Question

Describe the solubility of \(\mathrm{CaCO}_{3}\) in each of the following solutions compared to its solubility in water: (a) in \(0.10 \mathrm{M} \mathrm{NaCl}\) solution; \((\mathrm{b})\) in \(0.10 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) solution; (c) \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\); (d) \(0.10 \mathrm{M}\) HCl solution. (Answer same, less soluble, or more soluble.)

Short answer

The solubility of CaCO3 in the given solutions compared to water is as follows: (a) same in 0.10 M NaCl solution, (b) less soluble in 0.10 M Ca(NO3)2 solution, (c) less soluble in 0.10 M Na2CO3 solution, and (d) more soluble in 0.10 M HCl solution.

Step-by-step solution

(a) Solubility in 0.10 M NaCl solution:

In this case, both sodium (Na+) and chloride (Cl-) ions will be present in the solution. However, neither of these ions reacts with CaCO3. The solubility of CaCO3 in this case will remain unaffected, as the common ion effect doesn't come into play. Therefore, the solubility of CaCO3 in 0.10 M NaCl solution will be the same as in water.

(b) Solubility in 0.10 M Ca(NO3)2 solution:

In this case, the calcium (Ca2+) ions are present in the solution. The presence of calcium ions will cause a decrease in the solubility of CaCO3 due to the common ion effect. As the concentration of Ca2+ ions increases, the solubility product constant (Ksp) is reached more quickly, and less CaCO3 will dissolve. Hence, the solubility of CaCO3 in 0.10 M Ca(NO3)2 solution will be less soluble compared to its solubility in water.

(c) Solubility in 0.10 M Na2CO3 solution:

In this case, the carbonate (CO3 2-) ions are present in the solution. Similar to the previous case, the presence of carbonate ions will cause a decrease in the solubility of CaCO3 due to the common ion effect. As the concentration of CO3 2- ions increases, the solubility product constant (Ksp) is reached more quickly, and less CaCO3 will dissolve. Therefore, the solubility of CaCO3 in 0.10 M Na2CO3 solution will be less soluble compared to its solubility in water.

(d) Solubility in 0.10 M HCl solution:

In this case, the hydrogen ions (H+) present in the HCl solution will react with the carbonate ions (CO3 2-) from CaCO3, forming carbonic acid (H2CO3) and then forming water (H2O) and carbon dioxide (CO2). The reaction is as follows: \( CaCO_{3}(s) + 2H^{+}(aq) \rightarrow Ca^{2+}(aq) + H_{2}O(l) + CO_{2}(g) \) Since carbon dioxide escapes as a gas, the precipitation reaction is driven to the right, and more CaCO3 will dissolve in the acidic solution. Hence, the solubility of CaCO3 in 0.10 M HCl solution will be more soluble compared to its solubility in water.

Key concepts

Common Ion Effect

The common ion effect is an important concept in chemistry that impacts the solubility of compounds. It occurs when a compound is dissolved in a solution that already contains one of the ions from the compound itself. This results in decreased solubility. For instance, if you have a solution of \(CaCO_3\), and then add more calcium ions (Ca^2+) or carbonate ions (CO_3^{2-}), the solubility of \(CaCO_3\) will reduce. This happens because the excess ions push the equilibrium towards the formation of the solid, undissolved \(CaCO_3\).
The common ion effect is a great illustration of Le Chatelier's Principle, which states that the addition of a common ion drives the equilibrium to reduce the changes imposed on it. Understanding this can help in predicting and controlling the solubility of compounds in various chemical processes.

Solubility Product Constant

The solubility product constant, often denoted as \(K_{sp}\), is a value that indicates the extent to which a compound can dissolve in water. It's particularly useful for sparingly soluble salts.\[CaCO_3\] is one such salt, with a low \(K_{sp}\) value, meaning it doesn't dissolve very readily in water.
The \(K_{sp}\) expression for \(CaCO_3\) is written as:
\[K_{sp} = [Ca^{2+}][CO_3^{2-}]\]
Here, \([Ca^{2+}]\) and \([CO_3^{2-}]\) are the molar concentrations of the ions in a saturated solution. If you introduce additional ions that are already present in the system, the equilibrium is reached earlier with less \(CaCO_3\) dissolving. Conversely, if you remove ions through a subsequent reaction, for instance in an acidic solution, more \(CaCO_3\) dissolves.

Calcium Carbonate

Calcium carbonate ( CaCO_3) is a compound that is only sparingly soluble in water, meaning that only a small amount can dissolve at a given temperature. It is commonly found in rocks as mineral calcite and aragonite and in various organic forms in pearls and the shells of marine organisms.
In solutions, it can break down into calcium ions ( Ca^2+) and carbonate ions ( CO_3^{2-}). The solubility of CaCO_3 can be affected by changes in the ionic environment of the solution. Under normal conditions, no extra ions are present to change solubility significantly. However, changes like the addition of a common ion or adjustment of pH can notably change the amount of CaCO_3 that dissolves.

Acidic Solution

When CaCO_3 is dissolved in an acidic solution, for example, 0.10 M HCl, the solubility increases. This is because the hydrogen ions (H^+) in the acidic solution react with the carbonate ions (CO_3^{2-}) from CaCO_3. This reaction converts CO_3^{2-} into carbonic acid (H_2CO_3), which further breaks down into water and carbon dioxide gas.
The reaction is as follows:
\[CaCO_{3}(s) + 2H^{+}(aq) \rightarrow Ca^{2+}(aq) + H_{2}O(l) + CO_{2}(g)\]
The carbon dioxide escapes as a gas, driving the equilibrium towards the production of more dissolved Ca^{2+} ions in response. This results in more CaCO_3 dissolving as the equilibrium of the dissolution process shifts, enhancing solubility in acidic solutions compared to pure water.