• If the concentration of H⁺?of an acid or alkali is known, the?pH?can be calculated using the equation:

pH = -log [H⁺]

• Similarly, the?concentration?of?H⁺?of a solution can be calculated if the pH is known by rearranging the above equation to:

[H⁺] = 10^-pH

Strong acids

• Strong acids?are completely?ionised?in solution

HA (aq) → H⁺?(aq) + A^-?(aq)

• Therefore, the concentration of hydrogen ions ([H⁺]) is?equal?to the concentration of acid ([HA])
• The number of hydrogen ions ([H⁺]) formed from the ionisation of water is?very small?relative to the [H⁺] due to ionisation of the strong acid and can therefore be?neglected
• The?total?[H⁺] is therefore the same as the [HA]

Worked Example: pH calculations of a strong acid

Answer

Hydrochloric acid is a strong monobasic acid

HCl (aq) → H⁺?(aq) + Cl^-?(aq)

Answer 1

The pH of the solution is:pH = -log [H⁺]

= -log 1.6 x 10^-4

= 3.80

Answer 2

The hydrogen concentration can be calculated by rearranging the equation for pH

pH = -log [H⁺]

[H⁺] = 10^-pH

= 10^-3.1

= 7.9 x 10^-4?mol dm^-3

Strong alkalis

• Strong alkalis?are completely?ionised?in solution

BOH (aq) → B⁺?(aq) + OH^-?(aq)

• Therefore, the concentration of hydroxide ions ([OH^-]) is?equal?to the concentration of base ([BOH])
  • Even strong alkalis have small amounts of H⁺?in solution which is due to the ionisation of water
• The concentration of OH^-?in solution can be used to calculate the pH using the?ionic product of water

K_w?= [H⁺] [OH^-]

• Since?K_w?is 1.00 x 10^-14?mol²?dm^-6

• Once the [H⁺] has been determined, the pH of the strong alkali can be founding using pH = -log[H⁺]
• Similarly, the ionic product of water can be used to find the concentration of OH^-?ions in solution if [H⁺] is known

Worked Example: pH calculations of a strong alkali

Answer

Sodium hydroxide is a strong base which ionises as follows:

NaOH (aq) → Na⁺?(aq) + OH^-?(aq)?

Answer 1

The pH of the solution is:pH = -log [H⁺]

= -log 3.5 x 10^-11

= 10.5

Answer 2

• ?Step 1:?Calculate hydrogen concentration by rearranging the equation for pH

pH = -log [H⁺]

= 10^-pH

= 10^-12.3

= 5.01 x 10^-13?mol dm^-3

• Step 2:?Rearrange the?ionic product of water??to find the concentration of hydroxide ions

K_w?= [H⁺] [OH^-]

• Step 3:?Substitute the values into the expression to find the concentration of hydroxide ions

Since?K_w?is 1.00 x 10^-14?mol²?dm^-6

= 0.0199^?mol dm^-3

Weak acids

• The pH of?weak acids?can be calculated when the following is known:
  • The?concentration?of the acid
  • The?K_a?value of the acid

Worked Example: pH calculations of weak acids

Answer

Ethanoic acid is a weak acid which ionises as follows:

CH₃COOH (aq) ? H⁺?(aq) + CH₃COO^-?(aq)

• Step 1:?Write down the equilibrium expression to find?K_a

• Step 2:?Simplify the expression

The ratio of H⁺?to CH₃COO^-?ions is 1:1

The concentration of H⁺?and CH₃COO^-?ions are therefore the same

The expression can be simplified to:

• Step 3:?Rearrange the expression to find [H⁺]

• Step 4:?Substitute the values into the expression to find [H⁺]

= 1.32 x 10^-3?mol dm^-3

• Step 5:?Find the pH

pH = -log₁₀?[H⁺]

= -log₁₀?1.32 x 10^-3

= 2.88