CIE A Level Chemistry

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5.5.3 pH & [H+] Calculations

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Calculating [H+] & pH

  • 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

Equilibria - Worked Example - pH calculations of a strong acid, downloadable AS & A Level Chemistry revision notes

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

Kw = [H+] [OH-]

Calculating H & pH equation 1

  • Since Kw is 1.00 x 10-14 mol2 dm-6

Calculating H & pH equation 2

  • 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

Calculating H & pH equation 3

Worked Example: pH calculations of a strong alkali

Equilibria - Worked Example - pH calculations of a strong alkali, downloadable AS & A Level Chemistry revision notes

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

Kw = [H+] [OH-]

Calculating H & pH equation 4

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

Since Kw 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 Ka value of the acid

Worked Example: pH calculations of weak acids

Equilibria - Worked Example - pH calculations of weak acids, downloadable AS & A Level Chemistry revision notes

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 Ka

Calculating H & pH equation 6

  • 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:

Calculating H & pH equation 7
  • Step 3: Rearrange the expression to find [H+]

Calculating H & pH equation 8
  • Step 4: Substitute the values into the expression to find [H+]

Calculating H & pH equation 9

= 1.32 x 10-3 mol dm-3

  • Step 5: Find the pH

pH = -log10 [H+]

= -log10 1.32 x 10-3

= 2.88

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