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Consider the following neutralisation reaction:
2 H3PO4(aq) + 3 Ca(OH)2 (aq) → Ca3(PO4)2 + 6 H2O

Determine the volume of 0.75 M H3PO4 necessary to neutralise 35 ml of 0.45 M Ca(OH)2

  • chem -

    since 2 moles of H3PO4 reacts with 3 moles of Ca(OH)2, it takes 2/3 mole for each mole of Ca(OH)2

    So, you have .035L*.45mole/L = .01575 moles of Ca(OH)2.
    2/3 of that is 0.0105 moles of H3PO4.

    .0105 mole / (.75 mole/L) = .014L = 14ml

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