Question

The solubility of \(\mathrm{CaCO}_{3}\) is \(\mathrm{pH}\) dependent. (a) Calculate the molar solubility of \(\mathrm{CaCO}_{3}\left(K_{a p}=4.5 \times 10^{-9}\right)\) 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 (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)? (e) If the pH is buffered at \(7.5\), what is the molar solubility of \(\mathrm{CaCO}_{3}\) ?

Short answer

The molar solubility of CaCO3 is (a) \(6.71 \times 10^{-5} M\), neglecting the acid-base character of the carbonate ion. (b) The equilibrium constant for the reaction involving the carbonate ion and water is \(2.22 \times 10^{-8}\). (c) The molar solubility of CaCO3 considering the equilibrium expression is approximately \(3.07 \times 10^{-6} M\). (d) To find the molar solubility of CaCO3 at pH = 8.3, calculate pOH, find the [OH-] concentration, plug it into the Kb expression, and solve for y (HCO3^- ion concentration). (e) Similarly, for a buffered pH of 7.5, follow the same approach as in part (d) to determine the molar solubility of CaCO3.

Step-by-step solution

Part (a): Molar solubility of CaCO3 without considering acid-base character of carbonate ion

We have CaCO3 as a solid, and its dissociation can be represented by the following equation: \[CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)\] To determine the molar solubility of CaCO3 without considering the acid-base character of the carbonate ion, we can use the solubility product constant (Ksp) expression like: \[K_{sp} = [Ca^{2+}][CO_3^{2-}]\] Given the Ksp value as \(4.5 \times 10^{-9}\), let x be the molarity of the dissociated ions; thus, \[K_{sp} = (x)(x) = x^2\] Solving for x, we get \[x = \sqrt{K_{sp}} = \sqrt{4.5 \times 10^{-9}}\] \[x = 6.71 \times 10^{-5} M\] The molar solubility of CaCO3, neglecting the acid-base character of the carbonate ion, is \(6.71 \times 10^{-5} M\).

Part (b): Determining the equilibrium constant

Now, we consider the acid-base character of the carbonate ion and its reaction with water: \[CO_3^{2-} (aq) + H_2O (l) \rightleftharpoons HCO_3^{-} (aq) + OH^{-} (aq)\] The equilibrium expression for this reaction is given by: \[K_b = \frac{[HCO_3^{-}][OH^{-}]}{[CO_3^{2-}]}\] Since the question doesn't provide the Kb value, we'll have to determine it using the relationship between Ka, Kb, and Kw: \[K_w = K_a \times K_b\] Considering the reaction of HCO3^- with water: \[HCO_3^{-}(aq) + H_2O (l) \rightleftharpoons H_2CO_3(aq) + OH^{-}(aq)\] We're given Ka value for H2CO3 as 4.5 x 10^-7. The Kw value (the ion product constant for water) is 1 x 10^-14 at 25°C. Now, we can find Kb value: \[K_b = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{4.5 \times 10^{-7}}\] \[K_b = 2.22 \times 10^{-8}\] The equilibrium constant for the reaction involving the carbonate ion and water is \(2.22 \times 10^{-8}\).

Part (c): Molar solubility of CaCO3 using the equilibrium expression

With the assumption that the only sources of Ca^2+, HCO3^-, and OH^- ions are from the dissolution of CaCO3, we can use the equilibrium expression to calculate the molar solubility of CaCO3. Let y represent the molarity of the dissociated ions of CO3^2-, which are the only source of HCO3^- and OH^-. The Kb expression can be written as: \[K_b = \frac{(y)(y)}{([CO_3^{2-}] - y)}\] Now, substituting the Kb value and solving for y, we get: \[2.22 \times 10^{-8} = \frac{y^2}{([CO_3^{2-}] - y)}\] As the amount of dissociated ions y is very small compared to [CO3^2-], we can approximate this as: \[y \approx \sqrt{[CO_3^{2-}] \times K_b}\] Substitute the values and solve for y: \[y \approx \sqrt{(6.71 \times 10^{-5})(2.22 \times 10^{-8})}\] \[y \approx 3.07 \times 10^{-6} M\] The molar solubility of CaCO3, considering the equilibrium expression, is approximately \(3.07 \times 10^{-6} M\).

