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20 January, 05:01

What volume of 0.100 M Na3PO4 is required to precipitate all the lead (ii) ions from 150.0 mL of 0.250 M Pb (NO3) 2?

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  1. 20 January, 05:06
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    First, we write the reaction equation:

    3Pb (NO₃) ₂ + 2Na₃PO₄ → 6NaNO₃ + 3Pb₃ (PO₄) ₂

    Moles of Pb ions present:

    moles = concentration x volume

    = 0.15 x 0.25

    = 0.0375

    From the equation,

    moles Pb : moles Na₃PO₄

    = 3 : 2

    Moles of Na₃PO₄:

    2/3 x 0.0375

    = 0.025

    volume = moles / concentration

    = 0.025 / 0.1

    = 0.25 L

    = 250 ml
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