Question

You need to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. You have a 0.1539 M standard solution. You take a 25.00 mL sample of the original acid solution and dilute it to 250.0 mL. You then take a 10.00 mL sample of the dilute acid solution and titrate it with the standard solution. You need 12.23 mL of the standard solution to reach the endpoint. What is the concentration of the original acid solution?

Answer 1

Solution :-

Balanced reaction equation

2NaOH + H2SO4 ----- > Na2SO4 + 2H2O

Lets calculate the moles of the NaOH needed for the titriation of the H2SO4

Moles of NaOH = molarity * volume

                             = 0.1539 mol per L * 0.01223 L

                            = 0.0018822 mol NaOH

Now lets calculate the moles of H2SO4

0.0018822 mol NaOH * 1 mol H2SO4 / 2 mol NaOH = 0.0009411 mol H2SO4

So 10 ml solution of diluted H2SO4 have 0.0009411 mol H2SO4

Lets calculate the moles of the H2SO4 in 250 ml solution

0.0009411 mol * 250 ml / 10 ml = 0.0235275 mol H2SO4

Now lets calculate the molarity of the original H2SO4

Molarity of H2SO4 = moles / volume in liter

                                  = 0.0235275 mol / 0.0250 L

                                = 0.9411 ml H2SO4

So the molarity of the 25.0 ml H2SO4 solution is 0.9411 M