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What is the [pb2+] in a solution made by adding 100 g of pbcl2(s) (mm = 278.1 g/mol) to a 5.4 m solution of nacl and allowing it to come to equilibrium? ignore any changes to solution volume due to the addition of pbcl2(s). once solution reaches equilibrium, it is noticed that some solid pbcl2 remains undissolved. the ksp for pbcl2 is 1.6 × 10−5?

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                    PbCl₂ ⇄ Pb²⁺ + 2 Cl⁻
Initially                         0         5.4 M
Change          -x         + x        + 2x
Equilibrium                  x          5.4 + 2x

Ksp = [Pb²⁺] [Cl⁻]²
1.6 x 10⁻⁵ = x (5.4 + 2x)²
since x <<<< 5.4 so 2x + 5.4 = 5.4
1.6 x 10⁻⁵ = x (5.4)²
x = 5.48 x 10⁻⁷
[Pb²⁺] = 5.48 x 10⁻⁷ M

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