Question

What is the pH at the equivalence point in the titration of a 29.7 ml sample of a 0.464 M aqueous nitrous acid solution with a 0.433 M hydroxide solution? ucous sodium pH-

Answer 1

- CH Nog] = 0.464M - 9.9. ml [NaOH] = 0.433M Nach = ? At equivalence point Mili equiralent of acid = Mili equiralent of 1 base (Molarity) & Volume xn-factor = (Molanity) & Volum n factor 7 6:46 H x 2 1.1mLx 1 = 0.1 3 3 H x W L x! V Naoh = 0.464 x 29.7 ml 0.433 M 31.8 mL = Total = = 89.7 + 31-8 mb 61.5 mL . . :: Rxn: HNO + NaOH Ž NaNog + Ho 0.464 x 29.7 mb 6.433 x 31.8 mL 61.5 mL = 0.224M -0.224 M - 61.5 mL но 4 -0.224M -0.224 M +0.224 M - 0.0094 M -

→ only Nano is present at equiralence point a NaNo NaO + Nogo Conjugate base of Hwa. Hydrolysis reaction will occur. Nogo + H2O ² HNO + oHal Key= kwa I 0.2244 - ka = 4.5*10-1 c -y Key = 10-14 4.5x50-4 0.224-7 0.224 4.5 Keq for this reaction is very very small. → will reaction not go much in direction. forward y is very small, can be neglected. + TRę = 10-10 S Keq = Knydrolysis = [HNO] [OH-] [Nogo = y x y 0.9944 . 4 x 10-10 4.5 4.98 x 10 y = 2.3x10C = [OH-] por = -log Con-3 = -log [.23x10-6] = 6 - 10gQ.493) = 5.65 PH = 14-poh - 14-5.65 = 8.35