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The balanced equation for the neutralization reaction of aqueous H₂SO4 with aqueous KOH is shown. H₂SO₂(aq) + 2 KOH(aq) → 2H₂O(1) + K₂SO₂(aq) What volume of 0.210 M KOH is needed to react completely with 12.2 mL of 0.185 M H₂SO4?​

User PyNEwbie
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Final answer:

The balanced equation for the neutralization reaction of aqueous H₂SO4 with aqueous KOH is H₂SO₂(aq) + 2 KOH(aq) → 2H₂O(1) + K₂SO₂(aq). To find the volume of 0.210 M KOH needed to react completely with 12.2 mL of 0.185 M H₂SO4, we can use stoichiometry and the molarity equation.

Step-by-step explanation:

The balanced equation for the neutralization reaction of aqueous H₂SO4 with aqueous KOH is:

H₂SO₂(aq) + 2 KOH(aq) → 2H₂O(1) + K₂SO₂(aq)

To find the volume of 0.210 M KOH needed to react completely with 12.2 mL of 0.185 M H₂SO4, we need to apply stoichiometry and the molarity equation.

First, we can calculate the amount in moles of H₂SO4: moles H₂SO4 = volume H₂SO4 (L) × concentration H₂SO4 (M)

Next, we can use the balanced equation to determine the ratio of moles of H₂SO4 to moles of KOH. In this case, the ratio is 1:2.

Finally, we can use the calculated amount of moles of H₂SO4 to find the volume in liters of 0.210 M KOH needed, using the molarity equation. Remember to convert mL to L.

User Mahmoh
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