Question

Neutralization occurs when `15.0mL` of `KOH` react with `25.0 mL` of `HNO3`. If the molarity of `HNO3` is `0.750 M`, what is the molarity of the `KOH`?

Answer 1

`[KOH]=1.25*mol*L^-1....`

Explanation:

We interrogate the reaction....

`KOH(aq) + HNO_3(aq) rarr KNO_3(aq) + H_2O(l)`

`"Moles of nitric acid"=25.0*mLxx10^-3*L*mL^-1xx0.750*mol*L^-1=0.01875*mol`...

And an EQUIVALENT molar quantity of `"potassium hydroxide"` was present in the `15.0*mL` volume...and thus a concentration of...

`(0.01875*mol)/(15.0*mLxx10^-3*L*mL^-1)=??*mol*L^-1`