Problem 129 - Chemistry: An Atoms-Focused Approach

Question

Cobalt (II) and zinc (II) carbonate have nearly identical \(K_{\mathrm{sp}}\) values. Do they have the same molar solubility? Do they have the same solubility in \(\mathrm{g} / \mathrm{L} ?\)

Step-by-Step Solution

Answer

Explain your answer. Answer: Cobalt (II) carbonate and Zinc (II) carbonate have the same molar solubility due to their similar \(K_{\mathrm{sp}}\) values. However, they have different solubilities in g/L due to their different molar masses.

Step 1: Write the dissolution reactions

We write the dissolution reactions for both the compounds: 1. CoCO3(s) \(\rightarrow\) Co2+(aq) + CO32-(aq) 2. ZnCO3(s) \(\rightarrow\) Zn2+(aq) + CO32-(aq)

Step 2: Calculate molar solubility using Ksp

Let the solubility of each compound be 'S' mol/L. Therefore, \(K_{sp}\) = [Co2+][CO32-] = [S][S] = \(S^2\) for CoCO3 \(K_{sp}\) = [Zn2+][CO32-] = [S][S] = \(S^2\) for ZnCO3 Given that the \(K_{\mathrm{sp}}\) values are nearly identical, their solubility products ('S') will be the same. Therefore, they have the same molar solubility.

Step 3: Calculate solubility in g/L

To calculate solubility in g/L, we need to multiply the molar solubility by the compound's molar mass. Solubility of CoCO3 in g/L = \(S \times M(CoCO3)\) Solubility of ZnCO3 in g/L = \(S \times M(ZnCO3)\) Since S is the same for both compounds, and M(ZnCO3) > M(CoCO3), it follows that the solubility of ZnCO3 in g/L is greater than the solubility of CoCO3 in g/L. In conclusion, Cobalt (II) carbonate and Zinc (II) carbonate have the same molar solubility due to their similar \(K_{\mathrm{sp}}\) values. However, they have different solubilities in g/L due to their different molar masses.

Key Concepts

Ksp (Solubility Product)

The solubility product constant, denoted as \( K_{sp} \), is an essential concept in chemistry. It helps predict how much of a compound will dissolve in water to form a saturated solution. The \( K_{sp} \) value is specific to each compound and depends on the temperature.
The \( K_{sp} \) tells us about the maximum concentration of ions that can exist in a solution at equilibrium. For a simple ionic compound like cobalt (II) carbonate \( \text{CoCO}_3 \) or zinc (II) carbonate \( \text{ZnCO}_3 \), the dissolved ions reach an equilibrium with the solid compound.
For both cobalt and zinc carbonates, the \( K_{sp} \) expressions are:

• \([\text{Co}^{2+}][\text{CO}_3^{2-}] = S^2\) for cobalt carbonate
• \([\text{Zn}^{2+}][\text{CO}_3^{2-}] = S^2\) for zinc carbonate

Since the \( K_{sp} \) values for cobalt (II) carbonate and zinc (II) carbonate are nearly identical, the amount of each that will dissolve in molar terms is the same, but their solubility in grams per liter differs due to their different molar masses.

Molar Solubility

Molar solubility represents the number of moles of a substance that can dissolve to form a liter of saturated solution. It is closely related to the solubility product, \( K_{sp} \).
For cobalt (II) carbonate and zinc (II) carbonate, assuming the \( K_{sp} \) is the same, the molar solubility \( S \) can be calculated as follows:

• Given the equation \( S^2 = K_{sp} \),
• solve for \( S \) to find the molar solubility.

This means that at equilibrium, each compound can dissolve into its ions, \([\text{Co}^{2+}]\) and \([\text{Zn}^{2+}]\), to yield a concentration of \( S \) mol/L respectively.
This explains why both cobalt and zinc carbonates share the same molar solubility under identical conditions, making their comparative study in solutions intriguing.

Chemical Reactions

Chemical reactions involving dissolution provide insight into solubility. The dissolution reaction for both cobalt (II) carbonate and zinc (II) carbonate reads:

• \( \text{CoCO}_3 (s) \rightarrow \text{Co}^{2+} (aq) + \text{CO}_3^{2-} (aq) \)
• \( \text{ZnCO}_3 (s) \rightarrow \text{Zn}^{2+} (aq) + \text{CO}_3^{2-} (aq) \)

These reactions indicate that one mole of the solid compound dissolves to give one mole of each ion.
This process reaches an equilibrium state, which is when the concentration of the ions stops changing. At equilibrium, the solution is saturated, and no more solids dissolve.
Understanding these chemical reactions allows us to connect the concepts of \( K_{sp} \) and molar solubility, explaining how and why different compounds dissolve under similar circumstances.