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0.350 L of 0.490 M H2SO4 is mixed with 0.300 L of 0.260 M KOH. What concentration of sulfuric acid remains after neutralization?

User Aemdy
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Final answer:

To find the concentration of sulfuric acid remaining after neutralization with KOH, calculate moles of both reactants, use stoichiometry to determine how much H2SO4 can be neutralized, subtract reacted moles from initial moles, and divide by total volume. The remaining concentration is 0.204 M.

Step-by-step explanation:

To calculate the concentration of sulfuric acid (H2SO4) that remains after neutralization with potassium hydroxide (KOH), we must use a stoichiometric approach. Sulfuric acid is a strong diprotic acid, meaning it can donate two protons, and KOH is a strong base that can accept one proton.

The balanced chemical equation for the reaction is:
H2SO4 (aq) + 2KOH (aq) → K2SO4 (aq) + 2H2O (l)

First, calculate moles of H2SO4:
0.350 L × 0.490 M = 0.1715 moles H2SO4

Next, calculate moles of KOH:
0.300 L × 0.260 M = 0.0780 moles KOH

According to the equation, 2 moles of KOH are required to neutralize 1 mole of H2SO4, so we divide the moles of KOH by 2:
0.0780 moles KOH / 2 = 0.039 moles of H2SO4 that can be neutralized

Subtract the moles of H2SO4 that reacted:
0.1715 moles - 0.039 moles = 0.1325 moles of H2SO4 remain

The total volume of the mixture is the sum of the volumes of the solutions that were mixed:
0.350 L + 0.300 L = 0.650 L

Finally, calculate the remaining concentration of H2SO4:
0.1325 moles / 0.650 L = 0.204 M H2SO4

Therefore, the concentration of sulfuric acid that remains after neutralization is 0.204 M.

User Abhinavsinghvirsen
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0.137 mol / 0.650 L = 0.211 M H2SO4 is concentration of sulfuric acid.
User Tyrelle
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