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
anor277
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.

Related questions
See all questions in Titration Calculations
Impact of this question
1591 views around the world
You can reuse this answer
Creative Commons License