0.650 L of 0.420 M H2SO4 is mixed with 0.600 L of 0.280 M KOH. What concentration of sulfuric acid remains after neutralization?
To find the concentration of sulfuric acid that remains after neutralization, we need to determine the limiting reagent and calculate the amount of excess reactant.
First, let's identify the limiting reagent. The reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH) can be represented as follows:
H2SO4 + 2KOH -> K2SO4 + 2H2O
From the balanced chemical equation, we can see that one mole of sulfuric acid reacts with two moles of potassium hydroxide. Therefore, the moles of sulfuric acid and potassium hydroxide can be calculated as:
moles of H2SO4 = volume (L) × concentration (M) = 0.650 L × 0.420 M = 0.273 mol H2SO4
moles of KOH = volume (L) × concentration (M) = 0.600 L × 0.280 M = 0.168 mol KOH
Based on the stoichiometry of the reaction, we see that the mole ratio between H2SO4 and KOH is 1:2. Therefore, the ratio between the moles of H2SO4 and KOH is also 1:2.
To completely neutralize sulfuric acid, we need twice as many moles of potassium hydroxide. Since we have fewer moles of KOH than H2SO4, it is the limiting reagent. This means all the KOH will be consumed, and there will be some excess sulfuric acid remaining.
To calculate the moles of sulfuric acid that reacted, we multiply the moles of KOH by the mole ratio of H2SO4 to KOH:
moles of H2SO4 reacted = 0.168 mol KOH × (1 mol H2SO4 / 2 mol KOH) = 0.084 mol H2SO4
Now, we can calculate the moles of sulfuric acid remaining by subtracting the moles of H2SO4 reacted from the initial moles of H2SO4:
moles of H2SO4 remaining = 0.273 mol H2SO4 - 0.084 mol H2SO4 = 0.189 mol H2SO4
Finally, to find the concentration of sulfuric acid remaining, divide the moles of H2SO4 remaining by the final volume of the solution, which is the sum of the initial volumes of sulfuric acid and potassium hydroxide:
volume of final solution = volume of H2SO4 + volume of KOH = 0.650 L + 0.600 L = 1.250 L
concentration of H2SO4 remaining = moles of H2SO4 remaining / volume of final solution
= 0.189 mol H2SO4 / 1.250 L
= 0.151 M
Therefore, the concentration of sulfuric acid remaining after neutralization is 0.151 M.