Question

Seven 5.0 mL aliquots of a dilute solution of HCl were titrated with 0.1245 M sodiumhydroxide. The following volumes were needed to neutralize the acid solution:

Answer 1

In this question, we have to determine the molar concentration of the acid HCl, we have the following information:

5.0 mL of HCl

0.1245 M of NaOH

7 different volumes of NaOH, if we take the average value, we will have 58.8 mL of average

Now we can use the titration formula, which is:

MaVa = MbVb

Ma = molar concentration of acid

Va = volume of acid

Mb = molar concentration of base

Vb = volume of base

Now we add the values into the formula:

Ma * 5.0 = 0.1245 * 58.8

5Ma = 7.32

Ma = 7.32/5.0

Ma = 1.46 M

The molar concentration of HCl, based on the information provided and using the average value of volume for the base is 1.46 M