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

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

Expert Solution

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

queen_honey_blossom answered 2 years ago

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