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Construct an ICE table to calculate the molar solubility of [tex]$CaF_2$[/tex] at [tex]$25^{\circ} C$[/tex] in a solution containing [tex]$0.010 \text{ M} \, Ca \left( NO_3 \right)_2$[/tex]. The solubility product constant is [tex][tex]$K_{sp} = 3.9 \times 10^{-11}$[/tex][/tex].

Complete the table by typing in the correct answer, represented by the orange letters in the ICE table. (When writing values between 0 and 1, make sure to enter a 0 before the decimal place.)

A: [tex]$\square$[/tex] [tex]$\square$[/tex]
B: [tex]$\square$[/tex] (Write to three decimal places.)

C: [tex]$\square$[/tex]
D: [tex]$\square$[/tex]
E: [tex]$\square$[/tex] [tex]$\square$[/tex] (Write to three decimal places.)

\begin{tabular}{|c|c|c|c|}
\hline
& [tex]$CaF_2(s)$[/tex] & [tex]$Ca^{2+}(aq)$[/tex] & [tex]$2F^{-}(aq)$[/tex] \\
\hline
I & Solid & A M & B M \\
\hline
C & [tex]$-x$[/tex] & + C & + D \\
\hline
E & Less solid & EM & [tex]$2x$[/tex] \\
\hline
\end{tabular}



Answer :

GinnyAnswer
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]

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