Question

Titration and Molarity Problems 1. What is the molarity of a sodium hydroxide solution if 9 mL of the solution is titrated to the end point with 10mL of 0.20M Hydrochloric acid? NaOH + HCl → H2O + NaCl

Answer 1

Here , sodium hydroxide is a strong base and hydrochloric acid is a strong acid.

At the end point of titration, number of moles of base = number of moles of acid

Number of moles of any substance= molarity× volume (in liter)

So we can write, M1V1 = M2V2

M1= molarity of NaOH , V1 = volume of NaOH

M2= molarity of HCl and V2 = volume of HCl

Number of moles of NaOH =0.22 M

Mivi= M2V = MX9m2 = (0.20 m) X (10 ML) >> M = _0.20m) x (Tom) = 0.20 x 10 m (9mL) = 0.2am RE