Question

900mL of 1.80mol/L HCl solution are added to 300mL of 0.40mol/L naOH solution. the resulting hydronium concentration is

Answer 1

\[(.9L)1.8MHCl = 1.44 \ moles\ of \ {H^+}\]\[.3L(.40M\ NaOH) = .12\ moles \ OH^-\]The OH will neutralize the H.\[1.44 - .12 = 1.32 \ moles \ H^+\]\[{{1.32 \ moles \ H^+} \over Total \ volume(.3L+.9L)} = \Huge{The \ Answer!}\]

Answer 2

It should read :
\[(.9L)1.8MHCl = 1.62 \ moles\ of \ {H^+}\]