 # pH calculations for AQA A-level Chemistry 1. Calculating pH
2. Ionic product of water
3. Dissociation constant
4. Buffers

To simplify the measurement of the acidity or the alkalinity of a solution, pH scale is used. pH is a negative logarithm with a base of 10 of the hydrogen ion concentration in a solution. Most of the pH values are in between 0 and 14, yet occasionally some of them are outside of this range. Hydrogen ion concentration in a solution can be found by raising 10 to the power of negative pH. Although in many reactions we assume that water does not dissociate, a small fraction of H₂O molecules split into H⁺ and OH⁻ ions. The multiplication product of [H⁺] and [OH⁻] is called the ionic product of water. The value of this measure depends on the temperature, at room temperature and many general questions Kw = [H⁺][OH⁻] = 10⁻¹⁴. The ionic product of water can be simplified in a similar way to pH by obtaining pKw = −log₁₀Kw = 14. Dissociation constant is a measure of a the ability of an acid to dissociate. The dissociation constant is found by multiplying the concentrations of the two dissociated ions A⁻ and H⁺, one of which is hydrogen, and dividing the result by the concentration of the undissociated acid HA. The dissociation constant can be simplified in a similar way to pH by obtaining pKa = −log₁₀Ka. For acids that have [HA] much greater than [A⁻], the initial concentration of HA before the dissociation for the calculations can be taken. A buffer is a solution that responds to changes in pH or concentration by adjusting its composition to return to the initial pH value. Buffer solutions can be grouped to weak acid + its salt and weak base + its salt solutions. An example of a buffer is blood which ensures a relatively constant pH by using an equilibrium between H₂CO₃ and HCO₃⁻.

Buffers shift an equilibrium in reversible reactions to counteract an introduction of new ions. When hydrogen ions are added to a weak acid + salt buffer solution, the equilibrium shifts towards the left (undissociated acid); when a base is added to such buffer, the equilibrium shifts towards the right (dissociated ions). When hydrogen ions are added to a weak base + salt buffer solution, the equilibrium shifts towards the right (dissociated ions); when a base is added to such buffer, the equilibrium shifts towards the left (undissociated base). # 1

20 cm³ of 0.25 mol/dm³ HNO₃ solution is mixed with 10 cm³ of 0.75 mol/dm³ HNO₃ solution. What is the pH of the new solution produced?

n₁ = c₁ × V₁ = 0.020 × 0.25 = 0.0050 mol
n₂ = c₂ × V₂ = 0.010 × 0.75 = 0.0075 mol
n = n₁ + n₂ = 0.0050 + 0.0075 = 0.0125 mol
V = V₁ + V₂ = 0.020 + 0.010 = 0.030 dm³
c = n ÷ V = 0.0125 ÷ 0.030 = 0.42 mol/dm³
pH = −log₁₀[H⁺]
pH = −log₁₀0.42
pH = 0.38

0.38 # 2

Describe what happens when a few drops of sodium hydroxide solution are added to an ethanoic acid and sodium ethanoate buffer.

When a base is added to a weak acid + salt buffer solution, the equilibrium shifts towards the right (dissociated ions). Therefore, some of the ethanoic acid in the buffer will dissociate to compensate for the H⁺ ions lost due to OH⁻ ions reacting with them.

Some of the ethanoic acid will dissociate to compensate for the H⁺ ions lost due to OH⁻ ions. # 3

Which two of the following compounds can be combined together with water to make a buffer solution?

• CH₄
• HCOOH
• NaCl
• HCOONa
• CH₃COOH

Buffer solutions can be grouped to weak acid + its salt and weak base + its salt solutions. In this case an acid + its salt solution can be achieved by using methanoic acid and sodium methanoate.

HCOOH, HCOONa # 4

Provide an equation that defines the dissociation constant of hydrofluoric acid.

The dissociation constant is found by multiplying the concentrations of the two dissociated ions A⁻ and H⁺, one of which is hydrogen, and dividing the result by the concentration of the undissociated acid HA.

HF ⇌ H⁺ + F⁻

Ka = [H⁺][F⁻]/[HF] # 5

The hydrogen ion concentration in a solution is 2 × 10⁻⁴ mol/dm³. Find the concentration of hydroxide ions in this solution at room temperature.

[H⁺][OH⁻] = 10⁻¹⁴
2 × 10⁻⁴ × [OH⁻] = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴ ÷ (2 × 10⁻⁴) = 5 × 10⁻¹¹ mol/dm³

5 × 10⁻¹¹ mol/dm³ End of page