The molar solubility of lead hydroxide in a 0.240 M lead nitrate solution is _____M.

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Expert Solution

Pb(OH)2 <-> Pb+2 + 2OH-

[Pb+2] = 0.24 (from Pb(NO3)2)

Ksp= 1.2×10^–15

Ksp = [Pb+2][OH-]^2

1.2*10^-15 = (0.24)(2S)^2

S = ((1.2*10^-15)/(0.24)/4)^(1/2)

S = 3.53553*10^-8 M

queen_honey_blossom answered 1 year ago

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