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27 December, 16:37

What minimum volume of 0.200 m potassium iodide solution is required to completely precipitate all of the lead in 195.0 ml of a 0.194 m lead (ii) nitrate solution?

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  1. 27 December, 16:54
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    First, we write the reaction equation:

    2KI + PbNO₃ → K₂NO₃ + PbI₂

    The molar ratio of KI to PbNO₃ is 2 : 1

    Moles of PbNO₃ present:

    Moles = concentration (M) x volume (dm³)

    = 0.194 x 0.195

    = 0.038

    Moles of KI required = 2 x 0.038 = 0.076 moles

    concentration = moles / volume

    volume = moles / concentration

    = 0.076 / 0.2

    = 0.38 L = 380 ml
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