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 \(\mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q)\) (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 preceding expression? What is the \(\mathrm{pH} ?\) (d) If the \(\mathrm{pH}\) is buffered at 8.2 (as is historically typical for the ocean), what is the molar solubility of \(\mathrm{CaCO}_{3} ?\) (e) If the \(\mathrm{pH}\) is buffered at \(7.5,\) what is the molar solubility of \(\mathrm{CaCO}_{3} ?\) How much does this drop in \(\mathrm{pH}\) increase solubility? solution remains \(0.50 \mathrm{~L},\) calculate the \(\mathrm{pH}\) of the resulting solution.

Short answer

The molar solubility of \(\mathrm{CaCO}_{3}\) is approximately \(6.7 \times 10^{-5} \mathrm{M}\) and its solubility does not change significantly due to pH change from 8.2 (historically typical for the ocean) to 7.5. The pH of the system is approximately 9.17.

Step-by-step solution

Expression for the solubility product constant

The dissociation reaction for CaCO3 is given by: $$\mathrm{CaCO_{3}(s) \rightleftharpoons Ca^{2+}(aq) + CO_{3}^{2-}(aq)}$$ The expression for the solubility product constant, \(K_{sp}\), is: $$K_{sp} = [\mathrm{Ca^{2+}}] [\mathrm{CO_{3}^{2-}}]$$ Given \(K_{sp} = 4.5 \times 10^{-9}\). Let \(s\) be the molar solubility of \(\mathrm{CaCO_{3}}\), then \([\mathrm{Ca}^{2+}] = s\) and \([\mathrm{CO}_{3}^{2-}] = s\). Substitute these into the \(K_{sp}\) expression.

Solve for s

The equation becomes: $$4.5 \times 10^{-9} = s^2$$ Solve for \(s\): $$s = \sqrt{4.5 \times 10^{-9}}$$ $$s \approx 6.7 \times 10^{-5} \mathrm{M}$$ The molar solubility of \(\mathrm{CaCO}_{3}\) is approximately \(6.7 \times 10^{-5} \mathrm{M}\). #2. Determine the equilibrium constant for the reaction involving the dissolution of CaCO3 including the acid-base character of the carbonate ion#

Use Kb expression for CO3^2- ion

For the dissociation of \(\mathrm{CO}_{3}^{2-}\), the base hydrolysis equation is: $$\mathrm{CO_{3}^{2-} + H_2 O \rightleftharpoons HCO_{3}^{-} + OH^{-}}$$ If we write the expression for the \(\mathrm{CaCO}_{3}\) dissolution including the hydrolysis as: $$\mathrm{CaCO_{3}(s) + H_2 O(l) \rightleftharpoons Ca^{2+}(aq) + HCO_{3}^{-}(aq) + OH^{-}(aq)}$$ The base dissociation constant is given by \(K_{b} = [\mathrm{HCO}_{3}^{-}][\mathrm{OH}^{-}]/[\mathrm{CO}_{3}^{2-}]\). Let \(x\) be the equilibrium concentration of \(\mathrm{HCO}_{3}^{-}\), \(\mathrm{Ca}^{2+}\), and \(\mathrm{OH}^{-}\). Given that \([\mathrm{Ca}^{2+}] = [\mathrm{HCO}_{3}^{-}]\), we can write the product of the hydrolysis reaction as \([\mathrm{Ca}^{2+}(aq)][\mathrm{OH}^{-}(aq)] \). Thus, the equilibrium constant \(K\) is given by: $$K = \frac{[\mathrm{Ca}^{2+}][\mathrm{HCO}_{3}^{-}][\mathrm{OH}^{-}]}{[\mathrm{CaCO_{3}}]}= \frac{x^3}{[\mathrm{CaCO_{3}}]}$$ We can also express concentration of \(\mathrm{CaCO_{3}}\) in terms of \(x\). Since the molar solubility is approximately \(6.7 \times 10^{-5} \mathrm{M}\), we can write \([\mathrm{CaCO}_{3}] = (6.7 \times 10^{-5} - x)\) : $$K = \frac{x^3}{(6.7 \times 10^{-5} - x)}$$ From the previous part, we found \(K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{CO}_{3}^{2-}]\), and re-writing it in terms of \(x\), we get \(K_{sp} = x(6.7 \times 10^{-5} - x)\). Now, we can write \(K_{b}\) in terms of \(K_{sp}\): $$K = \frac{K_{sp}}{[\mathrm{CO}_{3}^{2-}]} = \frac{x(6.7 \times 10^{-5} - x)}{6.7 \times 10^{-5} - x}$$ \(K = x\) Now with the equilibrium constant \(K\) we can proceed to calculate the solubility of \(\mathrm{CaCO}_3\) including the acid-base equilibrium at different pH conditions. #3. Calculate the solubility of CaCO3 using the equilibrium constant and determine the pH#

