Question

question 5 (1 point)
write the equation for $k_{eq}$ for the following as a homogeneous equilibrium (only the aqueous phase)
$\ce{caco_{3(s)} + 2 hf_{(aq)} <-> caf_{2(aq)} + h_{2}o_{(liq)} + co_{2(aq)}}$
1 2
$k_{eq} = \frac{-----------}{__3__^2}$ use the order given in the reaction
the concentration of hf is squared because its coefficient is 2
a. $\ce{h_{2}o_{(liq)}}$ b. $\ce{oh^{-1}}$ c. $\ce{h_{3}o^{+1}}$ d. $\ce{so_{4}^{-2}}$ e. $\ce{mg^{+2}}$ f. $\ce{cl^{-1}}$
g. $\ce{hpo_{4}^{-2}}$ h. $\ce{h_{2}po_{4}^{-1}}$ i. $\ce{co_{3}^{-2}}$ j. $\ce{hco_{3}^{-1}}$ k. $\ce{h_{2}s}$ l. $\ce{hs^{-1}}$
m. $\ce{s^{-2}}$ n. $\ce{h_{2}o}$ o. $\ce{h_{2}co_{3}}$ p. $\ce{h_{3}po_{4}}$ q. $\ce{hc_{2}h_{3}o_{2}}$ r. $\ce{c_{2}h_{3}o_{2}^{-1}}$
s. $\ce{po_{4}^{-3}}$ t. $\ce{so_{4}^{-2}}$ u. $\ce{caco_{3(s)}}$ v. $\ce{hf_{(aq)}}$ w. $\ce{caf_{2(aq)}}$ x. $\ce{co_{2}}$

Explanation:

Step1: Recall Equilibrium Constant Rules

For a homogeneous equilibrium (aqueous phase here, and we ignore solids and pure liquids in \(K_{eq}\) expression), the formula is \(K_{eq}=\frac{[\text{products (aqueous/gaseous)}]}{[\text{reactants (aqueous/gaseous)}]}\), with exponents from stoichiometric coefficients. The reaction is \( \text{CaCO}_3(\text{s}) + 2\text{HF}(\text{aq}) ightleftharpoons \text{CaF}_2(\text{aq}) + \text{H}_2\text{O}(\text{liq}) + \text{CO}_2(\text{aq}) \). Solids (\(\text{CaCO}_3(\text{s})\)) and pure liquids (\(\text{H}_2\text{O}(\text{liq})\)) are excluded. So products (aqueous) are \(\text{CaF}_2(\text{aq})\) and \(\text{CO}_2(\text{aq})\)? Wait, no, wait the \(K_{eq}\) structure given: numerator has two terms (positions 1 and 2) and denominator is \([\text{HF}]^2\) (position 3 is HF). Wait, the numerator should be the concentrations of the products (aqueous or gaseous) raised to their coefficients, and denominator reactants (aqueous or gaseous) raised to their coefficients. Wait, the reaction's products in aqueous/gaseous: \(\text{CaF}_2(\text{aq})\), \(\text{CO}_2(\text{aq})\) (since \(\text{H}_2\text{O}\) is liquid, excluded). Reactant in aqueous: \(\text{HF}(\text{aq})\) (coefficient 2), and \(\text{CaCO}_3\) is solid (excluded). Wait, but the \(K_{eq}\) structure is \(K_{eq}=\frac{[\_1\_][\_2\_]}{[\_3\_]^2}\), where \(_3\_\) is HF (V). Then \(_1\_\) and \(_2\_\) should be the products (aqueous or gaseous) from the reaction. The products are \(\text{CaF}_2(\text{aq})\) (W), \(\text{CO}_2(\text{aq})\) (X), and \(\text{H}_2\text{O}\) (liquid, excluded). Wait, but maybe the problem considers \(\text{H}_2\text{O}\) as liquid, but maybe in the options, let's check the options. Wait, the options for position 1, 2, 3: position 3 is HF (V). Position 1 and 2: products. Let's list the products (aqueous or gaseous) in the reaction: \(\text{CaF}_2(\text{aq})\) (W), \(\text{CO}_2(\text{aq})\) (X), and \(\text{H}_2\text{O}\) (liquid, but option A is \(\text{H}_2\text{O}(\text{liq})\), but pure liquids are excluded from \(K_{eq}\). Wait, maybe the problem has a mistake, but looking at the options, the numerator terms: let's check the options. Wait, the \(K_{eq}\) formula given is \(K_{eq}=\frac{[\_1\_][\_2\_]}{[\_3\_]^2}\), with \(_3\_\) being HF (V). So \(_1\_\) and \(_2\_\) should be the products (aqueous) from the reaction. The products are \(\text{CaF}_2(\text{aq})\) (W) and \(\text{CO}_2(\text{aq})\) (X)? Wait, but option X is \(\text{CO}_2\), and W is \(\text{CaF}_2(\text{aq})\). Alternatively, maybe the problem includes \(\text{CaF}_2(\text{aq})\) (W) and \(\text{CO}_2(\text{aq})\) (X), or maybe \(\text{CaF}_2(\text{aq})\) (W) and \(\text{H}_2\text{O}\) (but \(\text{H}_2\text{O}\) is liquid, which is not included in \(K_{eq}\) for homogeneous (aqueous) but maybe the problem is considering it? No, pure liquids and solids are not included. Wait, maybe the reaction's products are \(\text{CaF}_2(\text{aq})\), \(\text{CO}_2(\text{aq})\), and the reactant is \(\text{HF}(\text{aq})\). So numerator: \([\text{CaF}_2][\text{CO}_2]\), denominator: \([\text{HF}]^2\). So position 1: \(\text{CaF}_2(\text{aq})\) (W), position 2: \(\text{CO}_2\) (X), position 3: \(\text{HF}(\text{aq})\) (V). Let's check the options:

- Position 1: W. \(\text{CaF}_2(\text{aq})\)
- Position 2: X. \(\text{CO}_2\)
- Position 3: V. \(\text{HF}(\text{aq})\)

Wait, but let's confirm the \(K_{eq}\) expression. For the reaction, excluding solid (\(\text{CaCO}_3\)) and liquid (\(\text{H}_2\text{O}\)), the equilibrium constant is \(K_{eq}=\frac{[\text{CaF}_2][\text{CO}_2]}{[\text{HF}]^2}\). So:

- \(_1\_\): \(\text{CaF}_2(\text{aq})\) (W)
- \(_2\_\): \(\text{CO}_2\) (X)
- \(_3\_\): \(\text{HF}(\text{aq})\) (V)

Step2: Match with Options

- Position 1 (first numerator term): W. \(\text{CaF}_2(\text{aq})\)
- Position 2 (second numerator term): X. \(\text{CO}_2\)
- Position 3 (denominator term): V. \(\text{HF}(\text{aq})\)

Wait, but let's check the options again. The options for position 1, 2, 3:

- For \(_1\_\): Looking at the options, W is \(\text{CaF}_2(\text{aq})\), X is \(\text{CO}_2\), A is \(\text{H}_2\text{O}(\text{liq})\) (liquid, excluded), so W and X are the products (aqueous/gaseous) and V is the reactant (aqueous) with coefficient 2.

Answer:

1. W. \(\text{CaF}_2(\text{aq})\)
2. X. \(\text{CO}_2\)
3. V. \(\text{HF}(\text{aq})\)