Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
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]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