Part (d): Molar solubility of CaCO3 at pH = 8.3

At pH = 8.3, we need to calculate the molar solubility of CaCO3. The relationship between pH and pOH could be used: \[pH + pOH = 14\] For a pH of 8.3, we get the pOH value: \[pOH = 14 - 8.3 = 5.7\] This means OH^- ion concentration is: \[ [OH^-] = 10^{-pOH} = 10^{-5.7} \] Since the only source of OH- ions is from the dissolution of CaCO3, we can write the Kb expression using this [OH-] value: \[K_b = \frac{[HCO_3^{-}][OH^{-}]}{[CO_3^{2-}]}\] \[2.22 \times 10^{-8} = \frac{(y)(10^{-5.7})}{(6.71 \times 10^{-5} - y)}\] Now, we can solve for y (HCO3^- ion concentration) and then get the molar solubility of CaCO3 considering the pH = 8.3.

Part (e): Molar solubility of CaCO3 at buffered pH = 7.5

Similarly, for a buffered pH of 7.5, we can determine the molar solubility of CaCO3 using the same approach as in part (d). Calculate the pOH value, find the [OH-] concentration, plug it into the Kb expression, and solve for y (HCO3^- ion concentration). Then, we can find the molar solubility of CaCO3 at pH = 7.5.

Key concepts

Solubility Product Constant

Understanding the solubility product constant, or Ksp, is crucial when discussing the solubility of ionic compounds in water. The solubility product constant is an equilibrium constant that describes the degree to which a solid dissolves to form a saturated solution. For a general salt, AB, that dissolves as A+ and B- ions in water, the Ksp expression is written as:

Ksp = [A+][B-]

Here, [A+] and [B-] represent the molar concentrations of the dissolved ions in a saturated solution. It is important to note that Ksp is affected by temperature, but not by the presence of other ions in the solution, unless they affect the overall ionic strength of the solution.

For the solubility of calcium carbonate (CaCO3), in part (a) of the exercise we consider its dissociation into calcium ions (Ca2+) and carbonate ions (CO32-). By assuming equimolar dissolution, the Ksp expression becomes as follows:

Ksp = [Ca2+][CO32-] = x^2

By solving for x, we calculate the molar solubility of CaCO3. The value of x gives us the concentration of each ion in the saturated solution, which translates to the solubility of the compound. Thus, a lower Ksp indicates a less soluble compound, while a higher Ksp suggests greater solubility in water.

Acid-Base Equilibrium

When dealing with solubility in varying pH conditions, it is essential to consider acid-base equilibria. An acid-base equilibrium refers to the balance that occurs in a solution between acid species (proton donors) and base species (proton acceptors). In this case, the carbonate ion (CO32-) acts as a base when it reacts with water to form bicarbonate (HCO3-) and hydroxide ions (OH-).

In part (b) of the original exercise, we write the acid-base equilibrium expression for the reaction of CO32- in water as follows:

CO32- (aq) + H2O (l) ⇌ HCO3- (aq) + OH- (aq)

Using the equilibrium constant, Kb, the relationship can be expressed mathematically as:

Kb = [HCO3-][OH-] / [CO32-]

In this scenario, pH affects the position of equilibrium. A higher pH indicates a greater concentration of OH- ions which drives the equilibrium towards the right, increasing the production of HCO3- ions and thus affecting the overall solubility of CaCO3. The concept of acid-base equilibrium is crucial because it helps explain why the solubility of a salt might change with pH, as seen in parts (d) and (e) of the exercise, which discuss the effect of ocean pH and a buffered pH on the solubility of CaCO3.

Chemical Equilibrium

Chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the reverse reaction, so the concentrations of the reactants and products remain constant over time. It is an essential concept in understanding reactions like the dissolution of CaCO3, which can reach a dynamic equilibrium in a saturated solution.

For the solubility of CaCO3, equilibrium is established between the solid CaCO3 and its ions in solution. According to the principle of Le Chatelier, if the external conditions, such as pH in our case, are changed, the position of equilibrium will shift to counteract the change. This concept helps us understand why the molar solubility of CaCO3 varies with the pH of the solution, as seen in part (c).

The equilibrium expressions for both Ksp and Kb in the given exercise are applied to calculate the molar solubility of CaCO3 under different conditions. The relationship between solubility and equilibrium is significant because it allows us to predict how the solubility will change in response to environmental factors and how to manipulate these factors to achieve desired solubility levels.