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Suppose 0.850 L of 0.400 M H₂SO, is mixed with 0.800 L of 0.250 M KOH. What concentration of sulfuric acid remains

after neutralization?
_______ M H₂SO4

User Csblo
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1 Answer

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Step-by-step explanation:

The balanced chemical equation for the reaction between sulfuric acid and potassium hydroxide is:

H₂SO4(aq) + 2KOH(aq) → K₂SO4(aq) + 2H₂O(l)

From the equation, we can see that 1 mole of H₂SO4 reacts with 2 moles of KOH. We can use this information to determine the number of moles of H₂SO4 and KOH present in the mixture before neutralization:

moles of H₂SO4 = 0.850 L x 0.400 mol/L = 0.34 mol

moles of KOH = 0.800 L x 0.250 mol/L = 0.20 mol

Since KOH is the limiting reagent, it will be completely consumed in the reaction. The number of moles of H₂SO4 that reacts with the KOH is given by:

moles of H₂SO4 reacted = 2 x moles of KOH = 0.40 mol

The remaining moles of H₂SO4 after neutralization is:

moles of H₂SO4 remaining = moles of H₂SO4 - moles of H₂SO4 reacted

moles of H₂SO4 remaining = 0.34 mol - 0.40 mol

moles of H₂SO4 remaining = -0.06 mol

Since the moles of H₂SO4 remaining is negative, it means that all of the H₂SO4 has reacted with the KOH and there is an excess of KOH. Therefore, the concentration of sulfuric acid remaining after neutralization is 0 M.

User Plasty Grove
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