When 19.16 mL of NaOH is titrated with 0.82 M HCl, 13.48 mL of HCl are required to reach the equivalence point. What is the concentration (molarity) of the NaOH solution?

Answer 1

Write the balanced equation:

NaOH + HCl ==> NaCl + H2O

Find moles HCl used:

13.48 ml HCl x 1 L / 1000 ml x 0.82 mol/L = 0.01105 moles HCl

Find moles NaOH present by using the stoichiometry of the balanced equation:

0.01105 mols HCl x 1 mol NaOH / 1 mol HCl = 0.01105 mols NaOH present

Determine concentration (molarity) of the NaOH:

molarity = moles NaOH / liter solution

molarity = 0.01105 mols / 19.16 mls x 1000 mls / L = 0.5767 M = 0.58 M (2 sig. figs.)

You can also solve the problem as follows since the mole ratio is 1 HCl : 1 NaOH (not a recommended way)

(19.6 ml )(x M) = (13.48 mls)(0.82 M)

x = 0.58 M