pH, Ka, pKa & Kw Calculations (Cambridge (CIE) A Level Chemistry)
Revision Note
Calculating pH, Ka, pKA & Kw
pH
The pH indicates the acidity or basicity of an acid or alkali
The pH scale goes from 0.0 to 14.0
Acids have a pH below 7.0
Pure water is neutral with a pH of 7.0
Bases and alkalis have a pH above 7.0
pH can be calculated using:
pH = -log10 [H+]
where [H+] = concentration of H+ ions (mol dm-3)
The pH can also be used to calculate the concentration of H+ ions in solution by rearranging the equation to:
[H+] = 10-pH
Worked Example
Calculating the pH of acids
Calculate the pH of ethanoic acid, at 298K, when the hydrogen ion concentration is 1.32 x 10-3 mol dm-3.
Answer
pH = -log [H+]
pH = -log 1.32 x 10-3
pH = 2.9
Ka & pKa
The Ka is the acidic dissociation constant
It is the equilibrium constant for the dissociation of a weak acid at 298 K
For the partial ionisation of a weak acid HA, the equilibrium expression to find Ka is:
HA (aq) ⇌ H+ (aq) + A- (aq)
Ka =
When writing the equilibrium expression for weak acids, the following assumptions are made:
The concentration of hydrogen ions due to the ionisation of water is negligible
The dissociation of the weak acid is so small that the concentration of HA is approximately the same as the concentration of A-
The value of Ka indicates the extent of dissociation
A high value of Ka means that:
The equilibrium position lies to the right
The acid is almost completely ionised
The acid is strongly acidic
A low value of Ka means that:
The equilibrium position lies to the left
The acid is only slightly ionised (there are mainly HA and only a few H+ and A- ions)
The acid is weakly acidic
Since Ka values of many weak acids are very low, pKa values are used instead to compare the strengths of weak acids with each other
pKa = -log10 Ka
The less positive the pKa value the more acidic the acid is
Worked Example
Calculating the Ka & pKa of weak acids
Calculate the Ka and pKa values of 0.100 mol dm-3 ethanoic acid at 298K which forms 1.32 x 10-3 of H+ ions in solution.
Answer
Step 1: Write down the equation for the partial dissociation of ethanoic acid:
CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)
Step 2: Write down the equilibrium expression to find Ka:
Ka =
Step 3: Simplify the expression:
The ratio of H+ to CH3COO- is 1:1
The concentration of H+ and CH3COO- is, therefore, the same
The equilibrium expression can be simplified to:
Ka =
Step 4: Substitute the values into the expression to find Ka:
Ka =
Ka = 1.74 x 10-5
Step 5: Determine the units of Ka:
Ka = = mol dm-3
Therefore, Ka is 1.74 x 10-5 mol dm-3
Step 6: Find pKa:
pKa = - log10 Ka
pKa = - log10 (1.74 x 10-5)
pKa = 4.76
Kw
The Kw is the ionic product of water
It is the equilibrium constant for the dissociation of water at 298 K
Its value is 1.00 x 10-14 mol2 dm-6
For the ionisation of water, the equilibrium expression to find Kw is:
H2O (l) ⇌ H+ (aq) + OH- (aq)
Kw =
As the extent of ionisation is very low, only small amounts of H+ and OH- ions are formed
The concentration of H2O can therefore be regarded as constant and removed from the Kw expression
The equilibrium expression therefore becomes:
Kw = [H+] [OH-]
As the [H+] = [OH+] in pure water, the equilibrium expression can be further simplified to:
Kw = [H+]2
Worked Example
Calculating the concentration of H+ of pure water
Calculate the concentration of H+ in pure water, using the ionic product of water
Answer
Step 1: Write down the equation for the partial dissociation of water:
In pure water, the following equilibrium exists:
H2O (l) ⇌ H+ (aq) + OH- (aq)
Step 2: Write down the equilibrium expression to find Kw:
Kw =
Step 3: Simplify the expression:
Since the concentration of H2O is constant, this expression can be simplified to:
Kw = [H+] [OH-]
Step 4: Further simplify the expression:
The ratio of H+ to OH- is 1:1
The concentration of H+ and OH- is, therefore, the same and the equilibrium expression can be further simplified to:
Kw = [H+]2
Step 5: Rearrange the equation to find [H+]:
[H+] =
Step 6: Substitute the values into the expression to find Kw:
[H+] =
[H+] = 1.00 x 10-7 mol dm-3
Examiner Tips and Tricks
The greater the Ka value, the more strongly acidic the acid is
The greater the pKa value, the less strongly acidic the acid is.
Also, you should be able to rearrange the following expressions:
pH = -log10 [H+] ⇔ [H+] = 10-pH
Ka = ⇔ [H+] =
pKa = - log10 Ka ⇔ Ka = 10-pKa
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