Calculate solubility

Given \(K = x\), we can calculate the solubility of \(\mathrm{CaCO}_3\) by solving the equation \(K = \frac{x^3}{(6.7 \times 10^{-5} - x)}\). For the moment, we are asked to find the solubility in a case where the only sources of \(\mathrm{Ca}^{2+},\) \(\mathrm{HCO}_{3}^{-},\) and \(\mathrm{OH}^{-}\) ions are from the dissolution of \(\mathrm{CaCO}_{3}\). Under such conditions and considering the dissociation constants found before, this results in an insoluble substance, and the molar solubility will remain the same as in the first part, approximately \(6.7 \times 10^{-5} \mathrm{M}\).

Calculate pH

We can calculate the pH using the relation: $$\mathrm{pH} = 14 - \mathrm{pOH} = 14 - \mathrm{(-log_{10}(x))}$$ Where x is the concentration of \(\mathrm{OH}^{-}\). Taking the log of the value previously found for \(x = 6.7 \times 10^{-5}\): $$\mathrm{pH} \approx 14 - (- \log_{10}(6.7 \times 10^{-5}))$$ $$\mathrm{pH} \approx 9.17$$ The pH of the system is approximately 9.17. #4. Calculate the solubility at a pH of 8.2 (historically typical for the ocean) and at a pH of 7.5#

Calculate solubility at pH 8.2

At a pH of 8.2, we can determine the pOH: $$\mathrm{pOH} = 14 - \mathrm{pH} = 14 - 8.2$$ $$\mathrm{pOH} \approx 5.8$$ Next, we obtain the concentration of \(\mathrm{OH}^{-}\) ions: $$[\mathrm{OH}^{-}] = 10^{-\mathrm{pOH}} \approx 10^{-5.8}$$ The molar solubility of \(\mathrm{CaCO}_3\) at a pH of 8.2 remains approximately the same at \(6.7 \times 10^{-5} \mathrm{M}\). This is because the pH change does not have a significant effect on the solubility given the insoluble nature of calcium carbonate.

Calculate solubility at pH 7.5

At a pH of 7.5, we can again determine the pOH: $$\mathrm{pOH} = 14 - \mathrm{pH} = 14 - 7.5$$ $$\mathrm{pOH} \approx 6.5$$ Next, we obtain the concentration of \(\mathrm{OH}^{-}\) ions: $$[\mathrm{OH}^{-}] = 10^{-\mathrm{pOH}} \approx 10^{-6.5}$$ The molar solubility of \(\mathrm{CaCO}_3\) at a pH of 7.5 remains approximately the same at \(6.7 \times 10^{-5} \mathrm{M}\).

Increased solubility due to pH drop

To find the increase in solubility due to the pH drop, we can compare the molar solubility of \(\mathrm{CaCO}_3\) at pH 8.2 and pH 7.5: The difference in solubility is negligible, and the solubility does not change significantly due to pH change. This is because the pH drop is not enough to cause a pronounced change in the solubility of calcium carbonate.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, represented as Ksp, is crucial for understanding the solubility behavior of sparingly soluble salts like calcium carbonate (CaCO3). It is an equilibrium constant that represents the level at which a solid solute dissolves in solution to form its constituent ions.

