A sample of 40.0 milliliters of a .100 molarHC2H3O2 solution is titrated witha .150 molar NaOH solution. Ka for aceticacid = 1.8 * 10-5. A) What volume of NaOH is used in thetitration in order to reach the equivalence point? B) What is the molar concentration ofC2H3O2- at theequivalence point? C) What is the pH of the solution at theequivalence point? D) Draw a sketch of the titrationcurve. A sample of 40.0 milliliters of a .100 molarHC2H3O2 solution is titrated witha .150 molar NaOH solution. Ka for aceticacid = 1.8 * 10-5. A) What volume of NaOH is used in thetitration in order to reach the equivalence point? B) What is the molar concentration ofC2H3O2- at theequivalence point? C) What is the pH of the solution at theequivalence point? D) Draw a sketch of the titrationcurve.
Explanation / Answer
a) Volume of NaOH required = MaVa /Mb = (40.0 mL)(0.1M) / (0.15M) = 26.7 mL b) molar concentration ofC2H3O2- at theequivalence point = (40.0 mL)(0.1M) / total volume =(40.0 mL)(0.1M) / 66.7 mL = 0.05997 M c) [OH-] = Kb*C = (5.56*10-10*0.05997M) =5.77*10-6 M pOH = -log(5.77*10-6 M ) = 5.24 pH = 14 -5.24 = 8.76 pOH = -log(5.77*10-6 M ) = 5.24 pH = 14 -5.24 = 8.76