AQA A Level Chemistry

Revision Notes

5.6.2 pKₐ

pKₐ

  • The range of values of Ka is very large and for weak acids, the values themselves are very small numbers

Table of Ka values

Table of Ka values, downloadable AS & A Level Chemistry revision notes

  • For this reason it is easier to work with another term called pKa
  • The pKa  is the negative log of the Ka value, so the concept is analogous to converting [H+] into pH values

pKa = -logKa

  • Looking at the pKa values for the same acids:

Table of pKvalues

Table of pKa values, downloadable AS & A Level Chemistry revision notes

  • The range of pKa values for most weak acids lies between 3 and 7

Worked Example

Finding Ka and pKa

At 298 K, a solution of 0.100 mol dm-3 ethanoic acid has a hydrogen ion concentration of 1.32 x 10-3 mol dm-3.

Calculate the Ka & pKa of the acid.

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

Calculating pH, Ka, pKA & Kw equation 2

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:

Calculating pH, Ka, pKA & Kw equation 3

Step 4: Substitute the values into the expression to find Ka

Calculating pH, Ka, pKA & Kw equation 4
Calculating pH, Ka, pKA & Kw equation 5

= 1.74 x 10-5

Step 5: Determine the units of Ka

Calculating pH, Ka, pKA & Kw equation 6

Calculating pH, Ka, pKA & Kw equation 7

= mol dm-3

The value of Ka is therefore 1.74 x 10-5 mol dm-3

Step 6: Find pKa

pKa = – log10Ka

= – log10 (1.74 x 10-5)

= 4.76

Kₐ from pH

How are Ka values found?

  • If we neutralise a weak acid with a strong base such sodium hydroxide the equation for the reaction is:

HA (aq) + NaOH (aq) → NaA (aq) + H2O (l)

HA (aq) + OH (aq) → A(aq) + H2O (l)

  • When half of the acid has been neutralised the concentration of [HA] is equal to the concentration of [A]
  • Since the acid is poorly dissociated it is assumed all the Acomes from the product rather than HA dissociating
  • This simplifies the expression to

Ka at half neutralisation, downloadable AS & A Level Chemistry revision notes

  • In practice, to find Ka the pH of a weak acid is measured as it is neutralised by a strong base and a graph is plotted for pH versus volume of base added
  • This type of graph is known as a pH titration curve
  • A vertical tie-line is drawn to the curve at half the volume of base needed for neutralisation
  • At this point  pKa = pH, so by reading off the pH value from the y-axis, you find the Ka of the acid

 

 

 

 

 

Finding the Ka of a weak acid from a pH titration curve

Exam Tip

You can regard the symbol p as meaning -log10 of a value. You don’t need to include the 10 as ‘log’ means log base 10. If a natural logarithm (base e) is required it is given the symbol ln.

Other uses of p include pOH and pKw . The latter gives a useful shortcut in problem solving:

                                                 Kw = [H+][OH] = 1.00 x 10-14 mol2 dm-6 at 298 K

                                               –logKw = –log[H+] + (-log[OH]) = -log(1.00 x 10-14)

                                                   pKw = pH + pOH = 14.00

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