Question

A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. He has a 0.1354 M standard sodium hydroxide solution. He takes a 25.00 mL sample of the original acid solution and dilutes it to 250.0 mL. Then, he takes a 10.00 mL sample of the dilute acid solution and titrates it with the standard solution. The endpoint was reached after the addition of 16.78 mL of the standard solution. What is the concentration of the original sulfuric acid solution?

Answer 1

we have reaction H2SO4 + 2NaOH --> Na2SO4 + 2H2O

at end point NaOH moles = Molarity x volume of NaOH solution in L

= 0.1354 mol/L x 0.01678 L = 0.00227 mol

as per reaction H2SO4 moles reacted = ( 1/2) x NaOH moles = ( 1/2) ( 0.00227) = 0.001136 mol

Volume of H2SO4 solution taken = 10 ml = 0.01 L

H2SO4 molarity = moles / volume = ( 0.001136 mol) / ( 0.01L) = 0.1136 M

Now this Molarity is same as in 250 ml solution ( since 10 ml was portion of 250 ml solution)

now 25 ml solution Molarity can be found by dilution law M1V1 = M2V2

where M1 = initial Molarity = , V1 = 25 ml , M2 = 0.1136 M , V2 = 250 ml

we get M1 x 25 ml = 0.1136 M x 250 ml

M1 = 1.136 M

Thus original sulphuric acid concentraton = 1.136 M