Reaction 1: CaC2(s) + 2 H2O(l) ⇄ C2H2(g) + Ca(OH)2(s) Reaction 2: NaOCl(aq) + 2 HCl(aq) ⇄ Cl2(g) + N — Equilibrium Chemistry Question

STEPS:

1. Calculate the number of moles of calcium carbide ($\text{CaC}_2$) added:
* The sample mass of $\text{CaC}_2$ is $0.064\text{ g}$, and its molar mass is $64\text{ g/mol}$.
* Use this to calculate the moles of reactant added:
$$n_{\text{CaC}_2} = \frac{0.064\text{ g}}{64\text{ g/mol}} = \mathbf{0.0010\text{ mol}}$$
2. Determine the moles of calcium hydroxide ($\text{Ca(OH)}_2$) produced:
* According to Reaction 1, the reaction of $\text{CaC}_2$ with water is:
$$\text{CaC}_2(s) + 2\ \text{H}_2\text{O}(l) \rightleftharpoons \text{C}_2\text{H}_2(g) + \text{Ca(OH)}_2(s)$$
* The stoichiometry of this reaction shows a $1:1$ molar ratio between the reactant $\text{CaC}_2$ consumed and the product $\text{Ca(OH)}_2$ formed.
* Therefore, the complete reaction of $0.0010\text{ mol}$ of $\text{CaC}_2$ yields exactly $0.0010\text{ mol}$ of $\text{Ca(OH)}_2$.
3. Calculate the initial concentrations of $\text{Ca}^{2+}$ and $\text{OH}^-$ in the solution:
* Assume that the produced $\text{Ca(OH)}_2$ initially dissolves completely in the $1.0\text{ L}$ volume of distilled water:
$$\text{Ca(OH)}_2(aq) \rightarrow \text{Ca}^{2+}(aq) + 2\ \text{OH}^-(aq)$$
* Since $0.0010\text{ mol}$ of $\text{Ca(OH)}_2$ is dissolved in $1.0\text{ L}$ of water, the concentration of calcium ions is:
$$[\text{Ca}^{2+}] = \frac{0.0010\text{ mol}}{1.0\text{ L}} = \mathbf{1.0 \times 10^{-3}\text{ M}}$$
* Because each formula unit of calcium hydroxide dissociates to release two hydroxide ions, the hydroxide concentration is:
$$[\text{OH}^-] = 2 \times (1.0 \times 10^{-3}\text{ M}) = \mathbf{2.0 \times 10^{-3}\text{ M}}$$
4. Determine the reaction quotient ($Q$) for the dissolution of $\text{Ca(OH)}_2$:
* The solubility product equilibrium is:
$$\text{Ca(OH)}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\ \text{OH}^-(aq)$$
* Write the expression for the reaction quotient $Q$:
$$Q = [\text{Ca}^{2+}][\text{OH}^-]^2$$
* Substitute the calculated concentrations into the expression:
$$Q = (1.0 \times 10^{-3}) \times (2.0 \times 10^{-3})^2$$
$$Q = (1.0 \times 10^{-3}) \times (4.0 \times 10^{-6}) = \mathbf{4.0 \times 10^{-9}}$$
5. Compare the reaction quotient ($Q$) to the solubility product constant ($K_{sp}$):
* The calculated value of $Q$ is $4.0 \times 10^{-9}$.
* The given $K_{sp}$ of $\text{Ca(OH)}_2$ is $8.0 \times 10^{-6}$ (the official exam booklet lists it as $8.0 \times 10^{-8}$).
* Since $Q < K_{sp}$ ($4.0 \times 10^{-9} < 8.0 \times 10^{-6}$ or $8.0 \times 10^{-8}$), the ion concentrations have not reached the threshold required for saturation.
* Consequently, the solution is unsaturated, meaning no precipitate of $\text{Ca(OH)}_2$ will form, confirming Option D is correct.

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WHY_OTHERS_WRONG:

• Option A is incorrect: This option claims a precipitate will form because $Q > K_{sp}$. This is mathematically incorrect because our calculated reaction quotient of $4.0 \times 10^{-9}$ is several orders of magnitude smaller than the $K_{sp}$.
• Option B is incorrect: This option states that a precipitate will form because $Q < K_{sp}$. While the comparison $Q < K_{sp}$ is correct, a precipitate only forms when the ion concentrations exceed the solubility limit (i.e., when $Q > K_{sp}$). An unsaturated solution ($Q < K_{sp}$) does not yield a precipitate.
• Option C is incorrect: This option correctly states that a precipitate will not form but provides an incorrect mathematical justification ($Q > K_{sp}$). If $Q > K_{sp}$ were true, the solution would be supersaturated, and a precipitate would have to form to restore equilibrium.