What is the normality of 10 ml of an acid that took 25 ml of 0.25 N NaOH to reach pH 7?

Answer 1

It depends upon which acid is being titrated with NaOH because the equivalence point is NOT always at 7.0. However, again, I assume this is a simple case. Use mL(acid) x N(acid) = mL(NaOH) x N(NaOH). The only unknown is N(acid).