Question #9c03b

2 Answers
Feb 19, 2015

The molarity of the sulfuric acid solution will be #"17.1 M"#.

You are dealing with a neutralization reaction - in this case, a strong acid and a strong base react to produce salt and water. The balanced chemical equation looks like this

#2NaOH + H_2SO_4 -> Na_2SO_4 + 2H_2O#

SIDE NOTE - the reaction takes place in aqueous solution; taht being said, I won't go into the net ionic equation

Now look at the mole ratio you have between #NaOH# and #H_2SO_4# - 2 moles of the former react with 1 mole of the latter. A neutralization reaction implies that both the acid and the base are consumed, i.e. the two cancel each other out.

This means that the number of #H_2SO_4# moles depends on the number of moles of #NaOH#, which you can determine by using its molar mass, and on the mole ratio between the two compounds

#"112.0 g" * ("1 mole")/("40.0 g") = "2.80 moles NaOH"#

Apply the aforementioned mole ratio to see exactly how many moles of #H_2SO_4# were neutralized by this much #NaOH#

#"2.80 moles NaOH" * ("1 mole"H_2SO_4)/("2 moles NaOH") = "1.40 moles"# #H_2SO_4#

The last thing to do is use the formula for molarity

#C = n/V = ("1.40 moles")/(82.0*10^(-3)"L") = "17.1 M"#

Don't forget to transform mL into L, since molarity uses liters, not mililiters!

Feb 19, 2015

First convert the weight of #NaOH# to moles.

Relative atomic masses are #Na=23,O=16,H=1#
For a total molecular mass #NaOH=40#

So #112.0g# converts to #112.0/40=2.8mol#

Since sulfuric acid #H_2SO_4# has two #H^+#-ions it neutralises in a ratio of 2:1, so there must have been #2.8//2=1.4mol# of sulfuric acid.

The molarity of the solution= #(mol)/L.#

Molarity = #(1.4"mol")/(82.0mL)=(1.4mol)/(0.082L)~~17.1mol//L#

Which is a very strong solution -- you may (almost) call this concentrated sulfuric acid. Don't spill it on your skin!

Extra :
Let's call this a thought experient. You would NOT want to do this in practice, not even in a fume cupboard, as the result will get explosive due to the extreme heat developed in this reaction!