Question
When 46.0 ml of 0.18 M HNO3 and 58.0 ml of 0.23 M LiOH are mixed, the concentration of [OH-] after mixing is ?
Answers
DrBob222
mols HNO3 = M x L = ?
mols LiOH = M x L = ?
mols LiOH - mols HNO3 = mols LiOH in excess.
Then (OH^-) = mols excess OH/total volume in L.
mols LiOH = M x L = ?
mols LiOH - mols HNO3 = mols LiOH in excess.
Then (OH^-) = mols excess OH/total volume in L.