Problem 99 - Chemistry The Central Science

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 equilibrium expression from part (b)? \((\boldsymbol{d} )\)What is the molar solubility of \(\mathrm{CaCO}_{3}\) at the pH of the ocean \((8.3) ?(\mathbf{e})\) If the \(\mathrm{pH}\) is buffered at \(7.5,\) what is the molar solubility of \(\mathrm{CaCO}_{3} ?\)

Step-by-Step Solution

Answer

The short version of the answer is: (a) The molar solubility of CaCO₃ neglecting the acid-base character of the carbonate ion is \(2.12 \times 10^{-5}\) M. (c) The molar solubility using the equilibrium expression from part (b) is \(x = \sqrt[3]{(K_{sp} \times K_b)}\) where x is the concentration of Ca²⁺, HCO₃⁻, and OH⁻ ions. For parts (d) and (e), we need to calculate the molar solubility of CaCO₃ at given pH values (8.3 for the ocean and 7.5 for the buffered solution) by repeating the steps for calculating the OH⁻ concentration and using it in the equilibrium expression from part (c).

Step 1: Write the solubility equilibrium expression

For the dissolution of CaCO₃, we can write the solubility equilibrium expression as: \[K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{CO}_{3}^{2-}]\] Here, the given value of \(K_{sp} = 4.5 \times 10^{-9}\).

Step 2: Determine the molar solubility

Let 'x' represent the molar solubility of CaCO3. Since CaCO3 dissolves into Ca²⁺ and CO₃²⁻ ions in a 1:1 ratio, both the concentrations of Ca²⁺ and CO₃²⁻ ions will be 'x'. Using the solubility equilibrium expression, we get: \(K_{sp} = (x)(x)\) Solve for x: \[\begin{aligned} x &= \sqrt{K_{sp}} \\ x &= \sqrt{4.5 \times 10^{-9}} \\ x &= 2.12 \times 10^{-5} \, \text{M} \end{aligned}\] Hence, the molar solubility of CaCO3 neglecting the acid-base character of the carbonate ion is \(2.12 \times 10^{-5} \, \text{M}\). (b) Equilibrium constant for the given reaction

Step 1: Write the given reaction

The given reaction is: \[\mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q)\]

Step 2: Determine the equilibrium constant

We are given that CO₃²⁻ ion has a \(K_{b}\) expression. Thus, the reaction can be broken down into 2 steps: (1) \(\mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q)\) with an equilibrium constant of \(K_b\) (2) \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q)\) with an equilibrium constant of \(K_{sp}\) Since both reactions add up to the given reaction, the equilibrium constant for the given reaction is the product of the individual equilibrium constants, i.e., \(K_{sp} \times K_b\). (c) Molar solubility using equilibrium expression from part (b)

Step 1: Write the equilibrium expression

Using the given reaction and the equilibrium constant as \(K_{sp} \times K_b\), we can write the equilibrium expression as: \[(K_{sp} \times K_b) = [\mathrm{Ca}^{2+}][\mathrm{HCO}_{3}^{-}][\mathrm{OH}^{-}]\]

Step 2: Determine the molar solubility

Assuming the only sources of Ca²⁺, HCO₃⁻, and OH⁻ ions are from the dissolution of CaCO3, we can let their concentrations be 'x', 'x', and 'x', respectively. Using the equilibrium expression, we get: \[(K_{sp} \times K_b) = (x)(x)(x) = x^3\] Solve for x (keeping in mind units if \(K_b\) is given or calculated): \[x = \sqrt[3]{(K_{sp} \times K_b)}\] This will give us the molar solubility of CaCO3 using the equilibrium expression from part (b). (d) Molar solubility at pH of the ocean

Step 1: Calculate pOH from pH

The pH of the ocean is given as 8.3. We can calculate the pOH as: \(pOH = 14 - pH = 14 - 8.3 = 5.7\).

Step 2: Calculate OH⁻ concentration

Calculate the OH⁻ concentration as: \([\mathrm{OH}^{-}] = 10^{-pOH} = 10^{-5.7}\] Use this concentration in the equilibrium constant expression from part (c) to determine the molar solubility at the pH of the ocean. (e) Molar solubility at pH buffered at 7.5 Repeat steps 1 and 2 from part (d) using the given pH of 7.5. Then, use the calculated OH⁻ concentration in the equilibrium constant expression from part (c) to determine the molar solubility of CaCO3 at a pH buffered at 7.5.

Key Concepts

Ksp (Solubility Product Constant)

The solubility product constant, known as Ksp, is a value that helps to determine how much of a substance, like calcium carbonate (CaCO₃), can dissolve in water before the solution becomes saturated. Ksp is specific for a particular compound and temperature. When dealing with sparingly soluble compounds like CaCO₃, it indicates the maximum concentration of its ions that can exist in solution.
Ksp is calculated using the concentrations of the ions present when a compound dissolves. For CaCO₃, which dissociates into Ca²⁺ and CO₃²⁻ ions in a 1:1 ratio, the Ksp expression is: \[ K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{CO}_{3}^{2-}]\] Solving for these concentrations provides information about the extent to which CaCO₃ can dissolve in its natural environment. This concept is crucial in predicting behaviors of various solutes in solutions, especially minerals like CaCO₃ in water environments.

Acid-Base Equilibria

Acid-base equilibria describe the balance between acids and bases in solution, and how various substances may act as either. Carbonate ions (CO₃²⁻), in particular, can act as a base in aqueous solutions. This is an important consideration when examining the solubility of CaCO₃ in water.
When CO₃²⁻ ions are in water, they can participate in the following equilibrium reaction:\[\mathrm{CO}_{3}^{2-}(aq) + \mathrm{H}_{2}O(l) \rightleftharpoons \mathrm{HCO}_{3}^{-}(aq) + \mathrm{OH}^{-}(aq)\]This behavior impacts the overall solubility of CaCO₃, as the presence of OH⁻ ions affects the pH of the solution, ultimately influencing the solubility depending on the mixture's acidity or basicity.
The Kb value corresponds to how readily a base reacts with water, making it a vital piece of the solubility puzzle for CaCO₃. Understanding this equilibrium helps chemists and environmental scientists predict how changes in pH affect the solubility of minerals.

Oceanic pH Solubility

The ocean's pH has significant implications for the solubility of calcium carbonate. The natural pH of ocean water is around 8.3, making it slightly basic. This pH level influences the concentration of hydroxide ions (OH⁻) and carbonate chemistry, which are vital in the process of calculating how much CaCO₃ can dissolve.
To predict the solubility of CaCO₃ in the ocean, taking the pH into account is essential. By knowing the pH (in this case, 8.3), you can calculate the pOH as:\[ pOH = 14 - pH = 5.7 \]Subsequently, obtaining the hydroxide ion concentration:\[[\mathrm{OH}^{-}] = 10^{-5.7}\]This OH⁻ concentration is used alongside the equilibrium constants derived from earlier steps to find the solubility of CaCO₃.Some areas of the ocean may experience pH changes due to various factors such as pollution or natural phenomena, altering mineral solubilities which can impact marine ecosystems. Understanding solubility in varying pH conditions is crucial for predicting the behavior of marine minerals.