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#?

1 Answer
Jun 3, 2018

#[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#