Exactly 23.6 ml of a 0.131 N HCl solution was required for complete neutralization of 25.0 ml of an NaOH solution.What was normality of the NaOH?

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  1. HCl + NaOH ==> NaCl + H2O
    How many milliequivalents(m.e.) HCl were used? m.e. = mL x N
    A m.e. of HCl = m.e. NaOH = m.e. of anything.
    Then N = m.e. NaOH/mL NaOH.

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