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16 July, 05:53

Determine the concentrations of K2SO4, K+, and SO42 - in a solution prepared by dissolving 2.07 * 10-4 g K2SO4 in 2.50 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per million (ppm). Note: Determine the formal concentration of SO42-. Ignore any reactions with water.

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  1. 16 July, 06:36
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    [K2SO4] = 4,75x10⁻⁷M; [K⁺] = 9.50x10⁻⁷M; [SO4⁻²] = 4,75x10⁻⁷M

    SO4⁻²: 0.045ppm; K⁺: 0.037ppm

    [SO4⁻²] = 4,70x10⁻⁷ F

    Explanation:

    Determine the equation

    K2SO4 → 2K⁺ + SO4⁻²

    Each mole of potassium sulfate generates two moles of potassium cation and one mole of sulfate anion

    Molar mass K2SO4: 174.26 g/m

    Moles of K2SO4: grams / molar mass

    2.07x10⁻⁴g / 174.26 g/m = 1.18x10⁻⁶ moles

    Molarity: Moles of solute in 1 L of solution

    1.18x10⁻⁶ moles / 2.5 L = 4,75x10⁻⁷M (K2SO4)

    K⁺ : 4,75x10⁻⁷M. 2 = 9.50x10⁻⁷M

    SO4⁻²: 4,75x10⁻⁷ M

    1 mol of K2SO4 has 2 moles of K and 1 mol of SO4

    1.18x10⁻⁶ moles of K2SO4 has 1.18x10⁻⁶ moles of SO4 and 2.37x10⁻⁶ moles of K.

    1.18x10⁻⁶ moles of SO4⁻² are 1.13x10⁻⁴ grams (moles. molar mass)

    2.37x10⁻⁶ moles of K are 9.26x10⁻⁵ grams (moles. molar mass)

    These grams are in 2.5 L of water, so we need μg/mL to get ppm

    2.5 L = 2500 mL

    1.13x10⁻⁴ grams SO4⁻² are 113.35 μg (1 μg = 1x10⁶ g)

    9.26x10⁻⁵ grams K⁺ are 92.6 μg (1 μg = 1x10⁶ g)

    113.35 μg / 2500 mL = 0.045ppm

    92.6 μg / 2500 mL = 0.037ppm

    Formal concentration of SO4⁻²:

    Formality = Number of formula weight of solute / Volume of solution (L)

    (1.13x10⁻⁴ grams / 96.06 g) / 2.5 L = 4,70x10⁻⁷ F
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