0.350 L of 0.470 M H2SO4 is mixed with 0.300 L of 0.250 M KOH. What concentration of sulfuric acid remains after neutralization?
Your question is poorly worded in my opinion. By my definitions, if the H2SO4 is neutralized then there will not be ANY H2SO4 left. What the problem means, I think, is after the addition of that much KOH to that much H2SO4, what is the concn of H2SO4.
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
mols H2SO4 = M x L = estimated 0.16
mols KOH added = M x L = estimated 0.075
So 0.075 mols KOH uses 1/2 x 0.075 mols H2SO4 or estimated 0.037. Then 0.16-0.o37 = approx 0.12 mols H2SO4 remaining.
(H2SO4) = mols/total L.
HUH
To find the concentration of sulfuric acid remaining after neutralization, we need to determine the quantity of sulfuric acid that reacts with potassium hydroxide.
To do this, we can use the stoichiometry of the balanced chemical equation for the reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH):
H2SO4 + 2 KOH -> K2SO4 + 2 H2O
From the equation, we can see that 1 mole of sulfuric acid reacts with 2 moles of potassium hydroxide.
First, let's calculate the number of moles of sulfuric acid in the initial solution by multiplying the volume (in liters) by the concentration (in moles per liter):
Moles of H2SO4 = Volume of H2SO4 x Concentration of H2SO4
= 0.350 L x 0.470 M
= 0.1645 moles of H2SO4
Next, let's calculate the number of moles of potassium hydroxide in the solution by multiplying the volume (in liters) by the concentration (in moles per liter):
Moles of KOH = Volume of KOH x Concentration of KOH
= 0.300 L x 0.250 M
= 0.075 moles of KOH
Comparing the stoichiometry of the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of KOH. Therefore, the limiting reactant in this case is KOH because we have fewer moles of it.
Since 1 mole of sulfuric acid reacts with 2 moles of KOH, we can calculate the moles of sulfuric acid that reacted by multiplying the moles of KOH by the stoichiometric ratio:
Moles of H2SO4 reacted = Moles of KOH x (1 mole H2SO4 / 2 moles KOH)
= 0.075 moles KOH x (1 mole H2SO4 / 2 moles KOH)
= 0.0375 moles H2SO4
To find the moles of sulfuric acid remaining, we subtract the moles of sulfuric acid that reacted from the initial moles of sulfuric acid:
Moles of H2SO4 remaining = Initial moles of H2SO4 - Moles of H2SO4 reacted
= 0.1645 moles - 0.0375 moles
= 0.127 moles
Finally, we determine the concentration of sulfuric acid remaining by dividing the moles of H2SO4 remaining by the final volume of the solution (the sum of the volumes of H2SO4 and KOH):
Concentration of H2SO4 remaining = Moles of H2SO4 remaining / Total Volume
= 0.127 moles / (0.350 L + 0.300 L)
= 0.127 moles / 0.650 L
= 0.195 M
Therefore, the concentration of sulfuric acid remaining after neutralization is 0.195 M.