Question

a solution contains 8.97×10⁻³ m calcium acetate and 1.07×10⁻² m lead nitrate. solid ammonium phosphate is added slowly to this mixture.
a. what is the formula of the substance that precipitates first?
formula =

b. what is the concentration of phosphate ion when this precipitation first begins?
po₄³⁻ = 5 m

Explanation:

Step1: Identify possible precipitates

Possible precipitates are calcium phosphate ($Ca_3(PO_4)_2$) and lead phosphate ($Pb_3(PO_4)_2$). Ksp values (standard): $K_{sp}(Ca_3(PO_4)_2)=2.07×10^{-33}$, $K_{sp}(Pb_3(PO_4)_2)=8.0×10^{-43}$.

Step2: Calculate [PO₄³⁻] for Ca₃(PO₄)₂

$K_{sp}=[Ca^{2+}]^3[PO_4^{3-}]^2$. $[Ca^{2+}]=8.97×10^{-3}M$.
$[PO_4^{3-}]=\sqrt{\frac{K_{sp}}{[Ca^{2+}]^3}}=\sqrt{\frac{2.07×10^{-33}}{(8.97×10^{-3})^3}}≈\sqrt{\frac{2.07×10^{-33}}{7.21×10^{-7}}}≈\sqrt{2.87×10^{-27}}≈5.36×10^{-14}M$.

Step3: Calculate [PO₄³⁻] for Pb₃(PO₄)₂

$[Pb^{2+}]=1.07×10^{-2}M$.
$[PO_4^{3-}]=\sqrt{\frac{K_{sp}}{[Pb^{2+}]^3}}=\sqrt{\frac{8.0×10^{-43}}{(1.07×10^{-2})^3}}≈\sqrt{\frac{8.0×10^{-43}}{1.22×10^{-6}}}≈\sqrt{6.56×10^{-37}}≈8.10×10^{-19}M$.

Step4: Compare [PO₄³⁻] values

Lower [PO₄³⁻] means earlier precipitation: $8.10×10^{-19} < 5.36×10^{-14}$, so Pb₃(PO₄)₂ precipitates first.

Answer:

A. $Pb_3(PO_4)_2$
B. $8.1×10^{-19}$ (rounded to two significant figures)