An experiment was performed to determine the solubility and the solubility produ

Question

An experiment was performed to determine the solubility and the solubility product constant of lead(II) chloride (gmm = 278.10).

The equilibrium involved is PbCl2(s) -->Pb2+(aq) + 2Cl-(aq)

The dissolved chloride in the filtered solution was quantitatively precipitated as AgCl by the addition of excess AgNO3 solution. The solid AgCl was filtered from solution, dried, and weighed. The following data were obtained for a sample of the filtered, saturated solution.

temperature of solution 20.2 volume of PbCl2 solution analyzed, mL 25.00 mass of dry AgCl, g 0.2543

Calculate: number of moles

AgCl _____

Cl- _____

Pb2+ _____

[Cl-] in PbCl2 solution, M _____

[Pb2+] in PbCl2 solution, M _____

Ksp _____

solubility of PbCl2 in g per 100 mL _____

chemistry handbook value for solubility at 20 degrees 1.15 per 100 mL ____

Please explain how to do this. Thanks in advance

Explanation / Answer

Mass of AgCl Obtanied = 0.2543 g.

Molar mass of AgCl = 143.32 g / mol

Number of moles of AgCl = 0.2543 / 143.32 = 0.001774 mol

1 mol of AgCl = 1 mol of Cl-

0.001774 mol of AgCl = 0.001774 mol of Cl-

2 mol of Cl- = 1 mol of Pb2+

0.001774 mol of Cl- = 0.000887 mol of Pb2+

[Cl-] = 0.001774 * 1000 / 20.2 = 0.08782 M

[Pb2+] = 0.000887 * 1000 / 20.2 = 0.04391 M

Ksp = [Pb2+][Cl-]2 = (0.04391)(0.08782)2 = 3.386 x 10-4 M3

Solubility in g/100 mL = (0.000887 x 278.10 x 100) / 1000 = 0.0242 g/100 mL

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