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To determine the molar solubility of [tex]\( \text{CaF}_2 \)[/tex] in a solution of [tex]\( 0.010 \, \text{M} \, \text{Ca(NO}_3)_2 \)[/tex] at [tex]\( 25^\circ \text{C} \)[/tex], we need to use the solubility product constant [tex]\( K_{sp} \)[/tex] of [tex]\( \text{CaF}_2 \)[/tex].
[tex]\[ \text{K}_{sp} (\text{CaF}_2) = 3.9 \times 10^{-11} \][/tex]
The dissociation of [tex]\( \text{CaF}_2 \)[/tex] in water is given by:
[tex]\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{F}^- (aq) \][/tex]
Given that the initial concentration of [tex]\( \text{Ca}^{2+} \)[/tex] is [tex]\( 0.010 \, \text{M} \)[/tex] from the [tex]\( \text{Ca(NO}_3)_2 \)[/tex] salt, we'll build an ICE (Initial, Change, Equilibrium) table:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline & $\text{CaF}_2 (s)$ & $\text{Ca}^{2+} (aq)$ & $2 \text{F}^- (aq)$ \\ \hline \text{I (Initial)} & \text{Solid} & 0.010 \, \text{M (A)} & 0.000 \, \text{M (B)} \\ \hline \text{C (Change)} & -x & +x & +2x \\ \hline \text{E (Equilibrium)} & \text{Less solid} & 0.010 + x & 2x \\ \hline \end{tabular} \][/tex]
Next, we'll use the [tex]\( K_{sp} \)[/tex] expression to solve for [tex]\( x \)[/tex], which represents the molar solubility of [tex]\( \text{CaF}_2 \)[/tex]:
[tex]\[ \text{K}_{sp} = [\text{Ca}^{2+}] [\text{F}^-]^2 \][/tex]
Substituting the equilibrium concentrations into the [tex]\( K_{sp} \)[/tex] expression:
[tex]\[ 3.9 \times 10^{-11} = (0.010 + x) (2x)^2 \][/tex]
Since [tex]\( x \)[/tex] is very small compared to the initial concentration of [tex]\( \text{Ca}^{2+} \)[/tex], we can approximate [tex]\( 0.010 + x \approx 0.010 \)[/tex]:
[tex]\[ 3.9 \times 10^{-11} \approx (0.010) (2x)^2 \][/tex]
[tex]\[ 3.9 \times 10^{-11} = 0.010 \cdot 4x^2 \][/tex]
[tex]\[ 3.9 \times 10^{-11} = 4.0 \times 10^{-2} x^2 \][/tex]
[tex]\[ x^2 = \frac{3.9 \times 10^{-11}}{4.0 \times 10^{-2}} \][/tex]
[tex]\[ x^2 = 9.75 \times 10^{-10} \][/tex]
[tex]\[ x = \sqrt{9.75 \times 10^{-10}} \][/tex]
[tex]\[ x \approx 3.12 \times 10^{-5} \][/tex]
Let's round this to three decimal places:
[tex]\[ x \approx 0.000 \, 0312 \text{ M} \][/tex]
Now, we can fill in the ICE table:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline & $\text{CaF}_2 (s)$ & $\text{Ca}^{2+} (aq)$ & $2 \text{F}^- (aq)$ \\ \hline \text{I (Initial)} & \text{Solid} & 0.010 \, \text{M} (A) & 0.000 \, \text{M} (B) \\ \hline \text{C (Change)} & -x & +0.000 \, 0312 \text{ M} (C) & +0.000 \, 0624 \text{ M} (D) \\ \hline \text{E (Equilibrium)} & \text{Less solid} & 0.010 \text{ , 000} (E) & 0.000 \, 0624 \text{ M} (rounded to three decimal places) \\ \hline \end{tabular} \][/tex]
So, the correct values for the orange letters in the ICE table are:
