You titrate 100.0 mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the Ph when half the of the benzoic acid has been consumed? C6H5CO2H + NaOH -> Na+ + C6H5CO2- + H2O. Provide answer with solution.
Titration of 100 ml of 0.025 M C6H5COOH with 0.1 M NaOH a. At equivalence point [C6H5COO-] = 0.025 M × 100 = 2.5 ml
= Molar Mass of NaOH = 39.997
= 0.1 × 39.997
= 3.9997
= -log 3.9997/2.5
= -0.2041
= 14--0.2041
= 14.2041
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