Question

You need to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. You have a 0.1104 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 14.90 mL of the standard solution to reach the endpoint. What is the concentration of the original acid solution?

Answer 1

we know that

moles = concentration x volume (L)

so

moles of NaOH = 0.1104 x 14.90 x 10-3

moles of NaoH = 1.64496 x 10-3

now

the reaction is given by

H2S04 + 2NaOH ---> Na2S04 + 2H20

we can see that

moles of H2S04 = 0.5 x moles of NaOH

so

moles of H2S04 = 0.5 x 1.64496 x 10-3

moles of H2S04 = 8.2248 x 10-4

now

this amount of acid is present in 10 ml of diluted solution

concentration = moles x 1000 / volume (ml)

so

concentration of 250 ml solution = 8.2248 x 10-4 x 1000 / 10

concentration of 250 ml solution = 0.082248

now

moles of acid in 250 ml solution = conc x volume

so

moles of acid in 250 ml solution = 0.082248 x 250 x 10-3 = 20.562 x 10-3

now

the inital acid solution contains 20.562 x 10-3 moles and 25 ml volume

so

concentration = 20.562 x 10-3 x 1000 / 25

concentration = 0.82248

so

the concentration of original acid solution is 0.82248 M