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by Chem Man on November 3rd, 2009

Chem Man

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This is the last question on my homework and I can't seem to solve it. If 26.44 mL of a standard 0.1650 M NaOH solution is required to neutralize 25.08 mL of H2SO4, what is the molarity of the acid solution? Please show how to get the answer

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Answers. 1 helpful answer below.

  • by leetmeat on November 15th, 2009

    leetmeat

    A bit late but w/e.

    First figure out the reaction.
    2NaOH + H2NO4 => Na2SO4 + 2H2O
    So 2:1 ratio of NaOH to H2NO4
    convert mL to L on NaOH solution, and multiply it by it's molarity to find the mols of NaOH. Molarity is essentially the ratio of solvent to solute btw, which makes this calculation possible.

    26.44mL * .001 = .02644L

    .02644L * .1650M = .0043626 mol NaOH

    So .0043626 mols of NaOH were reacted, half it to find the mols of H2NO4 as the reaction is a 2:1 ratio.

    .0043626mol / 2 = .0021813 mol H2NO4

    From here, just do the first equation in reverse to find the molarity, divide by the liters of solution.

    .0021813mol H2NO4 / .02508L = .08697M

    So the acid solution is 25.08mL of .08697M H2SO4

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