The titration of a 23 ml sample of 0.110 M HC2H3O2 with 0.130 M NaOH. Determine

Question

The titration of a 23 ml sample of 0.110 M HC2H3O2 with 0.130 M NaOH. Determine the following? the initial pH, the volume of added base required to reach the equivalence point, the pH at 5.00 ml of added base, the ph at one-half of the equivalence point, the ph at equivalence point. pH at equivalence point = 4.74 the volume of added base required to reach the equivalence point = 19.5 ml the initial pH = 2.85 the pH after adding 6.00 mL of base beyond the equivalence point

Explanation / Answer

Ka = 1.8 x 0^-5 = x^2/ 0.110-x x = [H+] =0.00141 M pH = 2.85 moles acetic acid = 0.0210 x 0.110=0.00231 moles NaOH required = 0.00231 Volume NaOH = 0.00231/ 0.130 M=0.0178 L => 17.8 mL moles NaOH = 5.00 x 10^-3 L x 0.130 M=0.00065 moles acetic acid in excess = 0.00231 - 0.00065=0.00166 moles acetate formed = 0.00065 pKa = 4.74 pH = 4.74 + log 0.00065/ 0.00166=4.33 at one half equivalence point moles acid = moles acetate pH = 4.74 moles NaOH required to reach the equivalence point = 0.00231 volume NaOH = 0.00231/ 0.130 M=0.0178 L total volume = 0.0178+ 0.0210 =0.0388 L moles acetate formed = 0.00231 concentration acetate = 0.00231/ 0.0388 =0.0596 M CH3COO- + H2O CH3COOH + OH- Kb = Kw/Ka = 5.56 x 10^-10 = x^2 / 0.0596-x x = [OH-]=5.76 x 10^-6 M pOH = 5.24 pH = 8.76

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