Question

a solution contains 1.23×10⁻² m calcium acetate and 8.73×10⁻³ m lead nitrate. solid ammonium chromate is added slowly to this mixture. what is the concentration of lead ion when calcium ion begins to precipitate? pb²⁺ = \boxed{} m

Explanation:

Step1: List relevant Ksp values

Assume standard Ksp: $K_{sp}(CaCrO_4)=7.1×10^{-4}$, $K_{sp}(PbCrO_4)=2.8×10^{-13}$

Step2: Calculate $[CrO_4^{2-}]$ for Ca precipitation

$[CrO_4^{2-}]=\frac{K_{sp}(CaCrO_4)}{[Ca^{2+}]}=\frac{7.1×10^{-4}}{1.23×10^{-2}}≈5.77×10^{-2} M$

Step3: Calculate $[Pb^{2+}]$ at this $[CrO_4^{2-}]$

$[Pb^{2+}]=\frac{K_{sp}(PbCrO_4)}{[CrO_4^{2-}]}=\frac{2.8×10^{-13}}{5.77×10^{-2}}≈4.85×10^{-12} M$

Answer:

$4.85×10^{-12}$