How does pH relate to pKa?

pH measures the concentration of H⁺ (protons) in an aqueous solution. pK_a 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⁺ + SO₄^2-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 pK_a, the better the acid dissociatespK_a = -log K_a, Ka = the acid dissociation constant, describing the extent of acid dissociationEquilibrium Law = K_a = [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 K_a, meaning that it is reversible.Calculate the pH of 0.2 M ethanoic acid (K_a = 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