8 CER Lab ManualJudith C. Foster An experiment similar to the one described in t
ID: 1043367 • Letter: 8
Question
8 CER Lab ManualJudith C. Foster An experiment similar to the one described in this module was performed to determine the solubility and solubility product constant of leadl) chloride (gmm 278.10). The equilibrium involved is 4. PbCl2(s)Pb (aq) +2CI (aq) The dissolved chloride in the filtered solution was quantitatively precipitated as AgCl by the addition of excess AgNOs solution. The solid AgCl was filtered from solution, dried, and weighed. The following data were obtained for three samples of the filtered, saturated solution. Determination 20.2 25.00 20.4 22.00 20.1 20.10 temperature of solution, "C volume of PbCl2 solution analyzed, mL mass of dry AgC1, g 0.2543 0.2276 0.2051 Calculate the following: number of moles of AgCl(s) CT (aq) Pb2(aq) ICI 1 in PbCl, solution, M [Pb2+] in PbCl2 solution, M Ksp average Kap average molar solubility solubility of PbClz, in g per 100 mL chemistry handbook value for solubility at 20, g per 100 mLExplanation / Answer
Moles of
AgCl 0.2543 / 143.3 = 0.001774 0.2276/143.3 = 0.001588 0.2051 / 143.3 = 0.001431
Cl- 0.001774 0.001588 0.001431
Pb2+ 0.0008870 0.0007940 0.0007155
[Cl-] 0.001774/0.02500=0.07096 0.06352 0.05724
[Pb2+] 0.03548 0.03176 0.02862
Ksp 0.0001786 0.0001281 0.00009377
(=[Pb2+][Cl-]2)
Average Ksp [0.0001786 + 0.0001281 + 0.00009377] / 3 = 0.0001335
Average molar solubility [0.03548 + 0.03176 + 0.02862] / 3 = 0.03195 M
in g/100 mL 0.03195 * 278.1 * 100 / 1000 = 0.8885 g/100 mL
Hand book Value 0.99 g / 100 L
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