A sample of 40.0 milliliters of a .100 molarHC 2 H 3 O 2 solution is titrated wi

Question

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

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