Question #a0e03

The idea here is that the number of moles of potassium hydroxide used to reach the equivalence point will give you the number of moles of sulfuric acid present in the diluted solution.

Once you know the number of moles of sulfuric acid present in the diluted solution, you can determine the number of moles of sulfuric acid, and hence the concentration, of the stock solution.

So, sulfuric acid and potassium hydroxide react in a #1:2# mole ratio to produce aqueous potassium sulfate and water.

#"H"_ 2"SO"_ (4(aq)) + 2"KOH"_ ((aq)) -> "K"_ 2"SO"_ (4(aq)) + 2"H"_ 2"O"_ ((l))#

You know that the titration required

#67.02 color(red)(cancel(color(black)("mL solution"))) * "6.000 moles KOH"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.40212 moles KOH"#

This means that the diluted solution contained

#0.40212 color(red)(cancel(color(black)("moles KOH"))) * ("1 mole H"_2"SO"_4)/(2color(red)(cancel(color(black)("moles KOH")))) = "0.20106 moles H"_2"SO"_4#

Now, you know that the diluted solution was prepared by diluting #"10.00 mL"# of stock solution to a final volume of #"100.00 mL"# by adding water.

This means that the number of moles of sulfuric acid present in the diluted solution is equal to the number of moles of sulfuric acid present in the stock solution.

The stock solution will thus contain #0.20106# moles of sulfuric acid, which is equivalent to a concentration of

#["H"_ 2"SO"_4] = "0.20106 moles"/(10.00 * 10^(-3)color(white)(.)"L") = color(darkgreen)(ul(color(black)("20.11 mol L"^(-1))))#

The answer is rounded to four sig figs.

SIDE NOTE The molarity of the stock solution is a little too high. The molarity of a #98%# sulfuric acid solution is about #"18.4 mol L"^(-1)#, so make sure to double-check the values given to you by the problem.

https://erowid.org/archive/rhodium/chemistry/equipment/molarity.html

My guess would be that the volume of potassium hydroxide solution should be smaller.