Problem 70 - Chemistry The Molecular Science

Question

Calculate the solubility of \(\mathrm{ZnCO}_{3}, K_{\mathrm{sp}}=1.4 \times 10^{-11}\), in (a) water. (b) \(0.050-\mathrm{M} \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\). (c) \(0.050-\mathrm{M} \mathrm{K}_{2} \mathrm{CO}_{2}\)

Step-by-Step Solution

Answer

(a) 1.18 × 10⁻⁶ M; (b) 2.8 × 10⁻¹⁰ M; (c) 2.8 × 10⁻¹⁰ M.

Step 1: Write the Dissolution Equation for ZnCO3

The dissolution of zinc carbonate can be expressed as \( \text{ZnCO}_3 (s) \rightleftharpoons \text{Zn}^{2+} (aq) + \text{CO}_3^{2-} (aq) \). This indicates that solid ZnCO3 dissociates into zinc ions and carbonate ions in solution.

Step 2: Expression for Ksp

For the reaction \( \text{ZnCO}_3 (s) \rightleftharpoons \text{Zn}^{2+} (aq) + \text{CO}_3^{2-} (aq) \), the solubility product \( K_{sp} \) is given by \( K_{sp} = [\text{Zn}^{2+}][\text{CO}_3^{2-}] \). The value given is \( K_{sp} = 1.4 \times 10^{-11} \).

Step 3: Calculate Solubility in Pure Water

In pure water, the concentration of \( \text{Zn}^{2+} \) and \( \text{CO}_3^{2-} \) will be the same because for every mole of \( \text{ZnCO}_3 \) that dissolves, one mole each of \( \text{Zn}^{2+} \) and \( \text{CO}_3^{2-} \) are produced. Let the solubility (S) of \( \text{ZnCO}_3 \) be S. So, \([\text{Zn}^{2+}] = [\text{CO}_3^{2-}] = S\). Substitute these into the Ksp expression: \( 1.4 \times 10^{-11} = S^2 \). Solving for S gives \( S = \sqrt{1.4 \times 10^{-11}} \approx 1.18 \times 10^{-6} \text{ mol/L} \).

Step 4: Calculate Solubility in Zn(NO3)2 Solution

In a \(0.050\, \text{M} \) Zn(NO\(_3\))\(_2\) solution, there are already \(0.050\, \text{M} \) \( \text{Zn}^{2+} \) ions. Let the additional solubility of ZnCO\(_3\) be S. The equation is \( 1.4 \times 10^{-11} = (0.050 + S)S \). Assuming \( S \) is small compared to \(0.050\), it simplifies to \( 1.4 \times 10^{-11} \approx 0.050S \). Solving gives \( S = \frac{1.4 \times 10^{-11}}{0.050} \approx 2.8 \times 10^{-10} \text{ mol/L} \).

Step 5: Calculate Solubility in K2CO3 Solution

In a \(0.050\, \text{M} \) K\(_2\)CO\(_3\) solution, there are already \(0.050\, \text{M} \) \( \text{CO}_3^{2-} \) ions. Let the additional solubility of ZnCO\(_3\) be \( S \). So \( 1.4 \times 10^{-11} = S(0.050 + S) \). Assuming \( S \) is small compared to \(0.050\), \( 1.4 \times 10^{-11} \approx 0.050S \). Solving gives \( S = \frac{1.4 \times 10^{-11}}{0.050} \approx 2.8 \times 10^{-10} \text{ mol/L} \).

Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant, often symbolized as \( K_{sp} \), is a key concept in understanding the solubility of sparingly soluble ionic compounds. In simple terms, \( K_{sp} \) indicates the maximum extent to which a compound can dissolve in water to form its constituent ions before the solution becomes saturated. It is specific to the particular salt and is typically a small number for salts that are not very soluble in water.
For example, the dissociation of zinc carbonate (\( \text{ZnCO}_3 \)) in water can be represented by the equation \( \text{ZnCO}_3 (s) \rightleftharpoons \text{Zn}^{2+} (aq) + \text{CO}_3^{2-} (aq) \). The \( K_{sp} \) expression for this equilibrium is \( K_{sp} = [\text{Zn}^{2+}][\text{CO}_3^{2-}] \).

• It defines the concentrations of ions at equilibrium.
• For zinc carbonate, given \( K_{sp} = 1.4 \times 10^{-11} \), this tiny number tells us it doesn't dissolve very well.

To find the solubility of \( \text{ZnCO}_3 \), we look for the concentrations at which \( [\text{Zn}^{2+}] \) and \( [\text{CO}_3^{2-}] \) meet this \( K_{sp} \) value after the salt has dissolved.

Ionic Equilibrium

Ionic equilibrium concerns the balance between ions in a solution and the undissolved solid form of an ionic compound. When a slightly soluble salt like \( \text{ZnCO}_3 \) is added to water, it reaches a state where the rate of dissolution equals the rate of precipitation—the static point being ionic equilibrium. Understanding this concept is crucial for predicting how much of a compound will dissolve under a variety of conditions.
In a stagnant solution, concentrations of ions are not stable but fluctuate until equilibrium is reached. For each dissolving unit of \( \text{ZnCO}_3 \), one zinc ion (\( \text{Zn}^{2+} \)) and one carbonate ion (\( \text{CO}_3^{2-} \)) are released, balancing the system. As equilibrium is established:

• The concentrations of \( \text{Zn}^{2+} \) and \( \text{CO}_3^{2-} \) are equal if no other ions are present.
• This symmetrical balance lets us use simple algebra to find solubility, like solving \( S^2 = K_{sp} \), where \( S \) is the solubility in mol/L.

The more complex scenarios involve additional ions or alterations to the solution makeup, forcing a shift in equilibrium, which is a prelude to our next concept—the common ion effect.

Common Ion Effect

The common ion effect plays a crucial role in determining solubility, especially when a solution already contains an ion present in the salt. When \( \text{ZnCO}_3 \) is introduced into a solution already containing either \( \text{Zn}^{2+} \) or \( \text{CO}_3^{2-} \), it influences how much more \( \text{ZnCO}_3 \) will dissolve.
This effect arises because the presence of a common ion causes a shift in ionic equilibrium, according to Le Chatelier's principle:

• If you add \( \text{Zn}^{2+} \) ions from \( \text{Zn(NO}_3)_2 \), \( \text{ZnCO}_3 \) dissolves less due to increased ionic product exceeding \( K_{sp} \).
• The equilibrium shifts, precipitating more \( \text{ZnCO}_3 \) until \( K_{sp} \) is restored, significantly lowering solubility.
• In the presence of extra \( \text{CO}_3^{2-} \) ions, the scenario is mirrored, reducing solubility similarly.

For example, in a \( 0.050 \)-M \( \text{Zn(NO}_3)_2 \) or \( \text{K}_2\text{CO}_3 \) solution, solubility becomes \( 2.8 \times 10^{-10} \) mol/L, a significant drop from pure water, demonstrating the common ion effect in action.