The electrolysis of 250ml of a brine solution was carried out for a period of 20mins with a current of 2A. The resulting solution was titrated with 0.62M HCL. How many ml of HCL would be required to achieve natural point

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  1. 2NaCl + 2H2O ==> Cl2 + H2 + 2NaOH

    coulombs = amperes x seconds = approx 2400

    96,485 coulombs will form 40 g NaOH.
    g NaOH produced = 40 x 2400/96,485 = 0.995. How many mols is that? 0.995/40 = approx 0.25

    Then you know the titration is
    NaOH + HCl ==> NaCl + H2O
    Approx (you need to do the exact number) 0.25 mol NaOH will require approx 0.25 mol HCl.
    M HCl = mols HCl/L HCl. You know mols and M, solve for L

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