Question

The solubility of \(\mathrm{CaCO}_{3}\) is pH dependent. (a) Calculate the molar solubility of \(\mathrm{CaCO}_{3}\left(K_{s p}=4.5 \times 10^{-9}\right)\) neglecting the acid-base character of the carbonate ion. (b) Use the \(K_{b}\) expression for the \(\mathrm{CO}_{3}^{2-}\) ion to determine the equilibrium constant for the reaction $$ \mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons $$ (c) If we assume that the only sources of \(\mathrm{Ca}^{2+}, \mathrm{HCO}_{3}^{-},\) and \(\mathrm{OH}^{-}\) ions are from the dissolution of \(\mathrm{CaCO}_{3},\) what is the molar solubility of \(\mathrm{CaCO}_{3}\) using the equilibrium expression from part (b)? (d) What is the molar solubility of \(\mathrm{CaCO}_{3}\) at the \(\mathrm{pH}\) of the ocean (8.3)\(?(\mathbf{e})\) If the \(\mathrm{pH}\) is buffered at \(7.5,\) what is the molar solubility of \(\mathrm{CaCO}_{3} ?\)

Short answer

The molar solubility of \(\mathrm{CaCO}_3\) neglecting the acid-base character of the carbonate ion is calculated using the given solubility product (\(K_{sp} = 4.5 \times 10^{-9}\)) and found to be \(x = 2.12 \times 10^{-5}\) M. When considering the acid-base character of the carbonate ion, the molar solubility of \(\mathrm{CaCO}_3\) is found using the Kb expression, and the specific pH values are considered. At a pH of 8.3 (ocean pH), the molar solubility of \(\mathrm{CaCO}_3\) is found to be higher than at a pH of 7.5 (buffered pH).

Step-by-step solution

Calculate the molar solubility neglecting the acid-base character of the carbonate ion

We are given the solubility product, \(K_{sp}\), for the dissolution of \(\mathrm{CaCO}_3\): $$\mathrm{CaCO_3} \leftrightarrow \mathrm{Ca^{2+}}+\mathrm{CO_3^{2-}}$$ and \(K_{sp} = 4.5\times10^{-9}\). Let the molar solubility of \(\mathrm{CaCO}_3\) be \(x\). Then, at equilibrium, the concentrations of \(\mathrm{Ca^{2+}}\) and \(\mathrm{CO_3^{2-}}\) would be \(x\) and \(x\) respectively. We can write the \(K_{sp}\) expression as $$K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{CO_3^{2-}}] = x^2$$ Solving for \(x\), we get the molar solubility.

Use the Kb expression for the CO3^2- ion to determine the equilibrium constant for the reaction

The Kb expression for the \(\mathrm{CO_3^{2-}}\) ion is given by: $$\mathrm{CO_3^{2-}}+\mathrm{H_2O} \rightleftharpoons \mathrm{HCO_3^-}+\mathrm{OH^-}$$ The equilibrium constant expression for the reaction of \(\mathrm{CaCO}_3\) with water can be represented as: $$K = \frac{[\mathrm{HCO_3^-}][\mathrm{OH^-}]}{[\mathrm{CO_3^{2-}}]}$$ We can replace \([\mathrm{HCO_3^-}]\) and \([\mathrm{OH^-}]\) with \(x\) since the molar solubility of \(\mathrm{CaCO}_3\) is assumed to be the only source of these ions. Since the concentration of the water is considered to be constant, we can include it in the equilibrium constant. Thus, the new equilibrium constant can be found.

Calculate the molar solubility of CaCO3 using the equilibrium expression from part (b)

Using the new equilibrium constant, we can write the equilibrium expression for the reaction involving \(\mathrm{CaCO}_3\) and water: $$K = \frac{x^2}{[\mathrm{CO_3^{2-}}]}$$ We can once again replace \([\mathrm{CO_3^{2-}}]\) with \(x\) since the molar solubility of \(\mathrm{CaCO}_3\) is assumed to be the only source of carbonate ions. Solving this equation for \(x\) will give us the molar solubility of \(\mathrm{CaCO}_3\) under these conditions.

Calculate the molar solubility of CaCO3 at the pH of the ocean (8.3)

To find the molar solubility of \(\mathrm{CaCO}_3\) at the pH of the ocean (8.3), we first need to calculate the concentration of \(\mathrm{CO_3^{2-}}\) at this pH. Using the relationship between pH, pOH, and pKw: $$\mathrm{pH}+\mathrm{pOH}=\mathrm{pK_w} \Rightarrow \mathrm{pOH}=\mathrm{pK_w}-\mathrm{pH}$$ We can find the concentration of \(\mathrm{OH^-}\) ions at pH 8.3 using pOH and then find the concentration of \(\mathrm{CO_3^{2-}}\) using the Kb expression for the \(\mathrm{CO_3^{2-}}\) ion. Finally, using the equilibrium expression derived in Step 3, we can find the molar solubility of \(\mathrm{CaCO}_3\) at this pH.

