A 38.74*mL volume of 0.50*mol*L^-1 sodium hydroxide solution reaches a stoichiometric endpoint with 19.37xx10^-3*mol of sulfuric acid dissolve in 50.0*mL of solution. What is the concentration with respect to "sulfuric acid"?

1 Answer
May 1, 2017

The first step (AS ALWAYS) is to write a stoichiometric equation......and I think your friend is right.

Explanation:

H_2SO_4(aq) + 2KOH(aq) rarr K_2SO_4(aq) + 2H_2O(l)

This is an acid base reaction, and as you know sulfuric acid is diprotic. And this is probably the source of the confusion: do I divide by 2; or do I multiply? We have all made these sorts of mistakes.

"Moles of KOH"=38.74xx10^-3Lxx0.500*mol*L^-1=19.37xx10^-3*mol.

Given the stoichiometric equation, there were (19.37xx10^-3*mol)/2, i.e. half an equiv of sulfuric acid in the original solution.

And thus "Concentration"_(H_2SO_4)=(19.37xx10^-3*mol)/(2xx50.00xx10^-3*L)

=0.194*mol*L^-1 with respect to sulfuric acid.

Note that normally we would titrate a volume of 50.00*mL with 38.74*mL titrant, not the reverse.