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 show work

Explanation / Answer

Cl- + AgNO3 ----> AgCl + NO3-

molecular weight of AgCl=143.32

so number of moles of AgCl=0.2543/143.32

=1.77*10^-3 moles

moles of Cl- = moles opf AgCl

=1.77*10^-3

moles of Pb2+ = 0.5*moles of Cl-

=0.5*1.77*10^-3

=8.85*10^-4 moles

[Cl-]=1.77*10^-3*1000/25

=0.0708 M

[Pb2+]=0.5*[Cl-]

=0.5*0.0708

=0.354 M

Ksp=[Pb2+][Cl-]^2

=(8.85*10^-4)*(1.77*10^-3)^2

=2.77*10^-9

let the solubility be s.so,

4s^3=2.77*10^-9

or s=8.85*10^-4

so solubility per 100ml=0.1*8.85*10^-4*278.1

=0.0246 grams

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