How many milliliters of 0.612M hydrochloric acid (HCl(aq)) must be added to 15.00mL of 0.557M sodium hydroxide (NaOH(aq)) to reach the equivalence point? Report your answer to 2 decimal places.

Question

How many milliliters of 0.612M hydrochloric acid (HCl(aq)) must be added to 15.00mL of 0.557M sodium hydroxide (NaOH(aq)) to reach the equivalence point? Report your answer to 2 decimal places.

Answer 1

Looking at the balanced equation for the titration:

HCl + NaOH ==> NaCl + H2O (1:1 mol ratio of HCl : NaOH)

Here are two methods to solve this:

V1M1 = V2M2

(x ml)(0.612 M) = (15.00 ml)(0.557 M)

x = 13.65 mls HCl

or

moles NaOH present = 15.00 ml 1 L /1000 ml x 0.557 mol/L = 0.008355 mols NaOH

moles HCl present = 0.008355 mols NaOH x 1 mol HCl/mol NaOH = 0.008355 mols HCl

volum HCl needed = 0.008355 mols HCl x 1 L / 0.612 mol = 0.01365 L = 13.65 mls HCl