# Comparing Acid/Base Strength

• Conduct an investigation to demonstrate the use of pH to indicate the differences between the strengths of acids and bases

## Using pH to Compare Strength of Acids

We should be familiar by now with what the terms strong and weak refer to in the context of acids and bases.

Consider two solutions:

 Solution 1:  1.0 mol L-1 HCl (strong acid) Completely dissociates [H+] = 1.0 mol L-1  pH = 0.00 Solution 2:  1.0 mol L-1 CH3COOH (weak acid) Partially dissociates [H+] = 0.020 mol L-1  pH = 1.70
• Since solution 1 is a stronger acid and produces a higher [H+], it results in a lower pH
• Comparison is only fair when both acids have equal starting concentration and are monoprotic

Consider two more solutions:

 Solution 1:  1.0 mol L-1 HCl (strong acid) Completely dissociates [H+] = 1.0 mol L-1  pH = 0.00 Solution 2:  1.0 mol L-1 H2SO4 (strong acid) First proton completely dissociates Second proton partially dissociates [H+] = 1.2 mol L-1  pH = -0.79
• The diprotic strong acid produces a higher hydrogen ion concentration and results in an overall lower pH
• Comparison is only fair when both acids have equal starting concentration and both the first proton of HCl and H2SO4 are strong

## Using Electrical Conductivity to compare [H+]

• How effective a substance is at conducting electricity (movement of electrons

• Ions can act as charge carriers

• ↑ [ion] = ↑ electrical conductance

• [ion] is affected by concentration and strength of acid

# Comparing Base Strength

• Comparing base strength is similar to the comparison of acid strength
• pH: Equimolar stronger bases will have higher pH
• Conductance: stronger bases have greater electrical conductivity due to a higher [OH-]

## Solubility and pH

• Although alkali metal (group 1) hydroxides are water-soluble and will completely dissociate

$$NaOH(s) \rightarrow Na^+(aq) + OH^-(aq)$$

• Alkali Earth metal (group 2) hydroxides are sparingly soluble in water
• The dissociation is limited by solubility
• Dissociation reaches an equilibrium when the solution has been saturated

$$Mg(OH)_2(s) \leftrightharpoons Mg^{2+} (aq) + 2OH^-(aq)$$