Calculate the molar solubility of CaCO3 at pH 7.5

Similar to Step 4, we need to find the concentration of \(\mathrm{CO_3^{2-}}\) ions at pH 7.5. Using the relationship between pH, pOH, and pKw, we can find the concentration of \(\mathrm{OH^-}\) ions at pH 7.5. Then, using the Kb expression for the \(\mathrm{CO_3^{2-}}\) ion, we can find the concentration of \(\mathrm{CO_3^{2-}}\). Finally, using the equilibrium expression derived in Step 3, we can find the molar solubility of \(\mathrm{CaCO}_3\) at pH 7.5.

Key concepts

Solubility Product (Ksp)

Understanding solubility product helps us unravel why certain compounds dissolve in solutions and to what extent. The solubility product, often represented as \(K_{sp}\), is an equilibrium constant that underscores the solubility of sparingly soluble salts like calcium carbonate \(\mathrm{CaCO}_3\). When \(\mathrm{CaCO}_3\) dissolves, it dissociates into calcium ions \(\mathrm{Ca^{2+}}\) and carbonate ions \(\mathrm{CO_3^{2-}}\). The equation for this is simple:

• \(\mathrm{CaCO_3} \leftrightarrow \mathrm{Ca^{2+}} + \mathrm{CO_3^{2-}}\)

The solubility product is expressed as:

• \(K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{CO_3^{2-}}]\)

If we know the \(K_{sp}\) value, we can find the molar solubility, represented as \(x\). This number tells us how much \(\mathrm{CaCO}_3\) can dissolve in water under equilibrium conditions. It's useful in predicting the behavior of salts in various environments and conditions.

Equilibrium Constant (Kb)

The equilibrium constant \(K_b\) in this context is crucial for understanding the acid-base relationship of the carbonate ion \(\mathrm{CO_3^{2-}}\) in water. When carbonate ions dissolve, they can react as bases, meaning they accept protons from water, creating bicarbonate \(\mathrm{HCO_3^-}\) and hydroxide ions \(\mathrm{OH^-}\):

• \(\mathrm{CO_3^{2-}} + \mathrm{H_2O} \rightleftharpoons \mathrm{HCO_3^-} + \mathrm{OH^-} \)

The expression for this reaction is given as:

• \(K_b = \frac{[\mathrm{HCO_3^-}][\mathrm{OH^-}]}{[\mathrm{CO_3^{2-}}]}\)

Understanding \(K_b\) allows us to calculate how changes in solubility are influenced by carbonate's interaction with water—a pivotal concept for chemical equilibria involving salts, bases, and weak acids. This helps us understand precipitation and dissolution in natural water systems.

pH and Solubility

The relationship between pH and solubility plays a significant role in determining how much of a compound like \(\mathrm{CaCO}_3\) will dissolve in a solution. Since pH affects the concentration of \(\mathrm{H^+}\) ions, it consequently influences the balance between different carbonate species in the water. At higher pH levels (more basic conditions), there are more \(\mathrm{OH^-}\) ions, which can affect the dissolution process. To calculate the molar solubility of \(\mathrm{CaCO}_3\) at specific pH levels, we first determine the \(\mathrm{OH^-}\) concentration using the relationship:

• \(\mathrm{pH} + \mathrm{pOH} = \mathrm{pK_w}\)

For example, if the pH of ocean water is 8.3, we can derive the \(\mathrm{OH^-}\) concentration, influencing the amount of \(\mathrm{CO_3^{2-}}\) present. This change affects \(\mathrm{CaCO}_3\)'s solubility, showcasing how marine chemistry adapts to varying hydrogen ion concentrations.

Ocean Chemistry

Ocean chemistry intricately connects with the solubility of calcium carbonate, influencing marine life's habitat and processes like carbon cycling. The oceans’ average pH is about 8.1, slightly alkaline, which sets a foundation for the availability and distribution of different carbonate species.In seawater, calcium carbonate exists in equilibrium with other forms of carbonates, like bicarbonates, determined partly by \(\text{pH}\). When oceans absorb carbon dioxide \(\text{CO}_2\), they undergo acidification, lowering the pH and altering carbonate chemistry. This change can cause a decrease in \(\mathrm{CO_3^{2-}}\) ions and possibly affect the solubility balance of \(\mathrm{CaCO}_3\), impacting the ecosystem.Researching this balance aids in understanding ecological impacts, from altering marine organisms' ability to form shells and skeletons to broader effects on the global carbon cycle. Careful monitoring of these dynamics helps predict responses to environmental changes.