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?
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.