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19 October, 17:59

If the [Na+] = 0.250 M in a Na3P solution, calculate the concentration of Phosphorous ions and the number of grams of Na3P required to make 1.50 L of the solution

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  1. 19 October, 18:25
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    [P³⁻] = 0.083 M

    Mass of Na₃P required: 12.5 g

    Explanation:

    We dissociate the salt, like this:

    Na₃P → 3Na⁺ + P³⁻

    It is a ionic salt that can be dissociated.

    As [Na⁺] is 0.25M, ratio is 3:1.

    For 3 moles of sodium cathion, I have 1 mol of salt

    So, for 0.25 moles of Na⁺, we have (0.25. 1) / 3 = 0.083M

    So ratio with the anion is 1:1. In conclussion, [P³⁻] = 0.083 M

    Finally, we have to know that molarity is mol/L so:

    M = mol/L → 0.083mol/L = moles / 1.50L

    To make 1.50 L of solution, we need 0.083 mol/L. 1.50L = 0.125 moles

    Let's find out the mass, with the molecular weight:

    0.125 mol. 99.97g / 1mol = 12.5 g
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