Question

Calculate the pH of a \(0.10-M \mathrm{CoCl}_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(1.0 \times 10^{-5} .\)

Short answer

The pH of the 0.10 M CoCl₃ solution is 3. We find this by first setting up an ICE table and using the given Kₐ value for Co(H₂O)₆³⁺ to calculate the H⁺ concentration, and finally applying the pH formula: \(pH = -\log_{10} [H^+] = -\log_{10} (1.0 \times 10^{-3})\).

Step-by-step solution

Write the ionization equilibrium

Write the ionization equilibrium for Co(H₂O)₆³⁺: \(Co(H_2 O)_6^{3+} \rightleftharpoons H^+ + Co(H_2 O)_5^{2+}\)

Set up an ICE table

Set up an Initial, Change, Equilibrium (ICE) table for the given equilibrium. The initial concentration of Co(H₂O)₆³⁺ is 0.10 M, while the initial concentration of H⁺ and Co(H₂O)₅²⁺ is 0. | | Co(H₂O)₆³⁺ | H⁺ | Co(H₂O)₅²⁺ | |-----|-----------|----|------------| | I | 0.10 | 0 | 0 | | C | -x | +x | +x | | E | 0.10 - x | x | x |

Write the expression for Kₐ

Write the expression for the Kₐ and plug in equilibrium concentrations from the ICE table: \( K_a = \frac{[H^+][Co(H_2 O)_5^{2+}]}{[Co(H_2 O)_6^{3+}]} = \frac{x * x}{0.10 - x} \)

Solve for x

We know that Kₐ is \(1.0 \times 10^{-5}\). Plug it into the expression and solve for x: \(1.0 \times 10^{-5} = \frac{x^2}{0.10 - x} \) As x is very small compared to 0.10, we can approximate it as: \(1.0 \times 10^{-5} = \frac{x^2}{0.10} \) Now, solve for x: \(x^2 = 1.0 \times 10^{-5} \times 0.10 \) \(x = \sqrt{1.0 \times 10^{-6}} \) \(x = 1.0 \times 10^{-3} \)

Calculate the pH

Now that we have found the concentration of H⁺ ions as \(1.0 \times 10^{-3} M\), we can calculate the pH of the solution using the pH formula: \(pH = -\log_{10} [H^+] = -\log_{10} (1.0 \times 10^{-3})\) The pH of the solution is 3.

Key concepts

Ionization Equilibrium

In chemistry, ionization equilibrium refers to the dynamic balance between the ionized and non-ionized form of a solute in solution. It is essential for understanding the behavior of acids and bases during dissolution.

For our exercise, the ionization equilibrium involves the complex ion \(Co(H_2O)_6^{3+}\), which partially dissociates into \(H^+\) ions and another complex ion, \(Co(H_2O)_5^{2+}\). This process is reversible, represented as a balanced chemical equation:

\Co(H_2 O)_6^{3+} ightleftharpoons H^+ + Co(H_2 O)_5^{2+}\.

Understanding ionization equilibrium helps predict the extent to which the dissociation occurs, which is crucial for calculating properties such as pH.

ICE Table

An ICE table stands for Initial, Change, and Equilibrium, and is a valuable tool in analyzing chemical equilibria. It helps visualize how concentrations of reactants and products change as the system reaches equilibrium.

Consider our previous equilibrium. Initially, we have 0.10 M \(Co(H_2O)_6^{3+}\) and 0 M for both \(H^+\) and \(Co(H_2O)_5^{2+}\). As the reaction proceeds, these concentrations change: \(Co(H_2O)_6^{3+}\) decreases by \(-x\), while \(H^+\) and \(Co(H_2O)_5^{2+}\) increase by \(+x\). At equilibrium, they adjust to reflect the final concentrations.

• Initial concentrations: 0.10 M for \(Co(H_2O)_6^{3+}\), 0 for the others.
• Change: defined by variable \(x\).
• Equilibrium: calculated after applying changes.

This setup is pivotal in solving equilibrium problems effectively.

Acid Dissociation Constant

The acid dissociation constant, represented as \(K_a\), quantifies the strength of an acid in solution. It tells us how well an acid donates protons (H⁺ ions) to the solution.

For our exercise, we use \(K_a\) for \(Co(H_2O)_6^{3+}\), which is given as \(1.0 \times 10^{-5}\). Using the equilibrium expression derived from our chemical equation, we can set up the \(K_a\) equation:

\[K_a = \frac{[H^+][Co(H_2 O)_5^{2+}]}{[Co(H_2 O)_6^{3+}]} = \frac{x^2}{0.10 - x}\]

This equation allows us to solve for \(x\), which represents the change in concentration of \(H^+\), helping assess how many protons are contributed by the acid at equilibrium.

pH Formula

The pH formula is an essential tool in chemistry used to determine the acidity or basicity of a solution. It is expressed as:

\(pH = -\log_{10} [H^+]\).

For our problem, after solving for \(x\), which represents the concentration of hydrogen ions \([H^+]\), we found \(x = 1.0 \times 10^{-3} M\). Substituting this into the pH formula, we find:

\(pH = -\log_{10} (1.0 \times 10^{-3}) = 3\).

Understanding the pH formula and how to apply it is crucial as it provides insight into the solution's acidic or basic nature, which is fundamental in various chemical applications.