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 $"sulfuric acid"$?

Md Ariful Islam

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Md Ariful Islam

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

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.

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