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10 August, 06:54

A solution is prepared by dissolving ammonium sulfate in enough water to make of stock solution. A sample of this stock solution is added to of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

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  1. 10 August, 10:26
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    Answer: molarity of ammonium ions = 0.274mol/L

    molarity of sulfate ions = 0.137mol/L

    Note: The complete question is given below

    A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00-mL sample of this stock solution is added to 50.00 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

    Explanation:

    Molar concentration = no of moles/volume in liters

    no of moles = mass/molar mass

    mass of ammonium sulfate = 10.8g, molar mass of ammonium sulfate, (NH₄) ₂SO₄ = (14+4) * 2 + 32 + (16) * 4 = 132g/mol

    no of moles = 10.8g/132g/mol = 0.0820moles

    Molarity of stock solution = 0.0820mol / (100ml/1000ml * 1L) = 0.0820mol/0.1L Molarity of stock solution = 0.820mol/L

    Concentration of final solution is obtained from the dilution formula,

    C1V1 = C2V2

    C1 = 0.820M, V1 = 10mL, C2 = ?, V2 = 60mL

    C2 = C1V1/V2

    C2 = 0.820*10/60 = 0.137mol/L

    molar concentration of ions = molarity of solution * no of ions

    molarity of ammonium ions = 0.137mol/L * 2 = 0.274mol/L

    molarity of sulfate ions = 0.137 mol/L * 1 = 0.137mol/L
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