Question

What is the molarity of an H2SO4 solution if 18.5 mL of 0.18 M NaOH are needed to neutralize 25.0 mL of the sample? H2SO4 + 2NaOH => Na2SO4 + 2H2O

Answer 1

The molarity is `0.066"mol/l"`

`c=n/v`

`n=cv`

So we can work out the no. moles `NaOH`:

`nNaOH=(18.5xx0.18)/(1000)=3.33xx10^(-3)`

From the equation we can see that the no. moles `H_2SO_4` must be half that amount so:

`nH_2SO_4=(3.33)/(2)xx10^(-3)=1.66xx10^(-3)`

`c=n/v` so:

`[H_2SO_4] = (1.66xx10^(-3))/(25.0xx10^(-3))=0.066"mol/l"`