An acid-base titration was performed. It took 21.50 mL of the base, NaOH, to titrate 25.00 mL of 1.013 M HCl, the acid. What was the Molarity of the NaOH?

Question

An acid-base titration was performed. It took 21.50 mL of the base, NaOH, to titrate 25.00 mL of 1.013 M HCl, the acid. What was the Molarity of the NaOH?

Answer 1

Write the equation for the reaction taking place:

NaOH + HCl ==> NaCl + H2O

Find the mole of HCl present:

25.00 ml HCl x 1 L / 1000 ml x 1.013 mol / L = 0.025325 mols HCl

Calculate moles NaOH used:

0.025325 mols HCl x 1 mol NaOH / mol HCl = 0.025325 mols NaOH used

Calculate molarity (moles/liter) of NaOH:

0.025325 mols NaOH / 21.50 ml x 1000 ml / L = 1.178 moles / L