pH Calculations (OCR A Level Chemistry A) : Revision Note
pH Calculations for Strong Acids & Bases
Strong acids
Strong acids are completely ionised in solution
This can also be described as dissociating in the ions
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 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
What is the pH of 0.01 mol dm-3 hydrochloric acid?
Answer
[HCl] = [H+] = 0.01 mol dm-3
pH = - log[H+]
pH = - log[0.01] = 2.00
Strong bases
Strong bases are completely ionised in solution
This can also be described as dissociating in the ions
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
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, simply by dividing Kw by the [H+]
Worked Example
Question 1: Calculate the pH of 0.15 mol dm-3 sodium hydroxide, NaOH
Question 2: Calculate the hydroxide concentration of a solution of sodium hydroxide when the pH is 10.50
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:
[H+] = Kw ÷ [OH-]
[H+] = (1 x 10-14) ÷ 0.15 = 6.66 x 10-14
pH = -log[H+]
= -log 6.66 x 10-14 = 13.17
Answer 2
Step 1: Calculate hydrogen concentration by rearranging the equation for pH
pH = -log[H+]
[H+]= 10-pH
[H+]= 10-10.50
[H+]= 3.16 x 10-11 mol dm-3
Step 2: Rearrange the ionic product of water to find the concentration of hydroxide ions
Kw = [H+] [OH-]
[OH-]= Kw ÷ [H+]
Step 3: Substitute the values into the expression to find the concentration of hydroxide ions
Since Kw is 1 x 10-14 mol2 dm-6,
[OH-]= (1 x 10-14) ÷ (3.16 x 10-11)
[OH-]= 3.16 x 10-4 mol dm-3
Worked Example
What is the pH of a solution of hydroxide ions of concentration 1.0 × 10−3 mol dm−3 ?
Kw = 1 × 10−14 mol2 dm-6
A. 3.00
B. 4.00
C. 10.00
D. 11.00
Answer
The correct option is D
Since Kw = [H+] [OH–], rearranging gives [H+] = Kw ÷ [OH–]
The concentration of [H+] is (1 × 10−14) ÷ (1.0 × 10−3) = 1.0 × 10−11 mol dm−3
[H+]= 10-pH
So the pH = 11.00
pH Calculations for Weak Acids
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
From the Ka expression we can see that there are three variables:
However, the equilibrium concentration of [H+] and [A-] will be the same since one molecule of HA dissociates into one of each ion
This means you can simplify and re-arrange the expression to
Ka x [HA] = [H+]2
[H+]2 = Ka x [HA]
Taking the square roots of each side
[H+] = √(Ka x [HA])
Then take the negative logs
pH = -log[H+] = -log√(Ka x [HA])
Worked Example
Calculate the pH of 0.100 mol dm-3 ethanoic acid, at 298 K, with a Ka value of 1.74 × 10-5 mol dm-3
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
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 = -log[H+]
= -log(1.32 x 10-3)
= 2.88
Limitations of Ka
We must make assumptions when calculating the pH of a weak acid
[H+] at equilibrium is equal to the [A-] at equilibrium because they have dissociated according to a 1:1 ratio
This is because the amount of H+ from the dissociation of water is insignificant
The amount of dissociation is so small that we assume that the initial concentration of the undissociated acid has remained constant
So initial [HA] is equal to the [HA] at equilibrium
The strength of acid
The stronger the acid the greater the concentration of hydrogen ions in the solution at equilibrium
This corresponds to a larger value for Ka
If the acid is stronger, the dissociation will be greater, therefore the difference in the values for initial [HA] and [HA] at equilibrium will be greater
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