A: [tex]\( 0.010 \)[/tex]
B: [tex]\( 0.000 \)[/tex]
C: [tex]\( 0.000 \, 0312 \)[/tex]
D: [tex]\( 0.000 \, 0624 \)[/tex]
E: [tex]\( 0.010 \)[/tex]
[tex]\[ \text{K}_{sp} (\text{CaF}_2) = 3.9 \times 10^{-11} \][/tex]
The dissociation of [tex]\( \text{CaF}_2 \)[/tex] in water is given by:
[tex]\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{F}^- (aq) \][/tex]
Given that the initial concentration of [tex]\( \text{Ca}^{2+} \)[/tex] is [tex]\( 0.010 \, \text{M} \)[/tex] from the [tex]\( \text{Ca(NO}_3)_2 \)[/tex] salt, we'll build an ICE (Initial, Change, Equilibrium) table:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline & $\text{CaF}_2 (s)$ & $\text{Ca}^{2+} (aq)$ & $2 \text{F}^- (aq)$ \\ \hline \text{I (Initial)} & \text{Solid} & 0.010 \, \text{M (A)} & 0.000 \, \text{M (B)} \\ \hline \text{C (Change)} & -x & +x & +2x \\ \hline \text{E (Equilibrium)} & \text{Less solid} & 0.010 + x & 2x \\ \hline \end{tabular} \][/tex]
Next, we'll use the [tex]\( K_{sp} \)[/tex] expression to solve for [tex]\( x \)[/tex], which represents the molar solubility of [tex]\( \text{CaF}_2 \)[/tex]:
[tex]\[ \text{K}_{sp} = [\text{Ca}^{2+}] [\text{F}^-]^2 \][/tex]
Substituting the equilibrium concentrations into the [tex]\( K_{sp} \)[/tex] expression:
[tex]\[ 3.9 \times 10^{-11} = (0.010 + x) (2x)^2 \][/tex]
Since [tex]\( x \)[/tex] is very small compared to the initial concentration of [tex]\( \text{Ca}^{2+} \)[/tex], we can approximate [tex]\( 0.010 + x \approx 0.010 \)[/tex]:
[tex]\[ 3.9 \times 10^{-11} \approx (0.010) (2x)^2 \][/tex]
[tex]\[ 3.9 \times 10^{-11} = 0.010 \cdot 4x^2 \][/tex]
[tex]\[ 3.9 \times 10^{-11} = 4.0 \times 10^{-2} x^2 \][/tex]
[tex]\[ x^2 = \frac{3.9 \times 10^{-11}}{4.0 \times 10^{-2}} \][/tex]
[tex]\[ x^2 = 9.75 \times 10^{-10} \][/tex]
[tex]\[ x = \sqrt{9.75 \times 10^{-10}} \][/tex]
[tex]\[ x \approx 3.12 \times 10^{-5} \][/tex]
Let's round this to three decimal places:
[tex]\[ x \approx 0.000 \, 0312 \text{ M} \][/tex]
Now, we can fill in the ICE table:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline & $\text{CaF}_2 (s)$ & $\text{Ca}^{2+} (aq)$ & $2 \text{F}^- (aq)$ \\ \hline \text{I (Initial)} & \text{Solid} & 0.010 \, \text{M} (A) & 0.000 \, \text{M} (B) \\ \hline \text{C (Change)} & -x & +0.000 \, 0312 \text{ M} (C) & +0.000 \, 0624 \text{ M} (D) \\ \hline \text{E (Equilibrium)} & \text{Less solid} & 0.010 \text{ , 000} (E) & 0.000 \, 0624 \text{ M} (rounded to three decimal places) \\ \hline \end{tabular} \][/tex]
So, the correct values for the orange letters in the ICE table are:
A: [tex]\( 0.010 \)[/tex]
B: [tex]\( 0.000 \)[/tex]
C: [tex]\( 0.000 \, 0312 \)[/tex]
D: [tex]\( 0.000 \, 0624 \)[/tex]
E: [tex]\( 0.010 \)[/tex]
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