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If 550 ml of some Pb(NO3)2 solution is mixed with 500 ml of 1.30 x 10-2...

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Chemistry

If 550 ml of some Pb(NO3)2 solution is mixed with 500 ml of 1.30x 10-2 M NaCl solution, what is the maximum concentration of thePb(NO3)2 solution added if no solid PbCl2 forms? (Assume Ksp = 2.00x 10-5 M at this temperature.) Enter the concentration in M.

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4.4 Ratings (893 Votes)

no of moles of NaCl = molarity * volume in L

                            = 1.3*10-2 *0.5

                           = 6.5*10-3 moles

total volume        = 550 + 500 = 1050ml = 1.050L

[Cl-]                   = no of moles/volume in L

                        = 6.5*10-3/1.050   = 6.19*10-3 M

PbCl2   -------> Pb+2 + 2Cl-

let x= mole/L of lead chloride that dissolve this gives us x moles/L of Pb+2 and 2x mols/L of cl-

KSp    = [Pb+2][Cl-]2

2*10-5       = x(2x+0.00619)2

x             = 0.157M
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