Answer:
[tex]\large \boxed{0.20 \, \% }[/tex]
Explanation:
(a) Calculate the [CO₃²⁻] needed to precipitate the Zn²⁺
Let x = [CO₃²⁻] when Zn²⁺ starts to precipitate.
The equation for the equilibrium is
ZnCO₃(s) ⇌ Zn²⁺(aq) + CO₃²⁻(aq); Ksp = 1.0 × 10⁻¹⁰
0.010 x
[tex]K_{sp} =\text{[Zn$^{2+}$][CO$_{3}^{2-}$]} = 0.010x = 1.0 \times 10^{-10}\\x = \dfrac{1.0 \times 10^{-10}}{0.010}\\\\\text{[CO$_{3}^{2-}$]} = x =\mathbf{1.0 \times 10^{-8}}[/tex]
The concentration of CO₃²⁻ when Zn²⁺ starts to precipitate is 1.0× 10⁻⁸ mol·L⁻¹.
2. Calculate [La³⁺] remaining in solution
La₂(CO₃)₂(s) ⇌ 2La³⁺(aq) + 3CO₃²⁻(aq)
x 1.0 × 10⁻⁸
[tex]K_{sp} =\text{[La$^{3+}$]$^{2}$[CO$_{3}^{2-}$]$^{3}$} =4.0 \times 10^{-34}\\x^{2} \times (1.0 \times 10^{-8})^{3} = 4.0 \times 10^{-34}\\x^{2} \times 1.0 \times 10^{-24} = 4.0 \times 10^{-34}\\x^{2} = \dfrac{4.0 \times 10^{-34}}{1.0 \times 10^{-24}}= 4.0 \times 10^{-10}\\\\\text{[La$^{3+}$]} = x = \mathbf{2.0 \times 10^{-5} \textbf{ mol/L}}[/tex]
3. Calculate the percentage of the original concentration
[tex]\text{Percent of original} = \dfrac{2.0 \times 10^{-5}}{0.010} \times \, 100 \, \% = \mathbf{0.20 \, \%}\\\text{ $\large \boxed{\mathbf{0.20 \, \% }}$ of the original La$^{3+}$ remains in solution.}[/tex]
