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A 29.00 mL sample of an unknown H3PO4 solution is titrated with a 0.130 M NaOH...

A 29.00 mL sample of an unknown H3PO4 solution is titrated with a 0.130 M NaOH solution. The equivalence point is reached when 27.28 mL of NaOH solution is added.What is the concentration of the unknown H3PO4 solution? The neutralization reaction is H3PO4(aq)+3NaOH(aq)→3H2O(l)+Na3PO4(aq)

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Answer #1

no. of mole = molarity volume of solution in liter

no. of mole of NaOH = 0.130 M 0.02728 liter = 0.0035464 mole

According to reaction 1 mole of H3PO4 react with 3 mole of NaOH therefore to react with 0.0035464 mole of NaOH required H3PO4 = 0.0035464 1 / 3 = 0.0011821333 mole

mole of H3PO4 = 0.0011821333

volume of H3PO4 = 29 ml =0.029 liter

Molarity = no. of mole / volume of solution in liter

Molarity of H3PO4 = 0.0011821333 / 0.029 = 0.0407 M

[H3PO4] = 0.0407 M

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