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23 January, 22:00

The concentration of Cu2 + ions in the water (which also contains sulfate ions) discharged from a certain industrial plant is determined by adding excess sodium sulfide (Na2S) solution to 0.800 L of the water. The molecular equation is Na2S (aq) + CuSO4 (aq) → Na2SO4 (aq) + CuS (s) Calculate the molar concentration of Cu2 + in the water sample if 0.0183 g of solid CuS is formed

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  1. 23 January, 22:13
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    0.000239 M

    Explanation:

    Na2S (aq) + CuSO4 (aq) → Na2SO4 (aq) + CuS (s)

    molar mass of = 159.609 g/mol

    molar mass of CuS = 95.611 g/mol

    if 159.609 g/mol yielded 95.611 g/mol of CuS

    then x g of CuSO4 that yielded 0.0183 g = 159.609 * 0.0183 / 95.611 = 0.0305 g

    molar concentration = mole / volume in liters

    number of mole of CuSO4 in the water sample = 0.0305 g / 159.609 g/mol = 0.000191 moles

    molar concentration of CuSO4 = 0.000191 moles / 0.8 L = 0.000239 M

    since when 1 mole of CuSO4 will dissociate to produce 1 mole of Cu2+

    the molar concentration Cu2 + = 0.000239 M
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