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Balanced Neutralization Reaction: H₂SO₄ and KOH in Aqueous Solution:

a) H₂SO₄ + KOH → K₂SO₄ + H2O
b) 2H₂SO₄ + 2KOH → 2K₂SO₄ + 2H₂O
c) H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
d) H₂SO₄ + 2KOH → K₂SO₄ + H₂O
e) 2H₂SO₄ + KOH → K₂SO₄ + H₂O

1 Answer

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Final answer:

The balanced equation for the neutralization reaction between H2SO4 and KOH in aqueous solution is H2SO4 + 2KOH → K2SO4 + 2H2O, where potassium sulfate (K2SO4) is the salt formed.

Step-by-step explanation:

The balanced chemical equation for the neutralization reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH) is:



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



In this reaction, sulfuric acid is a diprotic acid and can donate two protons (H+ ions). Potassium hydroxide is a strong base that fully dissociates in water to give OH− ions. Two moles of KOH are required to fully neutralize one mole of H2SO4 in order to produce water (H2O) and the salt potassium sulfate (K2SO4).

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