In an experiment to find the concentration of some dilute sulphuric acid, a student diluted it exactly 10 times to give a concentration suitable for titration. He made up some standard sodium hydroxide solution by dissolving 1.00g of it in 150cm^3 of solution. He then found that 25.0cm^3 of the sodium hydroxide solution required 23.5cm^3 of the diluted sulphuric acid for neutralisation. Calculate the concentration of the original dilute sulphuric acid in moldm^3

2NaOH + H2SO4 -- Na2SO4 +2H20

(H=1; O=16; Na=23

Question

In an experiment to find the concentration of some dilute sulphuric acid, a student diluted it exactly 10 times to give a concentration suitable for titration. He made up some standard sodium hydroxide solution by dissolving 1.00g of it in 150cm^3 of solution. He then found that 25.0cm^3 of the sodium hydroxide solution required 23.5cm^3 of the diluted sulphuric acid for neutralisation. Calculate the concentration of the original dilute sulphuric acid in moldm^3

2NaOH + H2SO4 --> Na2SO4 +2H20

(H=1; O=16; Na=23

anonymous · Answer

Is this for FLVS?

anonymous · Answer

what is flvs?

anonymous · Answer

florida virtual school

anonymous · Answer

no, its for my school (international school of madrid)

anonymous · Answer

oooooo ok

anonymous · Answer

you know how to do it? im stressing out xD

anonymous · Answer

no, but I wish i can help. Im not that good at chemistry

anonymous · Answer

hahahaha thanks anyways ;)

anonymous · Answer

Im better at math if that makes sense not equations in chemistry :). You welcome

aaronq · Answer

find how many moles of NaOH you used, which are equivalent to the moles of acid (half of the amount in this case because $H_2SO_4$ is diprotic (i.e. 2 acidic protons).

from the just apply the molarity formula