119 mL of HCl is titrated with 0.12 M NaOH. If the equivalence point is reached when 72 mL of NaOH is added, then what is the concentration of the HCl solution?

Question

7.3 M

8.64 M

.073 M

.864M

Answer 1

The chemical reaction taking place is an acid base neutralization reaction.

HCl + NaOH ==> NaCl + H2O

The equivalence point is the point at which the moles of acid (HCl) and moles of base (NaOH) are equal or equivalent. So, we first find the moles of of NaOH used:

72 ml x 1 L/1000 ml x 0.12 moles/L = 0.00864 moles NaOH used

Next, we find the moles of HCl present by using the mole ratio in the balanced equation of 1:1

0.00864 moles NaOH x 1 mol HCl/1 mol NaOH = 0.00864 moles HCl present

Finally, we find the concentration of the HCl by converting 119 ml to 0.119 liters and using the moles found:

0.00864 moles HCl/0.119 L = 0.0726 M = 0.073 M

NOTE: you could this same problem in a shorter, faster way but it isn't recommended for all problems of this nature unless the mole ratio of acid to base is 1:1. It would look like this...

(72 ml)(0.12 M) = (119 ml)(x M)

x = (72)(0.12)/119

x = 0.0726 M = 0.073 M