0.550 L of 0.420 M H2SO4 is mixed with 0.500 L of 0.270 M KOH. What concentration of sulfuric acid remains after neutralization?
To find the concentration of sulfuric acid remaining after neutralization, we can use the concept of stoichiometry. Here's how you can calculate it:
1. Write the balanced chemical equation for the reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH):
H2SO4 + 2KOH -> K2SO4 + 2H2O
This equation tells us that 1 mole of sulfuric acid reacts with 2 moles of potassium hydroxide to produce 1 mole of potassium sulfate and 2 moles of water.
2. Determine the number of moles of sulfuric acid and potassium hydroxide used.
Moles of H2SO4 = volume (in liters) x molarity = 0.550 L x 0.420 M = 0.231 moles
Moles of KOH = volume (in liters) x molarity = 0.500 L x 0.270 M = 0.135 moles
3. Identify the limiting reactant. The limiting reactant is the one in the reaction that will be completely consumed and determines the amount of product formed. To do this, compare the number of moles of each reactant with the stoichiometric ratio of the balanced equation.
From the balanced equation, we see that 1 mole of H2SO4 reacts with 2 moles of KOH. Therefore, we need 2 moles of KOH for every 1 mole of H2SO4. Since we have only 0.135 moles of KOH, which is less than half of the 0.231 moles of H2SO4, KOH is the limiting reactant.
4. Determine the number of moles of excess H2SO4.
Since we have established that KOH is the limiting reactant, all of it will be used up in the reaction. This means that the remaining H2SO4 will be in excess.
Moles of H2SO4 remaining = Moles of H2SO4 initially - Moles of H2SO4 used
= 0.231 moles - 0.135 moles
= 0.096 moles
5. Calculate the concentration of sulfuric acid remaining.
To find the concentration, divide the moles of H2SO4 remaining by the total volume of the solution.
Concentration (M) of H2SO4 remaining = Moles of H2SO4 remaining / Total volume (in liters)
= 0.096 moles / (0.550 L + 0.500 L)
= 0.096 moles / 1.050 L
≈ 0.0914 M
Therefore, the concentration of sulfuric acid remaining after neutralization is approximately 0.0914 M.
H2SO4 + 2KOH ==> K2SO4 + 2H2O
mols H2SO4 initially = M x L = about 0.23 but you need to do it more accurately.
mols KOH added = M x L = about 0.135 and that will neutralize 2*0.135 mols H2SO4 = about 0.27 mol H2SO4. You have less H2SO4 than that; therefore, no H2SO4 will remain.