0 M		1.0 M		0.10 M		0.010 M		0.0010 M		0.00010 M
pH		-1.00		0.00		1.00		2.00		3.00		4.00

Notice that the pH gets bigger as the concentration of [H₃O⁺(aq)] gets smaller. More concentrated acids have lower pH values!

Strong bases

Strong bases

Strong bases like the alkali metal hydroxides LiOH, NaOH, KOH and the alkaline earth hydroxides Mg(OH)₂ and Ca(OH)₂ are completely dissociated in water:

• LiOH → Li⁺(aq) + OH^-(aq)
• NaOH → Na⁺(aq) + OH^-(aq)
• KOH → K⁺(aq) + OH^-(aq)
• Mg(OH)₂ → Mg⁺(aq) + 2OH^-(aq)
• Ca(OH)₂ → Ca⁺(aq) + 2OH^-(aq)

So if a bottle says a solution contains 1.0 mol per litre of NaOH, it really contains 1.0 mol per litre of OH^-(aq). A 0.10 M solution of KOH has a concentration of OH^- of 0.10 M.

Because of the formula and the chemcial equations above, note that if a solution contains 10. mol per litre of Mg(OH)₂ or Ca(OH)₂, it contains 2 × 10. mol per litre of OH^-.

The pOH of a strong base can therefore be calculated straight from the concentration of the base. It can then be converted to the pH:

1. A 1.0 M solution of NaOH has [OH^-(aq)] = 1.0 M so:

   pOH = -log₁₀[OH^-(aq)] = -log₁₀(1.0) = 0.00
   pH = 14.00 - pOH = 14.00 - 0.00 = 15.00

2. A 0.10 M solution of KOH has [OH⁰(aq)] = 0.10 M so:

   pOH = -log₁₀[OH^-(aq)] = -log₁₀(0.10) = 1.00
   pH = 14.00 - pH = 14.00 - 1.00 = 13.00

3. A 0.10 M solution of Mg(OH)₂ has [OH⁰(aq)] = 2 × 0.10 M = 0.20 M so:

   pOH = -log₁₀[OH^-(aq)] = -log₁₀(0.20) = 0.70
   pH = 14.00 - pH = 14.00 - 0.70 = 13.30

Concentration of OH-		10.0 M		1.0 M		0.10 M		0.010 M		0.0010 M		0.00010 M
pOH		-1.00		0.00		1.00		2.00		3.00		4.00
pH		15.00		14.00		13.00		12.00		11.00		10.00

Notice that the pH gets smaller as the concentration of [OH^-(aq)] gets smaller. More concentrated bases have higher pH values!

Self test quiz



See also:


ChemCAL: Acids and Bases