How do you determine the molarity of 2.25 mole of sulfuric acid, #H_2SO_4#, dissolved in 725 mL of solution?

1 Answer
Apr 24, 2017

Answer:

With relative ease...........

Explanation:

Now, by definition, #"molarity"="moles of solute (mol)"/"volume of solution (L)"#.

Now the wording of your problem gave us that #2.25*mol# sulfuric acid were dissolved in water to give us #725*mL# of solution. Had it said that #2.25*mol# sulfuric acid were dissolved in #725*mL# of water, we would be uncertain as to the volume of the sulfuric acid solution. Do you appreciate the distinction I make here?

And thus, we simply have to address the quotient:

#(2.25*mol)/(725*mLxx10^-3L*mL^-1)=3.10*mol*L^-1# WITH RESPECT TO SULFURIC ACID.

Again, I am trying to be cautious in my wording, and chemists often speak of a #"formal concentration"# to indicate that of course in solution, the solute might speciate, and certainly sulfuric acid does, to #H_3O^+#, and #HSO_4^(-)#, and #SO_4^(2-)#. As a practical chemist, you would treat this solution as #[SO_4^(2-)]=3.10*mol*L^-1#, and #[H_3O^+]=6.20*mol*L^-1#. Capisce?