Question

In a titration, 22.5 mL of 1.8 M NaOH are required to neutralize 65.2 mL of HNO3 of unknown concentration. Determine the molarity of the HNO3?

Answer 1

`[HNO_3(aq)]~=0.6*mol*L^-1`

Explanation:

We interrogate the reaction..

`HNO_3(aq) + NaOH(aq) rarr NaNO_3(aq) + H_2O(l)`

And thus there is 1:1 equivalence between the nitric acid and the sodium hydroxide...

`"Moles of NaOH"=22.5*mLxx10^-3*L*mL^-1xx1.80*mol*L^-1=0.0405*mol.`

And given the stoichiometry, there were `0.0405*mol` nitric acid, contained in the original `65.2*mL` volume....and so...

`[HNO_3(aq)]=(0.0405*mol)/(65.2*mLxx10^-3*L*mL^-1)=0.621*mol*L^-1`