Explanation Given The concentration of Pb ( NO 3 ) 2 solution is 0.10 M . The volume of the sample is 10 mL . The volume of Pb ( NO 3 ) 2 solution added to the sample is 0.2 mL . The molar mass of I − is 126.9045 g / mol . The number of moles of Pb ( NO 3 ) 2 or Pb 2 + is calculated by the formula, Number of moles = Molar concentration × Volume of the solution in liters Substitute the value of molar concentration of Pb ( NO 3 ) 2 and the volume of the solution in the above formula. Number of moles = 0.10 M × 0.2 1000 L = 2 × 10 − 5 mol The total volume of the solution is, 10 mL + 0.2 mL = 10.2 mL Therefore, the final concentration of Pb 2 + after the addition of one drop of Pb ( NO 3 ) 2 to the sample is calculated by the formula, Molar concentration = Number of moles Total volume of the solution in liters Substitute the values of the number of moles of Pb 2 + and the total volume of the solution in the above formula. Molar concentration = 2 × 10 − 5 mol 10.2 1000 L = 1.96 × 10 − 3 M The dissolution equilibrium for PbI 2 is, PbI 2 ( s ) ⇌ Pb 2 + ( a q ) + 2 I − ( a q ) The value of the solubility product ( K sp ) for the dissolution equilibrium is 7

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Chapter 17, Problem 72E

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To determine: The minimum number of grams of I− needed to precipitate PbI2 by the addition of one drop (0.2 mL) of 0.10 M

Pb( NO3)2 to 10 mL of the sample.

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Chapter 17, Problem 71E

Chapter 17, Problem 73E

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Chapter 17 Solutions

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The minimum number of grams of I − needed to precipitate PbI 2 by the addition of one drop ( 0.2 mL ) of 0.10 M Pb ( NO 3 ) 2 to 10 mL of the sample.

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To determine: The minimum number of grams of I− needed to precipitate PbI2 by the addition of one drop (0.2 mL) of 0.10 M Pb( NO3)2 to 10 mL of the sample.

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Chapter 17 Solutions

To determine: The minimum number of grams of I− needed to precipitate PbI2 by the addition of one drop (0.2 mL) of 0.10 M

Pb( NO3)2 to 10 mL of the sample.