Here is our balanced neutralisation equation:
#H_2SO_4(aq) + 2KOH(aq) -> 2H_2O(l) + K_2SO_4 (aq)#
From this, we know that for every #1# mole of #H_2SO_4# that reacts, #2# moles of #KOH# reacts. In other words, the mole ratio of #H_2SO_4# to #KOH# is #1:2#.
To find the concentration of #H_2SO_4# left after neutralisation, we just need to find how much it's in excess.
#"0.650 L"# of #"0.400 M"# #H_2SO_4# is #"0.650 L" xx "0.400 mol/L" = 0.260# moles of #H_2SO_4#.
#"0.600 L"# of #"0.280 M"# #KOH# is #"0.600 L" xx "0.280 mol/L" = 0.168# moles of #KOH#.
We know that the mole ratio of #KOH# to #H_2SO_4# is #1:2#, so, to fully react with the #0.168# moles of #KOH#, #0.168/2=0.084# moles of #H_2SO_4# is needed.
We have #0.260# moles.
This means that only #0.084# moles out of #0.260# moles of #H_2SO_4# react. #0.260 - 0.084 = 0.176# moles of #H_2SO_4# don't react at all.
Now that we know how much #H_2SO_4# is left sitting there in the solution (#0.176#), we can use that to calculate its concentration in molarity.
#"molarity" = "moles" / "volume (L)"#
The number of moles is #0.176#, and the volume is #"0.650 L + 0.600 L = 1.25 L"#. Plugging these values into the equation, we get:
#"molarity" = "0.176 moles" / "1.25 L" = "0.141 M"#
