Question

The equilibrium constant \(K_{\mathrm{a}}\) for the reaction $$\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)$$ is \(1.0 \times 10^{-5}\) a. Calculate the pH of a 0.10\(M\) solution of \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}\) b. Will a 1.0\(M\) solution of cobalt(Il) nitrate have a higher or lower pH than a 1.0\(M\) solution of cobalt (III) nitrate? Explain. c. \(\mathrm{Co}^{3+}\) complex ions are generally low-spin cases, whereas \(\mathrm{Co}^{2+}\) complex ions are generally high-spin cases. Explain. If this is the situation, how many unpaired electrons are present in \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) and \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+} ?\)

Short answer

a. The pH of a 0.10 M solution of Co[H2O]6Cl3 is approximately 3. b. A 1.0 M solution of cobalt(II) nitrate will have a higher pH than a 1.0 M solution of cobalt(III) nitrate, as Co^2+ is less acidic than Co^3+ due to its lower charge. c. Co[H2O]6^3+ is a low-spin complex with 0 unpaired electrons, while Co[H2O]6^2+ is a high-spin complex with 4 unpaired electrons.

Step-by-step solution

Write the chemical equilibrium expression for Ka

We are given the reaction: \[ \mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \] The equilibrium constant expression for Ka is: \[ K_\mathrm{a} = \frac{[\mathrm{Co}(\mathrm{H}_2 \mathrm{O})_5 (\mathrm{OH})^{2+}] [\mathrm{H}_3 \mathrm{O}^+]}{[\mathrm{Co}(\mathrm{H}_2 \mathrm{O})_6^{3+}]} \]

The initial and equilibrium concentrations of the species

Initially, we have a 0.10 M solution of Co[H2O]6^3+. Therefore, the initial concentrations are: [\(\mathrm{Co}(\mathrm{H}_2 \mathrm{O})_6^{3+}\)] = 0.10 M [\(\mathrm{Co}(\mathrm{H}_2 \mathrm{O})_5 (\mathrm{OH})^{2+}\)] = 0 [\(\mathrm{H}_3 \mathrm{O}^+\)] = 0 At equilibrium, the concentrations of the species will change as follows: [\(\mathrm{Co}(\mathrm{H}_2 \mathrm{O})_6^{3+}\)] = 0.10 - x [\(\mathrm{Co}(\mathrm{H}_2 \mathrm{O})_5 (\mathrm{OH})^{2+}\)] = x [\(\mathrm{H}_3 \mathrm{O}^+\)] = x

Substitute the equilibrium concentrations in the Ka expression and solve for x

Let x be the concentration of H3O+ ions at equilibrium. Now substitute the equilibrium concentrations in the Ka expression and solve for x: \[ 1.0 \times 10^{-5} = \frac{x^2}{0.10 - x} \] Assume that x is much smaller than 0.10, so we can approximate it as: \[ 1.0 \times 10^{-5} \approx \frac{x^2}{0.10} \] Solve for x: \[ x \approx \sqrt{(1.0 \times 10^{-5})(0.10)} \approx 1.0 \times 10^{-3} \]

Calculate the pH of the solution

The concentration of H3O+ ions is approximately 1.0 × 10^(-3) M. Using the definition of pH: \[ \mathrm{pH} = -\log([\mathrm{H}_3 \mathrm{O}^+]) \] Therefore, the pH of the solution is: \[ \mathrm{pH} \approx -\log(1.0 \times 10^{-3}) \approx 3 \] So, the pH of the 0.10 M solution of Co[H2O]6Cl3 is approximately 3. #b. Will a 1.0M solution of cobalt(II) nitrate have a higher or lower pH than a 1.0M solution of cobalt (III) nitrate? Explain# Cobalt(II) nitrate has the formula Co(NO3)2, and cobalt(III) nitrate has the formula Co(NO3)3. Co^2+ (in cobalt(II) nitrate) is less acidic than Co^3+ (in cobalt(III) nitrate) because it has a lower charge, and thus less ability to polarize the water molecules in the coordination sphere. This means that cobalt(II) nitrate will not be as strong an acid as cobalt(III) nitrate and will have a higher pH. #c. Explain the spin states of Co^3+ and Co^2+ complex ions and find the number of unpaired electrons in Co[H2O]6^3+ and Co[H2O]6^2+# The cobalt(III) complex ions, like Co[H2O]6^3+, are generally low-spin because of their stronger ligand field (due to the higher charge on the central cobalt ion). The stronger ligand field causes the energy splitting between the t2g and eg orbitals within the d orbital to be large, which leads to pairing of the electrons in the t2g orbitals. In Co[H2O]6^3+, cobalt has 6 electrons in its 3d orbitals (Co^3+ has d^6 configuration). Since it is low-spin, all the electrons will pair, leading to 0 unpaired electrons. The cobalt(II) complex ions, like Co[H2O]6^2+, are generally high-spin because of their weaker ligand field (due to the lower charge on the central cobalt ion). The weaker ligand field causes the energy splitting between the t2g and eg orbitals within the d orbital to be small, which leads to electrons occupying the eg orbitals first before pairing in the t2g orbitals. In Co[H2O]6^2+, cobalt has 7 electrons in its 3d orbitals (Co^2+ has d^7 configuration). In the high-spin state, it has three unpaired electrons in the t2g orbitals and one unpaired electron in one of the eg orbital; thus, it has a total of 4 unpaired electrons.

