Question

You need to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. You have a 0.1689 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 17.49 mL of the standard solution to reach the endpoint. What is the concentration of the original acid solution?

Answer 1

The reaction between H₂SO₄ and NaOH is as follows

That means to react one mole of H2SO4 we need 2 Mole of NaOH.

The formula for this type of titration problem is

V₁M₁   = V₂M₂

N₁            N₂               

Here                  V₁ Volume of NaOH = 17.49 mL

                         M₁ Molarity of NaOH = 0.1689 M

             N₁ Number of moles of NaOH = 2

                       V₂ Volume of H₂SO₄ = 10 mL

                     M₂ Molarity of H₂SO₄ = x to be calculated

             N₂ Number of moles of H₂SO₄ = 1

By substituting the above values in the above equation get

               17.49X0.1689       =       10X x

                     2                                 1

17.49X0.1689 =10 X x X 2

Molarity of H₂SO₄ x = 17.49X0.1689    =0.1477 M

                                          2X10

How ever this is not that original solution it is 10 times (25 mL of H₂SO_{4 i}s diluted to 250 mL) diluted solution of original H₂SO₄ Solution

So the original H₂SO₄ solution morality 10 times of diluted solution = 0.1477X10 = 1.477 M