Question

Calculate the \(\mathrm{pH}\) 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 \(\mathrm{CoCl}_3\) solution can be calculated by considering the hydrolysis of \(\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\) ions. The balanced equation for this reaction is \(\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+} \left(aq\right) + \mathrm{H}_{2}\mathrm{O} \left(l\right) \rightleftarrows \mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\left(\mathrm{OH}\right)^{2+} \left(aq\right) + \mathrm{H_{3}O}^{+} \left(aq\right)\). Using the given \(K_{\mathrm{a}}\) value of \(1.0 \times 10^{-5}\) and assuming that \(x << 0.10\mathrm{M}\), we can approximate the concentration of \(\mathrm{H}^{+}\) ions as \(1.0 \times 10^{-3}\mathrm{M}\). Therefore, the pH of the solution is approximately 3.00.

Step-by-step solution

Write the balanced equation for the reaction

First, we need to recognize that the \(\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\) ions undergo hydrolysis in the solution, which means they react with water to release \(\mathrm{H}^{+}\) ions. The balanced equation for this reaction is: \[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+} \left(aq\right) + \mathrm{H}_{2}\mathrm{O} \left(l\right) \rightleftarrows \mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\left(\mathrm{OH}\right)^{2+} \left(aq\right) + \mathrm{H_{3}O}^{+} \left(aq\right)\]

Set up and solve the equilibrium expression

Now we need to set up the equilibrium expression for the reaction, using the given \(K_{\mathrm{a}}\) value: \[K_{\mathrm{a}} = \dfrac{[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\left(\mathrm{OH}\right)^{2+}][\mathrm{H_{3}O}^{+}]}{[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}]}\] Since we're given the initial concentration of \(\mathrm{CoCl}_{3}\) (which is the same as the initial concentration of \(\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\)), we can solve for the equilibrium concentration of \(\mathrm{H}^{+}\) ions. Let's call this concentration \(x\). At equilibrium, the concentrations are as follows: \([\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}] = 0.10 \mathrm{M} - x\) \([\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\left(\mathrm{OH}\right)^{2+}] = x\) \([\mathrm{H_{3}O}^{+}] = x\) Now substitute these values into the equilibrium expression: \[1.0 \times 10^{-5} = \dfrac{x^2}{0.10\mathrm{M} - x}\] We can simplify this expression by assuming that the amount of dissociation is small and \(x << 0.10\mathrm{M}\). Therefore, we can approximate that \(0.10\mathrm{M} - x \approx 0.10\mathrm{M}\). So, \[1.0 \times 10^{-5} \approx \dfrac{x^2}{0.10\mathrm{M}}\] Now, we can solve for \(x\). \[x = \sqrt{1.0 \times 10^{-5} \times 0.10\mathrm{M}} \approx 1.0 \times 10^{-3}\mathrm{M}\]

Calculate the pH

Now that we have calculated the concentration of \(\mathrm{H}^{+}\) ions in the solution, we can use this value to calculate the pH: \[pH = -\log{[\mathrm{H}^{+}]}\] \[pH = -\log{(1.0 \times 10^{-3}\mathrm{M})} \approx 3.00\] Hence, the pH of the 0.10 M \(\mathrm{CoCl}_3\) solution is approximately 3.00.

Key concepts

Equilibrium Expression

An equilibrium expression is a crucial part of understanding how reactions occur and reach a balance within a solution. It is a mathematical way to describe the concentrations of reactants and products when a reaction is in a state of equilibrium.
In this exercise, we are dealing with the hydrolysis reaction of cobalt ions in water. The balanced chemical equation for this reaction is:\[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\left(\mathrm{OH}\right)^{2+} + \mathrm{H_{3}O}^{+}\]To create an equilibrium expression, we use the expression for the acid dissociation constant \(K_{a}\):\[K_{a} = \dfrac{[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\left(\mathrm{OH}\right)^{2+}][\mathrm{H_{3}O}^{+}]}{[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}]}\]This expression helps us understand the ratio of products to reactants and provides insight into the tendency of the compound to donate protons in solution. It is essential when calculating pH, as it allows you to solve for the concentration of hydrogen ions \([\mathrm{H}^{+}]\), which directly affects the pH level.

Hydrolysis Reaction

Hydrolysis reactions are fascinating chemical processes where a molecule, usually a salt or acid, reacts with water to form new products.
In our example with cobalt chloride \(\mathrm{CoCl}_{3}\), the cobalt ion \(\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\) acts as a Lewis acid. It accepts electrons from water molecules, leading to the release of hydrogen ions \(\mathrm{H}^{+}\) into the solution.
This process can be summarized by the following chemical equation:\[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\left(\mathrm{OH}\right)^{2+} + \mathrm{H_{3}O}^{+}\]This reaction is important for pH calculations because the hydronium ion \(\mathrm{H_{3}O}^{+}\) contributes to the acidity of the solution. More \(\mathrm{H}^{+}\) ions in solution increase the acidity, lowering pH value.
Understanding hydrolysis reactions gives valuable insights into how metal ions interact with water, how they affect acidity, and thus predicting the behavior of such solutions in various chemical contexts.

Acid Dissociation Constant

The acid dissociation constant \(K_{a}\) is an invaluable parameter in chemistry that measures the strength of an acid in a solution.
It reflects the acid's ability to release protons \(\mathrm{H}^{+}\) to the surrounding water molecules. A higher \(K_{a}\) value indicates a stronger acid, meaning it is more willing to donate its protons.
In our exercise, the \(K_{a}\) for \(\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\) is provided as \(1.0 \times 10^{-5}\). This relatively low value suggests that \(\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}\) is a weak acid.
The acid dissociation constant plugs directly into the equilibrium expression:\[K_{a} = \dfrac{[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\left(\mathrm{OH}\right)^{2+}][\mathrm{H_{3}O}^{+}]}{[\mathrm{Co}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}^{3+}]}\]This mathematical relationship helps calculate the concentrations of ions, which ultimately allows us to find the pH. Grasping the concept of \(K_{a}\) is essential for solving problems involving weak acids and estimating their contributions to the overall acidity or basicity in the solution.

Logarithmic Function

Logarithmic functions are mathematical tools frequently used in chemistry, especially in the context of pH calculations.
pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration in a solution:\[pH = -\log{[\mathrm{H}^{+}]}\]This transformation turns large, cumbersome numbers into more manageable figures.
For example, when the \([\mathrm{H}^{+}]\) in our solution is \(1.0 \times 10^{-3}\ \mathrm{M}\), calculating the pH involves applying the logarithmic function:\[pH = -\log{(1.0 \times 10^{-3})} \approx 3.00\]Logarithmic functions help simplify the interpretation of acidity by presenting hydrogen ion concentrations in a scale that's easier to understand and visually represent. Since hydrogen ion concentrations can vary vastly across solutions, the logarithmic scale condenses these variations into a practical range that chemists can work with effectively. Understanding how to manipulate and interpret logarithmic scales will significantly aid in performing accurate pH calculations and in general data analysis within scientific research.