0.150 L of 0.420 M H2SO4 is mixed with 0.100 L of 0.270 M KOH. What concentration of sulfuric acid remains after neutralization?
convert to moles
see how much H2SO4 remains after being neutralized by twice that many moles of KOH.
Convert that back to moles/L
To find the concentration of sulfuric acid remaining after neutralization, we need to determine the moles of sulfuric acid and potassium hydroxide reacted in the solution.
First, let's calculate the moles of H2SO4 based on its molarity and volume:
Moles of H2SO4 = Molarity × Volume
= 0.420 M × 0.150 L
= 0.063 moles
Next, let's calculate the moles of KOH using the same method:
Moles of KOH = Molarity × Volume
= 0.270 M × 0.100 L
= 0.027 moles
Since the balanced chemical equation for the neutralization reaction between H2SO4 and KOH is 1:1, we can see that 0.027 moles of H2SO4 reacted with 0.027 moles of KOH.
Now, we need to determine how much of the H2SO4 remains unreacted. Since 0.027 moles of H2SO4 and 0.027 moles of KOH react completely, the remaining moles of H2SO4 can be calculated by subtracting the reacted moles from the initial moles:
Remaining moles of H2SO4 = Initial moles of H2SO4 - Moles of H2SO4 reacted
= 0.063 moles - 0.027 moles
= 0.036 moles
Finally, we need to find the concentration of the remaining sulfuric acid by dividing the remaining moles by the total volume of the solution.
Remaining concentration of H2SO4 = Remaining moles of H2SO4 / Total volume of the solution
= 0.036 moles / (0.150 L + 0.100 L)
= 0.036 moles / 0.250 L
= 0.144 M
Therefore, the concentration of sulfuric acid remaining after neutralization is 0.144 M.