Question

Calculate the molar concentration (molarity) of the HCl soluition and the HC2H3O2 solution using the following information: 50mL of distilled water were mixed with 10 mL of the acid. A titration was performed using 0.100 M of NaOH solution. Find the average molar concentration for the two trials.

HCl: Equivalance point: 11.17 mL of NaOH added.
Acetic Acid: Equivalence point: 10.26 mL of NaOH added.

Answer 1

HCl:

Moles of NaOH = 11.17 mL x 0.100 moles/ 1 L

=0.0117 L x 0.100 moles/ 1 L

= 0.00117 moles

1 mole NaOH reacts with 1 mole HCl

So moles of HCl is 0.00117 moles

Molarity of HCl = moles/ volume in liters

= 0.00117 moles / (0.050+0.010) L

=0.0195 M

(2)

Acetic acid:

Moles of Acetic acid = 10.26 mL x 0.100 moles/ 1 L

=0.01026 L x 0.100 moles/ 1 L

= 0.001026 moles

1 mole NaOH reacts with 1 mole Acetic acid

So moles of Acetic acid is 0.001026 moles

Molarity of Acetic acid = moles/ volume in liters

= 0.001026 moles / (0.050+0.010) L

= 0.0171 M