Question

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]`?

Answer 1

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

Explanation:

`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...........