[VISUAL] Based on the information above, which of the following expressions represents the equilibri — Equilibrium Chemistry Question

STEPS:

1. Analyze the target chemical reaction and its equilibrium expression:
The target reaction is:
$$\text{La}^{3+} + \text{CO}_3^{2-} \rightleftharpoons \text{LaCO}_3^+$$
The equilibrium constant expression for this target reaction is:
$$K = \frac{[\text{LaCO}_3^+]}{[\text{La}^{3+}][\text{CO}_3^{2-}]}$$

2. Write out the given equilibrium reactions and their expressions:
* Reaction 1: $\text{La}^{3+} + \text{OH}^- + \text{HCO}_3^- \rightleftharpoons \text{LaCO}_3^+ + \text{H}_2\text{O} \quad (K_{eq} = K_1)$
$$K_1 = \frac{[\text{LaCO}_3^+]}{[\text{La}^{3+}][\text{OH}^-][\text{HCO}_3^-]}$$
* Reaction 2: $\text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-} \quad (K_{eq} = K_a)$
$$K_a = \frac{[\text{H}^+][\text{CO}_3^{2-}]}{[\text{HCO}_3^-]}$$
* Reaction 3: $\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \quad (K_{eq} = K_w)$
$$K_w = [\text{H}^+][\text{OH}^-]$$

3. Determine how to combine the equations to yield the target reaction:
We can manipulate and add the three given reactions to get the target net ionic equation:
* Keep Reaction 1 as written: We need $\text{La}^{3+}$ as a reactant and $\text{LaCO}_3^+$ as a product.
$$\text{La}^{3+} + \text{OH}^- + \text{HCO}_3^- \rightleftharpoons \text{LaCO}_3^+ + \text{H}_2\text{O} \quad (\text{new } K = K_1)$$
* Reverse Reaction 2: We need $\text{CO}_3^{2-}$ on the reactant side and need to cancel $\text{HCO}_3^-$. Reversing a reaction takes the reciprocal of its equilibrium constant ($1/K_eq$).
$$\text{H}^+ + \text{CO}_3^{2-} \rightleftharpoons \text{HCO}_3^- \quad \left(\text{new } K = \frac{1}{K_a}\right)$$
* Keep Reaction 3 as written: We need to cancel out the extra water, hydronium, and hydroxide ions introduced in the first two steps.
$$\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \quad (\text{new } K = K_w)$$

4. Sum the manipulated equations and simplify:
Add all three manipulated equations together:
$$\left(\text{La}^{3+} + \cancel{\text{OH}^-} + \cancel{\text{HCO}_3^-}\right) + \left(\cancel{\text{H}^+} + \text{CO}_3^{2-}\right) + \cancel{\text{H}_2\text{O}} \rightleftharpoons \left(\text{LaCO}_3^+ + \cancel{\text{H}_2\text{O}}\right) + \cancel{\text{HCO}_3^-} + \left(\cancel{\text{H}^+} + \cancel{\text{OH}^-}\right)$$
Canceling the species that appear on both sides of the equation leaves exactly the target reaction:
$$\text{La}^{3+} + \text{CO}_3^{2-} \rightleftharpoons \text{LaCO}_3^+$$

5. Apply the mathematical rules of combining equilibrium constants:
When reactions are added, their individual equilibrium constants are multiplied:
$$K = K_1 \times \left(\frac{1}{K_a}\right) \times K_w = \frac{K_1 K_w}{K_a}$$
This matches Option D exactly.

6. Verify the final expression algebraically:
Substitute the concentration expressions of each constant into Option D to verify:
$$\frac{K_1 K_w}{K_a} = \frac{\left(\frac{[\text{LaCO}_3^+]}{[\text{La}^{3+}][\text{OH}^-][\text{HCO}_3^-]}\right) \left([\text{H}^+][\text{OH}^-]\right)}{\left(\frac{[\text{H}^+][\text{CO}_3^{2-}]}{[\text{HCO}_3^-]}\right)}$$
$$\frac{K_1 K_w}{K_a} = \frac{[\text{LaCO}_3^+][\text{H}^+][\text{OH}^-]}{[\text{La}^{3+}][\text{OH}^-][\text{HCO}_3^-]} \times \frac{[\text{HCO}_3^-]}{[\text{H}^+][\text{CO}_3^{2-}]} = \frac{[\text{LaCO}_3^+]}{[\text{La}^{3+}][\text{CO}_3^{2-}]} = K$$

*

WHY_OTHERS_WRONG:

• Option A is incorrect: This expression ($K = K_1 K_a K_w$) represents the product of adding all three reactions as originally written. Summing the three unmodified reactions would yield the net equation: $\text{La}^{3+} + \text{OH}^- \rightleftharpoons \text{LaCO}_3^+ + 2\ \text{H}^+ + \text{CO}_3^{2-}$, which is not the target reaction.
• Option B is incorrect: This expression ($K = \frac{K_1 K_a}{K_w}$) would result from keeping Reaction 1 and Reaction 2 as written, and reversing Reaction 3. This fails to place the carbonate ion on the reactant side and does not cancel the intermediate species correctly.
• Option C is incorrect: This expression ($K = \frac{K_1}{K_a K_w}$) would result from keeping Reaction 1 as written, while reversing both Reaction 2 and Reaction 3. Reversing Reaction 3 would leave water on the product side and extra ions on the reactant side, preventing them from canceling.