Question

You determine the concentration of a solution of HCl by titration with NaOH. The titration reaction is:

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)

Using a volumetric pipet, you transfer 10.00 mL of the HCl solution into an Erlenmeyer flask, then dilute it with ~50 mL of water and add 3 drops of phenolphthalein. The endpoint is reached after you have added 44.00 mL of 0.1250 M NaOH solution from a buret.

Calculate the molarity of the original HCl solution.

a. 0.005500 M

b. 0.09167 M

c. 0.5500 M

d. 0.1100 M

Answer 1

The volume of diluted HCl = 10 +50 = 60mL

Let the molarity of HCl solution = M

volume of NaOH at equivalence = 44.00mL

molarity of NaCl = 0.1250M

For the neutralization titrations

M1V1/n1 = M2V2/n2

substituting the given values

60mL x M /1 = 44.00x 0.125 /1

Thus molarity of acid = 0.09167 M

The molarity of diluted acid = 0.09167 M

molarity before dilution is calculated as

M1V1 = M2 V2

10mLx M = 0.09167M x 60mL

Thus molarity of original HCl solution = 0.550M

Thus option C is correct