What is the number of ions in 0.20 mol of (NH4)3PO4?

with the working out please

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  • david
    Lv 7
    4 months ago
    Favorite Answer

    look at the formula === 3 NH4(+) ions and 1 PO4(3-) ion

    total 4 ions in each formula unit, F.U.

    0.20 mole X 6.022x10^23 FU/mole X 4ions/F.U. = 4.8 x 10^23 ions

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  • 4 months ago

    One ammonium phosphate molecule contains four ions , three ammonium ions and one phosphate ion . Now .20 mol ammonium phosphate 0.20x 6.023x10 23 number of ( NH4 )3 PO4 molecule . So the total number of ion should be 0.20x 6.023x10 23 x 4 .

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  • 4 months ago

    Number of ions .....

    You would think that if you dissolved triammonium phosphate in water you would get four ions for each molecule. Triammonium phosphate isn't even a "thing." You would never find any on your stock-room shelf. If you could form any at all, it would quickly decompose. The only stable "ammonium phosphate" is NH4H2PO4.

    But let's say you had some triammonium phosphate and you put it in water. The NH4+ and PO4^3- would hydrolyze to make a variety of species in solution:

    NH3, NH4+, PO4^3-, HPO4^2-, H2PO4^-, H3PO4, and of course, H+ and OH-.

    The relative number of each of these species varies, depending on the pH. For instance, there will be very little PO4^3- since most of it will hydrolyze to form HPO4^2- and H2PO4^-.

    Suppose you had A3X which completely ionized to form four stable ions per molecule.

    A3X --> 3A+ + X^3-

    The "working out" involves using the unit-factor method (aka dimensional analysis) where you start with what is given (0.20 mol of the salt) and multiply by two conversion factors to take you from there to the number of ions. Each conversion factor is "built" from an equivalency. (1 mol of the salt ≡ 4 ions), (6.022x10^23 ions ≡ 1 mol of ions).

    0.20 mol A3X x (4 mol ions / 1 mol A3X) x (6.022x10^23 ions / 1 mol ions) = 4.8x10^23 ions

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  • 4 months ago

    Following up on the answer provided by hcbiochem, you have 4 ions per mole of compound, and so 0.2 mol of compound = 0.8 moles of ions. Each mole represents 6.02x10^23 ions so you have...

    0.8 moles of ions x 6.02x10^23 ions/mole = 4.8x10^23 ions.

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  • 4 months ago

    (NH₄)₃PO₄, Ammonium phosphate, is 149.09 g/mol

    in water, each molecule forms 4 ions, one PO₄–, three NH₄+

    6.022e23 molecules/mole Avogadro constant

    6.022e23 molecules/mole x 0.2 mol = 1.204e23 molecules

    1.204e23 molecules x 4 ions/molecule = 4.8e23 ions

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  • 4 months ago

    When dissolved in water, each the (NH4)3PO4 will form 4 ions (3 NH4+ and 1 PO43-).

    So, 0.20 mol X (4 mol ions/mole) = 0.80 mol ions.

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