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What is the equilibrium concentration of Pb2+ ion in a solution of PbCl2 that is 0.075 M in chloride ion
[From: ][author: ][Date: 11-05-20][Hit: ]
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PbCl2(s) Pb2+(aq) + 2Cl1-(aq) K = 1.7e-5
Please help
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K = [Pb] [Cl]^2
1.7e-5 = [Pb] [0.075]^2
[Pb] = 1.7e-5 / (0.005625)
Pb+2 = 0.0030 Molar
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