Weak acids Weak acids only slightly dissociate when dissolved in water, giving an equilibrium mixture HA + H2O (l) H3O+ (aq) + A- (aq) HA (aq) H+ (aq) + A- (aq) We can simplify this to [H+ (aq)][A- (aq)] [HA (aq)] Ka= Weak acids dissociation expression The Ka for ethanoic acid is 1.7 x 10-5 mol dm-3 . The larger ka the stronger the acid [H+ (aq)][CH3CH2CO2 – (aq)] [CH3CH2CO2H(aq)] Ka= CH3CH2CO2H(aq) H+ (aq) + CH3CH2CO2 – (aq) Example 4 Write equation for dissociation of propanoic acid and its ka expression Calculating pH of a weak acid To make the calculation easier two assumptions are made to simplify the Ka expression: 1) [H+ (aq)]eqm = [A- (aq)] eqm because they have dissociated according to a 1:1 ratio 2) As the amount of dissociation is small we assume that the initial concentration of the undissociated acid has remained constant. So [HA (aq) ] eqm = [HA(aq) ] initial [H+ (aq)][A- (aq)] [HA (aq)] Ka= Simplifies to [H+ (aq)]2 [HA (aq)] initial Ka= Example 5 What is the pH of a solution of 0.01M ethanoic acid (ka is 1.7 x 10-5 mol dm-3 )? CH3CO2H(aq) H+ (aq) + CH3CO2 – (aq) [H+ (aq)][CH3CO2 – (aq)] [CH3CO2H(aq)] Ka= [H+ (aq)]2 [CH3CO2H(aq)] initial Ka= [H+ (aq)]2 0.01 1.7x 10-5 = [H+ (aq)]2 = 1.7 x 10-5 x 0.01 [H+ (aq)] = √ 1.7 x 10-7 = 4.12 x 10-4 pH = – log [H+ ] = -log (4.12 x10-4 ) pH =3.38 Example 6 What is the concentration of propanoic acid with a pH of 3.52 (ka is 1.35 x 10-5 mol dm-3 )? CH3CH2CO2H(aq) H+ (aq) + CH3CH2CO2 – (aq) [H+ (aq)][CH3CH2CO2 – (aq)] [CH3CH2CO2H(aq)] Ka= [H+ (aq)]2 [CH3CH2CO2H(aq)] initial Ka= [0.000302]2 [CH3CH2CO2H(aq)] initial 1.35 x 10-5 = [CH3CH2CO2H(aq)] = 9.12 x 10-8 /1.35 x 10-5 [H+ ] = 1 x 10-3.52 = 0.000302M [CH3CH2CO2H(aq)] = 6.75 x 10-3 M pKa Sometimes Ka values are quoted as pKa values pKa = -log Ka so Ka = 10-pKa

5.1.3 Acids, bases and buffers

(c) (i) the acid dissociation constant, Ka, for the extent of acid dissociation (see also 2.1.4 b) (ii) the relationship between Ka and pKa M0.1, M0.2, M0.4, M2.3, M2.4, M2.5

(g) calculations of pH, Ka or related quantities, for a weak monobasic acid using approximations M0.1, M0.4, M2.1, M2.2, M2.3, M2.4, M2.5 Approximations for weak acid calculations: [HA]equilibrium ~ [HA]undissociated i.e. [HA] >> [H+] [H+] equilibrium ~ [A–] equilibrium i.e. negligible dissociation of H2O. Learners will not be required to solve quadratic equations.(h) limitations of using approximations to Ka related calculations for ‘stronger’ weak acids M0.3 Including reasons why [HA]equilibrium ~ [HA]undissociated may no longer be valid. HSW6 Understanding of the circumstances under which Ka approximations break down.