pH of Weak Acids 1) The acid dissociation constant (Ka) for benzoic acid is 6.3 x 10-5. Find the pH of a 0.35 M solution of benzoic acid. 2) Find the pH of a 0.275 M hypochlorous acid solution. Ka = 3.0 x 10-8. 3) Find the pH of a solution that contains 0.0925 M nitrous acid (Ka = 4.5 x 10-4) and 0.139 M acetic acid (Ka = 1.8 x 10-5). 1) initial change equilibrium HC7H3O2(aq)  H+(aq) 0.35 M 0M -xM +xM (0.35 – x) M xM C7H3O2-(aq) 0M +x M xM + Note that: (0.35 – x) M ≈ 0.35 M so x2 = 6.3 x 10-5 Ka = [H+][ C7H3O2-] = (x)(x) = (x)(x) = [HC7H3O2] (0.35 – x) (0.35) (0.35) x2 = (6.3 x 10-5) (0.35) = 2.205 x 10-5 x = 4.7 x 10-3 M x = moles/L formed pH = - log (4.7 x 10-3) = 2.33 2) initial change equilibrium HClO(aq) 0.275 M -xM (0.275 – x) M  H+(aq) 0M +xM xM + ClO-(aq) 0M +x M xM Note that: (0.275 – x) M ≈ 0.275 M so Ka = [H+][ ClO-] = (x)(x) = (x)(x) = x2 = 3.0 x 10-8 [HClO] (0.275 – x) (0.275) (0.275) x2 = (3.0 x 10-8) (0.275) = 8.25 x 10-9 x = 9.08 x 10-5 M pH = - log (9.08 x 10-5) = 4.042 3) First the amount of H+ from each acid must be calculated. initial change equilibrium HNO2(aq) 0.0925 M -xM (0.0925 – x) M  H+(aq) 0M +xM xM + NO2-(aq) 0M +x M xM Note that: (0.0925 – x) M ≈ 0.0925 M so Ka = [H+][ NO2-] = (x)(x) = (x)(x) = x2 = 4.5 x 10-4 [HNO2] (0.0925 – x) (0.0925) (0.0925) x2 = (4.5 x 10-4) (0.0925) = 4.1625 x 10-5 x = 6.45 x 10-3 M x = moles/L formed initial change equilibrium HC2H3O2(aq)  0.139 M -xM (0.139 – x) M H+(aq) 0M +xM xM + C2H3O2-(aq) 0M +x M xM Note that: (0.139 – x) M ≈ 0.139 M so Ka = [H+][ C2H3O2-] = (x)(x) = (x)(x) = x2 = 1.8 x 10-5 [H C2H3O2] (0.139 – x) (0.139) (0.139) x2 = (1.8 x 10-5) (0.139) = 2.502 x 10-6 x = 1.58 x 10-3 M Then add the results together and use that value to find the pH. 6.45 x 10-3 M + 1.58 x 10-3 M = 8.03 x 10-3 M pH = - log (8.03 x 10-3) = 2.095

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