A #25mL# volume of sodium hydroxide solution of #0.150molL^-1# concentration is stoichiometrically equivalent to a #15.0mL# of sulfuric acid. What is #[H_2SO_4]#?

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Answer 1 · 2017-02-24T06:40:40

To find #[H_2SO_4]#, we need (i), a stoichiometrically balanced equation:

#H_2SO_4(aq) + 2NaOH rarr Na_2SO_4(aq) + 2H_2O(l)#.

And (ii) equivalent quantities of sodium hydroxide,

#"Moles of NaOH"=25.0*mLxx10^-3*L*mL^-1xx0.150*mol*L^-1=3.75xx10^-3*mol#

Given the equation, we KNOW that there was a half an equiv of sulfuric acid present in the original volume of #H_2SO_4(aq)#.

And thus #[H_2SO_4]=(1/2xx3.75xx10^-3*mol)/(15.0xx10^-3L)~=0.13*mol*L^-1#

All I have done here is to use the relationship:

#"Concentration"# #-=# #"Moles of solute"/"Volume of solution."#

And thus,

#"Concentration"xx"Volume"="Moles of solute"#, etc. etc...........

A #25*mL# volume of sodium hydroxide solution of #0.150*mol*L^-1# concentration is stoichiometrically equivalent to a #15.0*mL# of sulfuric acid. What is #[H_2SO_4]#?