Key concepts

Equilibrium Constant

In the realm of chemical reactions, the equilibrium constant, denoted as \(K\), is a crucial factor. It defines the ratio of the concentration of products to reactants at equilibrium. For any given reaction, this is expressed as:\[K = \frac{\text{[concentration of products]}}{\text{[concentration of reactants]}}\]In our specific case, the reaction involves a transition of cobalt complexes in water, where the equilibrium constant \(K_a\) has a value of \(1.0 \times 10^{-5}\). This small value indicates that, at equilibrium, the reactants will be favored over the products. This means only a small amount of the cobalt complex will dissociate to form hydronium ions \([\text{H}_3\text{O}^+]\). This balance between products and reactants isn't just significant for academic purposes; it plays a key role in many processes, from industrial reactions to biological systems.Understanding equilibrium constants help in predicting the direction of the reaction and calculating concentrations at equilibrium. These constants depend on temperature, and any change can shift the balance of reactions.

pH Calculation

The concept of pH is central to chemistry, biology, and environmental science. pH measures the hydrogen ion concentration in a solution and provides insight into its acidity or basicity. It's calculated using the equation:\[pH = -\log([\text{H}_3\text{O}^+])\]In this exercise, the concentration of hydronium ions \([\text{H}_3\text{O}^+]\) is calculated to be \(1.0 \times 10^{-3} M\). Plugging this into our equation, we get a pH value of approximately 3, which indicates an acidic solution. For instance, a solution of \(0.10 M\) Co[H2O]6Cl3 has a similar acidic pH due to the formation of H3O+ ions from cobalt complexes.Knowing the pH helps us understand the nature of the solution and its potential reactions. In real-world scenarios, this understanding is applicable in contexts like agriculture (soil pH affects plant growth) and medicine (pH stability in the human body is essential for health).

Coordination Compounds

Coordination compounds are fascinating structures where central metal ions are surrounded by molecules or ions known as ligands. These compounds have diverse properties and applications, from catalysis to material science.In the provided exercise, we deal with cobalt coordination complexes. Cobalt complexes can exist in two oxidation states: Co(III) and Co(II). The difference in charge influences the coordination sphere:

• Co(III) Complex: Such as \([\text{Co}([\text{H}_2\text{O}])_6]^{3+}\), where \(Co^{3+}\) forms a stronger ligand field, influencing its spin state.
• Co(II) Complex: Like \([\text{Co}([\text{H}_2\text{O}])_6]^{2+}\), which generally has a weaker ligand field.

Coordination compounds play a critical role in color chemistry, magnetic properties, and reactivity. Their unique configuration allows them to be used in diverse fields, such as transition metal catalysts in chemical reactions and as agents in medical diagnostics.

Spin States

The spin state of a metal ion in a coordination compound refers to how the electrons are arranged in its d-orbitals. Electrons can either pair up (low-spin) or remain unpaired (high-spin), depending on the strength of the ligand field.

• Low-Spin State: Co(III) complexes, such as \([\text{Co}([\text{H}_2\text{O}])_6]^{3+}\), typically exhibit low-spin configurations. This is due to strong ligand fields causing large energy splitting between orbitals, leading to electron pairing. For example, in Co(III) with a \(d^6\) configuration, all electrons pair up, resulting in zero unpaired electrons.
• High-Spin State: Co(II) complexes, like \([\text{Co}([\text{H}_2\text{O}])_6]^{2+}\), usually display high-spin states. The weaker ligand fields have smaller energy differences between orbitals, leading to unpaired electrons occupying higher orbitals. Therefore, in Co(II), which has a \(d^7\) configuration, there are typically four unpaired electrons.

These differences in spin states affect the magnetic and optical properties of the compounds. Understanding these concepts can assist in tailoring materials for specific technological applications, such as magnetic storage devices or light-emitting materials.