(a)Calculate the PH at equivalence point in titrating 0.100 M solutions of ( a)
ID: 760337 • Letter: #
Question
(a)Calculate the PH at equivalence point in titrating 0.100 M solutions of ( a) chlorous acid (hclo2) with 0.080MNaOH. the ka for chlorous acid is 1.1*10^-2 (b)calculate the ph at the equivalence point in titrating 0.100M solutions of benzoic acid (C6H5COOH) WITH 0.080MNaOH. THE Ka for benzoic acid is 6.3*10^-5 show the steps please a complete answer will be awarded all the karma points including some bonus point for the correct answer please.Explanation / Answer
0.1M HClO2 x 1L = 0.1moles HClO2 0.1moles H+ + 0.09moles OH- --> 0.09moles H2O + 0.1moles H+ since there are no volumes in the problem, i'm assuming 1L for all solutions. 0.1moles HClO2 / 2L = 0.05M HClO2 Ka HClO2 = 0.011 0.011 = [H+][ClO2-] / [HClO2] 0.011 = x^2 / 0.05 - x 5.5x10^-4 - 0.011x = x^2 x^2 + 0.011x - 5.5x10^-4 = 0 x = [H+] = 0.0186M pH = 1.7 0.1M HC7H5O2 + 0.09M OH- again, this will leave you with 0.05M HC7H5O2 Ka HC7H5O2 = 6.3x10^-5 6.3x10^-5 = [H+][H7H5O2-] / [HC7H5O2] 6.3x10^-5 = x^2 / 0.05-x 3.15x10^-6 - 6.3x10^-5x = x^2 x^2 + 6.3x10^-5x - 3.15x10^-6 = 0 x = [H+] = 0.00178M pH = 2.75
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