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 

K_a & pK_a

• The K_a 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 K_a is:

HA (aq) ⇌ H⁺ (aq) + A^- (aq)

K_a = $\frac{\left[H+\right]\left[A^{-}\right]}{\left[\mathrm{HA}\right]}$

• 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 can be approximated by its initial value

    • This is calculated from the number of moles divided by the volume of solvent

• The value of K_a indicates the extent of dissociation

  • A high value of K_a means that:

    • The equilibrium position lies to the right

    • The acid is almost completely ionised

    • The acid is strongly acidic

  • A low value of K_a 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 K_a values of many weak acids are very low, pK_a values are used instead to compare the strengths of weak acids with each other

pK_a = -log₁₀ K_a

• The less positive the pK_a value the more acidic the acid is

K_w

• The K_w 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 K_w is:

H₂O (l) ⇌ H⁺ (aq) + OH^- (aq)

K_w = $\frac{\left[H^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[H_{2}O\right]}$

• 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 K_w expression

• The equilibrium expression therefore becomes:

K_w = [H⁺] [OH^-]

• As the [H⁺] = [OH⁺] in pure water, the equilibrium expression can be further simplified to:

K_w = [H⁺]²