An acid solution is 0.100 M in HCl and 0.200 M in H₂SO₄. What volume of a 0.150 M KOH solution would completely neutralize all the acid in 500.0 mL of this solution?

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Determine the moles of HCl in the solution using the formula: \( \text{moles} = \text{concentration} \times \text{volume} \).

Calculate the moles of HCl: \( 0.100 \, \text{M} \times 0.500 \, \text{L} = \text{moles of HCl} \).

Determine the moles of H₂SO₄ in the solution using the same formula.

Calculate the moles of H₂SO₄: \( 0.200 \, \text{M} \times 0.500 \, \text{L} = \text{moles of H₂SO₄} \).

Use the stoichiometry of the neutralization reactions to find the total moles of KOH needed, considering that HCl provides 1 mole of H⁺ per mole and H₂SO₄ provides 2 moles of H⁺ per mole.

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주요 개념

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Acid-Base Neutralization

Acid-base neutralization is a chemical reaction where an acid reacts with a base to produce water and a salt. In this context, HCl and H2SO4 are strong acids that will react with KOH, a strong base. The stoichiometry of the reaction is crucial for determining the amounts of reactants needed for complete neutralization.

Molarity and Volume Relationships

Molarity (M) is a measure of concentration defined as moles of solute per liter of solution. To find the total moles of acid present in the solution, we multiply the molarity of each acid by its volume. This relationship allows us to calculate how much KOH is required to neutralize the acids based on their respective concentrations and the total volume of the solution.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on balanced equations. For the neutralization of HCl and H2SO4 by KOH, we need to consider the stoichiometric coefficients, which indicate the ratio of moles of acid to moles of base required for complete reaction. This is essential for determining the exact volume of KOH needed.

An acid solution is 0.100 M in HCl and 0.200 M in H2SO4. What volume of a 0.150 M KOH solution would completely neutralize all the acid in 500.0 mL of this solution?