Question

A beaker contains a solution of two metal cations: 0.010 M La+3 and 0.010 M Zn2+.

Answer 1

Answer:

$\large \boxed{0.20 \, \% }$

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

$K_{sp} =\text{[Zn ^{2+} ][CO _{3}^{2-} ]} = 0.010x = 1.0 \times 10^{-10}\\x = \dfrac{1.0 \times 10^{-10}}{0.010}\\\\\text{[CO _{3}^{2-} ]} = x =\mathbf{1.0 \times 10^{-8}}$

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

$K_{sp} =\text{[La ^{3+} ] ^{2} [CO _{3}^{2-} ] ^{3} } =4.0 \times 10^{-34}\\x^{2} \times (1.0 \times 10^{-8})^{3} = 4.0 \times 10^{-34}\\x^{2} \times 1.0 \times 10^{-24} = 4.0 \times 10^{-34}\\x^{2} = \dfrac{4.0 \times 10^{-34}}{1.0 \times 10^{-24}}= 4.0 \times 10^{-10}\\\\\text{[La ^{3+} ]} = x = \mathbf{2.0 \times 10^{-5} \textbf{ mol/L}}$

3. Calculate the percentage of the original concentration

$\text{Percent of original} = \dfrac{2.0 \times 10^{-5}}{0.010} \times \, 100 \, \% = \mathbf{0.20 \, \%}\\\text{ \large \boxed{\mathbf{0.20 \, \% }} of the original La ^{3+} remains in solution.}$