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"
