Problem 92 - Chemistry: An Atoms-Focused Approach

Question

Identify any of the following solids that are more soluble in acidic solution than in neutral water: \(\mathrm{CaCl}_{2}, \mathrm{Ba}\left(\mathrm{HCO}_{3}\right)_{2}\) \(\mathrm{PbSO}_{4}, \mathrm{Cu}(\mathrm{OH})_{2} .\) Explain your choices.

Step-by-Step Solution

Answer

Answer: Ba(HCO3)2 and Cu(OH)2 will have increased solubility in an acidic solution compared to a neutral one.

Step 1: Recall solubility equilibrium and the solubility product constant (Ksp)

A solubility equilibrium represents the equilibrium between a solid and its dissolved ions, and it can be represented as a balanced chemical equation. Solubility product constant (Ksp) is the equilibrium constant for this chemical equation, and it represents the solubility of a solid. It's important to mention that, when adding an acidic solution (source of \(\mathrm{H}^{+}\) ions), the solubility of solids can change due to a reaction with ions in the solid.

Step 2: Analyze the given solids

Write down the solubility equilibrium for each of the given solids and examine which ions can react with \(\mathrm{H}^{+}\) ions in an acidic solution: 1. \(\mathrm{CaCl}_{2}:\) \(\mathrm{CaCl}_{2(s)} \rightleftharpoons \mathrm{Ca}^{2+} (aq) + 2 \mathrm{Cl}^{-} (aq)\). Both \(\mathrm{Ca}^{2+}\) and \(\mathrm{Cl}^{-}\) do not form complexes with \(\mathrm{H}^{+}\) ions. Therefore, the solubility of \(\mathrm{CaCl}_{2}\) remains unchanged in an acidic solution. 2. \(\mathrm{Ba}\left(\mathrm{HCO}_{3}\right)_{2}:\) \(\mathrm{Ba}\left(\mathrm{HCO}_{3}\right)_{(2s)} \rightleftharpoons \mathrm{Ba}^{2+} (aq) + 2 \mathrm{HCO}_{3}^{-}\) \((aq)\). Bicarbonate ions (\(\mathrm{HCO}_{3}^{-}\)) react with \(\mathrm{H}^{+}\) ions in an acidic solution and form carbonic acid (\(\mathrm{H}_{2}\mathrm{CO}_{3}\)): \(\mathrm{H}^{+} (aq) + \mathrm{HCO}_{3}^{-} (aq) \rightarrow \mathrm{H}_{2}\mathrm{CO}_{3(aq)}\). This reaction will shift the equilibrium to the right side, increasing the solubility of \(\mathrm{Ba}\left(\mathrm{HCO}_{3}\right)_{2}\) in an acidic solution. 3. \(\mathrm{PbSO}_{4}:\) \(\mathrm{PbSO}_{4(s)} \rightleftharpoons \mathrm{Pb}^{2+} (aq) + \mathrm{SO}_{4}^{2-} (aq)\). Sulfate ions (\(\mathrm{SO}_{4}^{2-}\)) do not form complexes with \(\mathrm{H}^{+}\) ions, so the solubility of \(\mathrm{PbSO}_{4}\) remains unchanged in an acidic solution. 4. \(\mathrm{Cu}(\mathrm{OH})_{2}:\) \(\mathrm{Cu}(\mathrm{OH})_{2(s)} \rightleftharpoons \mathrm{Cu}^{2+} (aq) + 2 \mathrm{OH}^{-} (aq)\). Hydroxide ions (\(\mathrm{OH}^{-}\)) react with \(\mathrm{H}^{+}\) ions in an acidic solution to form water: \(\mathrm{H}^{+} (aq) + \mathrm{OH}^{-} (aq) \rightarrow \mathrm{H}_{2}\mathrm{O}_{(l)}\). This reaction will shift the equilibrium to the right side, increasing the solubility of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in an acidic solution.

Step 3: Conclusion

After analyzing the given solids, we can conclude that only \(\mathrm{Ba}\left(\mathrm{HCO}_{3}\right)_{2}\) and \(\mathrm{Cu}(\mathrm{OH})_{2}\) have increased solubility in acidic solutions compared to neutral water. This is because their respective anions, \(\mathrm{HCO}_{3}^{-}\) and \(\mathrm{OH}^{-}\), react with \(\mathrm{H}^{+}\) ions, which shifts the solubility equilibria and increases solubility.

Key Concepts

solubility product constant (Ksp)

The solubility product constant, or Ksp, is a crucial concept in understanding solubility equilibrium. It is an equilibrium constant that applies to the dissolution of slightly soluble salts. When a solid dissolves in water, it breaks down into its constituent ions and the Ksp indicates how far this dissolution process can go before reaching equilibrium.
The Ksp is determined through the balanced equation of the dissolution, providing a precise mathematical representation. For instance, when a salt like \( \text{PbSO}_4 \) dissolves, the equation is \( \text{PbSO}_{4(s)} \rightleftharpoons \text{Pb}^{2+} (aq) + \text{SO}_4^{2-} (aq) \). The Ksp represents the product of the concentrations of these ions raised to the power of their coefficients in the balanced equation:

• \( K_{sp} = [\text{Pb}^{2+}][\text{SO}_4^{2-}] \)

When the solution is saturated, the multiplication of these ion concentrations provides the exact Ksp value. Under different conditions, such as a change in pH, these values might alter the solubility and thus are crucial in predicting the salt's behavior in solution.

reaction with H+ ions

The presence of \( \text{H}^+ \) ions in a solution can cause reactions that significantly affect solubility equilibrium. \( \text{H}^+ \) ions can react with various anions present in a dissolved salt, leading to different products that alter the solubility equilibrium. This type of reaction is particularly important when the dissolved ions can combine with \( \text{H}^+ \) ions to form less ionic compounds or molecules.
For example, hydrochloric acid can donate \( \text{H}^+ \) ions which can react with hydroxide ions (\( \text{OH}^- \)) in a solution:

• The reaction \( \text{H}^+ (aq) + \text{OH}^- (aq) \rightarrow \text{H}_2\text{O}_{(l)} \) produces water.

This reaction effectively removes \( \text{OH}^- \) ions from the solution, thus driving the dissolution reaction of \( \text{Cu(OH)}_2 \) to the right and increasing its solubility. Such reactions demonstrate how \( \text{H}^+ \) ions can influence the solubility of a compound, especially ones that involve anions capable of reacting with protons to form neutral species.

solubility in acidic solution

Solubility can be notably affected by the acidity of the solution. A solid that is only slightly soluble in neutral water may exhibit increased solubility in an acidic environment due to the interactions between \( \text{H}^+ \) ions and the salt's anions.
Take the example of \( \text{Ba(HCO}_3\text{)}_2 \), where its bicarbonate ion \( \text{HCO}_3^- \) readily reacts with \( \text{H}^+ \) ions to form \( \text{H}_2\text{CO}_3 \) (carbonic acid):

• \( \text{H}^+ (aq) + \text{HCO}_3^- (aq) \rightarrow \text{H}_2\text{CO}_3 (aq) \).

This reaction shifts the equilibrium by decreasing the bicarbonate ion concentration, prompting more solid to dissolve to re-establish equilibrium. Similarly, \( \text{Cu(OH)}_2 \) is more soluble in acidic water due to the neutralization of \( \text{OH}^- \) ions by \( \text{H}^+ \) ions, driving further dissolution.
This principle of increased solubility in acidic solutions is utilized across numerous applications, from geochemistry to designing effective drug formulations, ensuring that compounds dissolve effectively where they are intended to act.