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*mL25*mL volume of sodium hydroxide solution of 0.150*mol*L^-10.150*mol*L^-1 concentration is stoichiometrically equivalent to a 15.0*mL15.0*mL of sulfuric acid. What is [H_2SO_4][H_2SO_4]?