The general formula for Ksp, using CaCO3 as an example, is given by: $$K_{sp} = [\mathrm{Ca^{2+}}] [\mathrm{CO_{3}^{2-}}]$$
In which [Ca²⁺] and [CO₃^2-] are the molar concentrations of the ions when the system reaches equilibrium.

To calculate the molar solubility, denoted as s for CaCO3, we set up the equation based on the stoichiometry of the dissociation, where one mole of CaCO3 dissociates into one mole of Ca²⁺ and one mole of CO₃^2-. This leads to the equilibrium expression of Ksp and allows us to solve for s, which is the molar solubility of the compound. Typically, this process involves setting up an equation such as: $$K_{sp} = s^2$$
And solving for s by taking the square root of the Ksp value. Knowing the molar solubility of a substance helps predict how much of the substance can dissolve in a given volume of solvent at equilibrium.

Base Dissociation Constant (Kb)

The base dissociation constant, Kb, measures the strength of a base in solution. It is particularly important in determining the pH-dependent solubility of substances like CaCO3 when the base component, in this case the carbonate ion (CO₃^2-), is capable of accepting a proton (H+).

The equation for the base dissociation reaction of the carbonate ion in water is: $$\mathrm{CO_{3}^{2-} + H_2 O \rightleftharpoons HCO_{3}^{-} + OH^{-}}$$
The base dissociation constant is then represented by the following relation: $$K_{b} = \frac{[\mathrm{HCO}_{3}^{-}][\mathrm{OH}^{-}]}{[\mathrm{CO}_{3}^{2-}]}$$
This constant is essential for understanding how the concentration of hydroxide ions (OH^-) and its conjugate base (HCO₃^-) are related to the carbonate ion concentration in a base hydrolysis reaction. Understanding the base dissociation constant can help us predict the shift in equilibria and solubility under various pH conditions, as well as in calculating the exact equilibrium constant when additional reactions, like hydrolysis, are considered.

pH Effect on Solubility

pH plays a significant role in the solubility of compounds like CaCO3, especially when the compound includes a weak acid or base group that can undergo protonation or deprotonation. In the case of CaCO3, the carbonate ion (CO₃^2-) can react with water to form bicarbonate (HCO₃^-) and hydroxide ions (OH^-), altering the solubility.

The presence of H+ or OH- ions can shift the equilibrium due to the common ion effect or through the formation of additional species. As a result, a change in pH can lead to increased solubility due to the change in concentration of the constituent ions.

For example, when the pH of the solution decreases (becomes more acidic), the concentration of H+ increases, which can drive the reaction towards the formation of bicarbonate and dissolve more CaCO3. Conversely, an increase in pH (becoming more basic) could lead to a decreased solubility due to the formation of hydroxide ions from the base dissociation of carbonate ions. Therefore, understanding the effect of pH on solubility is necessary for accurate predictions and calculations involving chemical equilibria in different pH conditions.

Equilibrium Constant Calculation

Calculating an equilibrium constant allows us to understand the extent of a reaction and predict the concentrations of reactants and products at equilibrium. This is especially important in solubility equilibria where multiple reactions, such as dissociation and hydrolysis, may occur simultaneously.

To determine the effective equilibrium constant for a reaction, we use the concentrations of the ions at equilibrium. For a given solubility equilibrium, the process typically involves setting up an ice table (Initial concentration, Change in concentration, Equilibrium concentration) to track the changes and reach the equilibrium concentrations of all species involved.

The equilibrium constant can be affected by changes in pH, temperature, and the presence of common ions. Understanding how to manipulate and calculate the equilibrium constant, particularly in the context of the solubility of substances, is key for correctly calculating the solubility in various environments and understanding how different factors play a role in the dissolution process.