What is the molar solubility of lead(II) chromate in 0.055 M Na2S2O3? For PbCrO4, Ksp = 2.0 x 10^-16; for Pb(S2O3)3^4-, Kf = 2.2 x 10^6. 2.7 x 10^-7 M 6.1 x 10^-5 M 1.7 x 10^-4 M 4.9 x 10^-5 M 4.4 x 10^-12 M
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Let x be the molar solubility of PbCrO4. Then, the solubility product expression is: Ksp = [Pb²⁺][CrO₄²⁻] = (x)(x) = x² Given Ksp = 2.0 x 10⁻¹⁶, we can solve for x: x² = 2.0 x 10⁻¹⁶ x = √(2.0 x 10⁻¹⁶) x = 1.41 x 10⁻⁸ Now, we need to consider the complexation Show more…
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