How does pH relate to pKa?
pH measures the concentration of H+ (protons) in an aqueous solution. pKa measures the ability for an acid to dissociate in an aquesous solution. Their relationship can be described my the Henderson-hasselbalch equation.Consider: HA (aq) <-> [H+(aq)] + [A-(aq)]where HA = acid solution, [H+] = concentration of H+, [A-] = concentration of conjugate basepHdepends on the concentration of H+ in solutionThe lower the pH, the higher the [H+], the more acidic the solution ispH = - log [H+]Example: pH of strong acidsStrong acids dissociates completely, meaning that the reaction is not reversible.Calculate the pH of a solution with 0.25 M sulfuric acidH2SO4 -> 2 H+ + SO42-Find the ratio of H2SO4 : H+ = 1 : 2 ———> 1 mole of H2SO4 gives 2 moles of H+Calculate the [H+]: [H+] = 0.25 x 2 M = 0.5 MApply the pH equation: pH = - log 0.5 = 0.3 (1 d.p.) {pH value is always to 1 d.p.}pKaCorrelates positively to pH, the lower the pH, the lower the pKa, the better the acid dissociatespKa = -log Ka, Ka = the acid dissociation constant, describing the extent of acid dissociationEquilibrium Law = Ka = [H+] [A-] / [HA]Hehenderson-Hasselbalch equationRelates pH and pKa in weak acid solutions onlyIt is only an assumption and should NOT be apply to extreme pHs.pH = pKa ([H+] [A-]/ [HA])Example: pH of weak acidsWeak acids ONLY dissociates to an extent, depending on its Ka, meaning that it is reversible.Calculate the pH of 0.2 M ethanoic acid (Ka = 1.78 x 10-5)CH3COOH <-> CH3COO- + H+Applying the equilibrium law: Ka = [H+] [A-] / [HA] ———> 1.78 x 10-5 = [H+] [CH3COO-] / 0.2Assumption: CH3COOH at equilibrium is approximately the same as that of the original (0.2 M)Find the ratio of CH3COO- : H+ = 1 : 1 ———> [H+] = [CH3COO-]Rearrangement: 1.78 x 10-5 x 0.2 = [H+]2, so H+ = 1.89 x 10-3Apply pH equation: pH = -log (1.89 x 10-3) = 2.7 (1 d.p.)Examples References: http://alevelchem.com/aqa_a_level_chemistry/unit3.4/s3403/04.htm
