Problem 124 - Chemistry

Question

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

Step-by-Step Solution

Answer

The \(\mathrm{pH}\) of the \(0.10 \mathrm{M}\) \(\mathrm{CoCl}_{3}\) solution can be calculated in three steps: writing the dissociation reaction for \(\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\), applying initial concentrations to solve for \([\mathrm{H}^{+}]\), and using this concentration to find the \(\mathrm{pH}\). Upon completing these steps, we find that the \(\mathrm{pH}\) of the solution is approximately \(3\).

Step 1: Write down the dissociation reaction and equilibrium expression

We start by writing the dissociation reaction for \(\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\): \[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+} \rightleftharpoons \mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\mathrm{OH}^{2+} + \mathrm{H}^{+}\] Now, we can write down the equilibrium expression using the given \(K_{\mathrm{a}}\) value: \[K_\mathrm{a} = \dfrac{[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\mathrm{OH}^{2+}][\mathrm{H}^{+}]}{[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}]}\]

Step 2: Apply initial concentrations and solve for \([\mathrm{H}^{+}]\)

Initially, the concentration of \(\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\) is \(0.10 \mathrm{M}\). Before dissociation occurs, we have \([\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\mathrm{OH}^{2+}] = [\mathrm{H}^{+}] = 0\). Letting \(x\) represent the change in concentration of \(\mathrm{H}^{+}\), the equilibrium concentrations will be: \[[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\mathrm{OH}^{2+}] = [\mathrm{H}^{+}] = x\] \[[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}] = 0.10 - x\] Since we expect \(x\) to be much smaller than \(0.10\), we can make the simplification that \(0.10 - x \approx 0.10\). Then, we can substitute these concentrations back into the equilibrium expression and solve for \(x\): \[K_\mathrm{a} = \dfrac{x \cdot x}{0.10} = 1.0 \times 10^{-5}\] \[x^2 = 1.0 \times 10^{-5} \cdot 0.10\] \[x = \sqrt{1.0 \times 10^{-6}} = 1 \times 10^{-3}\] Thus, we find that \([\mathrm{H}^{+}] \approx 1 \times 10^{-3} \mathrm{M}\).

Step 3: Calculate the \(\mathrm{pH}\) of the solution

Finally, we can use the concentration of \([\mathrm{H}^{+}]\) to calculate the \(\mathrm{pH}\) using the formula: \[\mathrm{pH} = -\log{[\mathrm{H}^{+}]}\] \[\mathrm{pH} = -\log{(1 \times 10^{-3})} = 3\] The \(\mathrm{pH}\) of the \(0.10 \mathrm{M}\) \(\mathrm{CoCl}_{3}\) solution is approximately \(3\).

Key Concepts

Equilibrium Expression

In chemistry, the concept of equilibrium expression is a key in understanding reactions that involve reversible chemical processes. When dealing with an acidic solution, it is important to first write down the dissociation reaction. This represents how a complex ion dissociates in solution.
For the given problem, we look at the dissociation reaction of \[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\] into its components:

• \[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\mathrm{OH}^{2+}\]
• \[\mathrm{H}^{+}\]

Writing this mathematically, the reaction is represented as: \[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+} \rightleftharpoons \mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\mathrm{OH}^{2+} + \mathrm{H}^{+} \]
The equilibrium expression for this reaction, which is an important tool for finding the concentration of ions in equilibrium, is given using the Acid Dissociation Constant, \(K_a\):\[K_\mathrm{a} = \dfrac{[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\mathrm{OH}^{2+}][\mathrm{H}^{+}]}{[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}]}\]This expresses the relationship between concentrations of the products and reactants at equilibrium.
The dissociation constants allow chemists to predict how far the reactions proceed, and also aid in calculating concentrations of different ions in equilibrium based on their initial conditions.

pH Calculation

The calculation of pH is essential in determining the acidity level of a solution, and it directly relates to the concentration of hydrogen ions \([\mathrm{H}^{+}]\) in a solution. pH is calculated using the formula: \[\mathrm{pH} = -\log{[\mathrm{H}^{+}]}\]
Let's apply this to the problem scenario. After substituting values into the equilibrium expression and solving, we found that

• the concentration of \([\mathrm{H}^{+}]\) is approximately \(1 \times 10^{-3} \mathrm{M}\).

Substitute this value into the pH formula, resulting in:\[\mathrm{pH} = -\log{(1 \times 10^{-3})} = 3\]This signifies that the solution is acidic since its pH value is below 7, indicating a higher concentration of hydrogen ions compared to what is found in pure water.
This calculation is important as it gives a quick way to understand how acidic or basic a solution is, allowing for adjustments to be made in various applications such as chemistry labs, pharmaceuticals, or environmental science fields where pH stability is crucial.

Acid Dissociation Constant

The Acid Dissociation Constant, denoted as \(K_a\), is a fundamental concept in understanding the strength of an acid in solution. It quantifies the tendency of an acid to donate a proton (\(\mathrm{H}^{+}\)) in an aqueous solution.
Larger \(K_a\) values indicate stronger acids which dissociate more fully in solution, while smaller \(K_a\) values suggest weaker acids.

• In our example, \(K_a = 1.0 \times 10^{-5}\) for \(\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\).

This small \(K_a\) value suggests that it is a weak acid, meaning it does not fully dissociate in solution, leading to an equilibrium whereby both the forward and reverse reactions occur but are not heavily one-sided toward products or reactants.
Understanding \(K_a\) helps predict and calculate the pH of acid solutions, understand reaction equilibria, and design buffer solutions. It's particularly crucial for determining the behavior of acids in biological systems where pH changes can be critical to life processes.