Question

You need to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. You have a 0.1015 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 19.16 mL of the standard solution to reach the endpoint. What is the concentration of the original acid solution?

Answer 1

H2SO4 + 2NaOH = 2H2O + Na2SO4

so...

M = 0.1015 M of NaOH

V = 25 mL

for the acid:

V = 25 mL from V = 250 mL ( same concentration)

then ,we take only 10 mL ... this is 10/250

so

mmol of base = MV = 0.1015*19.16 = 1.94474 mmol of NaOH

this is in the 10 mL solution...

mmol of H2SO4 = 1.94474/2 = 0.97237 mmol of H2SO4 in solution (10 mL)

this has a 1/10 concentration

so

[H2SO4]diluted = mmol/mL = 0.97237/10 = 0.097237 M

[H2SO4]original = 250/25 * 0.097237 = 0.97237 M