Problem 98 - Chemistry and Chemical Reactivity

Question

Which of the following barium salts should dissolve in a strong acid such as HCl: \(\mathrm{Ba}(\mathrm{OH})_{2}, \mathrm{BaSO}_{4},\) or \(\mathrm{BaCO}_{3} ?\)

Step-by-Step Solution

Answer

Ba(OH)2 and BaCO3 will dissolve in HCl.

Step 1: Understanding the solubility of Ba(OH)2 in HCl

Barium hydroxide, \( \mathrm{Ba(OH)_2} \), is a strong base. When \( \mathrm{Ba(OH)_2} \) is added to a strong acid like \( \mathrm{HCl} \), it will react to form water and a soluble salt (\( \mathrm{BaCl_2} \)). This is a classic example of an acid-base reaction where a base dissolves in an acid.

Step 2: Considering the solubility of BaSO4 in HCl

Barium sulfate, \( \mathrm{BaSO_4} \), is known for being highly insoluble in water and does not typically react with strong acids like \( \mathrm{HCl} \). The sulfate ion is quite stable and does not form a more soluble compound upon reacting with hydrochloric acid.

Step 3: Analyzing the solubility of BaCO3 in HCl

Barium carbonate, \( \mathrm{BaCO_3} \), reacts with strong acids like \( \mathrm{HCl} \) to produce water and carbon dioxide along with a soluble salt (\( \mathrm{BaCl_2} \)). The reaction can be represented as: \[ \mathrm{BaCO_3 (s) + 2 HCl (aq) \rightarrow BaCl_2 (aq) + H_2O (l) + CO_2 (g)} \]. This indicates that \( \mathrm{BaCO_3} \) will dissolve in \( \mathrm{HCl} \).

Key Concepts

Barium Hydroxide and HCl Reaction

Barium hydroxide, denoted as \( \mathrm{Ba(OH)_2} \), participates in an acid-base reaction when introduced to hydrochloric acid (\( \mathrm{HCl} \)). This reaction is a classic example of how a strong base interacts with a strong acid.
The equation for the reaction is as follows:

• \( \mathrm{Ba(OH)_2(s) + 2HCl(aq) \rightarrow BaCl_2(aq) + 2H_2O(l)} \)

In this process, barium hydroxide neutralizes the acid, producing water and barium chloride (\( \mathrm{BaCl_2} \)). Barium chloride is a soluble salt, which means it dissolves in the solution. This transformation effectively leads to the dissolution of barium hydroxide in acids like \( \mathrm{HCl} \).
Understanding this concept clarifies why \( \mathrm{Ba(OH)_2} \) can easily dissolve in strong acids, making it a great candidate for neutralizing acidic solutions.

Barium Sulfate Insolubility

Barium sulfate, chemically expressed as \( \mathrm{BaSO_4} \), is characteristically known for its insolubility. It does not dissolve easily in water or even in strong acids such as hydrochloric acid (\( \mathrm{HCl} \)). This widespread property can be attributed to the stable nature of the sulfate ion (\( \mathrm{SO_4^{2-}} \)).

• Unlike some other salts, \( \mathrm{BaSO_4} \) doesn't react with \( \mathrm{HCl} \) to form more soluble compounds.
• The strong ionic bonds in \( \mathrm{BaSO_4} \) contribute to its persistent insolubility.

Even in strong acidic conditions, \( \mathrm{BaSO_4} \) remains unreactive, maintaining its solid state without dissolving. This particular quality makes barium sulfate invaluable in various practical applications, like radiology, where its transparency to X-rays is beneficial despite its insolubility.

Barium Carbonate and HCl Reaction

Barium carbonate (\( \mathrm{BaCO_3} \)) behaves differently when mixed with strong acids like \( \mathrm{HCl} \). This compound readily reacts with hydrochloric acid in a notable manner that results in the evolution of carbon dioxide gas.

• The reaction is: \[ \mathrm{BaCO_3(s) + 2HCl(aq) \rightarrow BaCl_2(aq) + H_2O(l) + CO_2(g)} \]
• Here, \( \mathrm{BaCO_3} \) dissolves as it turns into soluble barium chloride while releasing \( \mathrm{CO_2} \) as a byproduct.

The release of carbon dioxide as a gas visibly evidences the chemical change occurring. This reaction shows why barium carbonate can dissolve in strong acids; the formation of \( \mathrm{BaCl_2} \) and subsequent liberation of carbon dioxide enable \( \mathrm{BaCO_3} \) to become part of the solution effectively. Such reactions are essential for processes needing the transformation of insoluble carbonates into their soluble counterparts.