Question

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"`?

Answer 1

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.