Question

A solution is prepared by dissolving 0.56 g benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) in enough water to make 1.0 \(\mathrm{L}\) of solution. Calculate \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\right],\left[\mathrm{H}^{+}\right]\) \(\left[\mathrm{OH}^{-}\right],\) and the pH of this solution.

Short answer

The concentrations and pH of the solution are as follows: \[[C_{6}H_{5}CO_{2}H] \approx 4.58 \times 10^{-3}\ mol / L \] \[[C_{6}H_{5}CO_{2^{-}}] \approx 1.22 \times 10^{-4}\ mol / L \] \[[H^{+}] \approx 1.22 \times 10^{-4}\ mol / L \] \[[OH^{-}] \approx 8.20 \times 10^{-11}\ mol / L \] \[pH \approx 3.91\]

Step-by-step solution

1. Calculate the initial concentration of benzoic acid

First, we need to find the initial concentration of benzoic acid by calculating the number of moles present in 0.56 g of benzoic acid and dividing it by the volume of the solution in L. The molar mass of benzoic acid is: Molar mass of C₆H₅CO₂H = 12.01 (C) * 7 + 1.01 (H) * 6 + 16.00 (O) * 2 = 122.12 g/mol Now we can find the initial concentration of benzoic acid: \[Initial\ concentration\ of\ benzoic\ acid = \frac{mass}{volume \times molar\ mass} = \frac{0.56\ g}{1.0\ L \times 122.12\ g/mol} = 4.58 \times 10^{-3}\ mol / L \]

2. Perform ICE table analysis

Now we can perform the ICE (Initial, Change, Equilibrium) analysis using the initial concentration to determine the equilibrium concentrations. Benzoic acid undergoes the reaction: \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}\rightleftharpoons\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2^{-}}+\mathrm{H}^{+}\) At equilibrium, the change in the concentrations of benzoic acid, benzoate ion, and hydrogen ion is represented by -x, x, and x, respectively. ICE table: C₆H₅CO₂H → C₆H₅CO₂⁻ + H⁺ Initial: 4.58 x 10⁻³ 0 0 Change: -x x x Equilibrium: 4.58 x 10⁻³ - x x x Now, we can write the Kₐ expression and solve for x: \[K_{\mathrm{a}} = \frac{[C_{6}H_{5}CO_{2^{-}}][H^{+}]}{[C_{6}H_{5}CO_{2}H]}\]

3. Calculate x

Plug the equilibrium concentrations into the Kₐ expression and solve for x: \[6.4 \times 10^{-5} = \frac{x^2}{4.58 \times 10^{-3} - x}\] Since the value of Kₐ is much smaller than the initial concentration of benzoic acid, we can assume that x is small compared to the initial concentration. Thus, we can simplify the equation to: \[x^2 \approx 6.4 \times 10^{-5} \times 4.58 \times 10^{-3}\] Solving for x, we get: \[x = [\mathrm{H^{+}}] \approx 1.22 \times 10^{-4}\ mol / L \]

4. Calculate equilibrium concentrations and pH

Now we can find the equilibrium concentrations of all species: \[[C_{6}H_{5}CO_{2}H] = 4.58 \times 10^{-3} - x \approx 4.58 \times 10^{-3}\ mol / L \] \[[C_{6}H_{5}CO_{2^{-}}] = [\mathrm{H^{+}}] \approx 1.22 \times 10^{-4}\ mol / L \] To find the [OH⁻], we know that \(K_w = [H^{+}] [OH^{-}]\), where \(K_w = 1.0 \times 10^{-14}\) at 25°C. So, \[[OH^{-}] = \frac{1.0 \times 10^{-14}}{[H^{+}]} = \frac{1.0 \times 10^{-14}}{1.22 \times 10^{-4}} \approx 8.20 \times 10^{-11}\ mol / L \] Finally, to find the pH of the solution, we can use the formula: \[pH = -\log_{10} [H^{+}] = -\log_{10} (1.22 \times 10^{-4}) \approx 3.91\]

5. Summary of results

The concentrations and pH of the solution are as follows: \[[C_{6}H_{5}CO_{2}H] \approx 4.58 \times 10^{-3}\ mol / L \] \[[C_{6}H_{5}CO_{2^{-}}] \approx 1.22 \times 10^{-4}\ mol / L \] \[[H^{+}] \approx 1.22 \times 10^{-4}\ mol / L \] \[[OH^{-}] \approx 8.20 \times 10^{-11}\ mol / L \] \[pH \approx 3.91\]

Key concepts

Equilibrium Concentration

When dealing with weak acids like benzoic acid, equilibrium concentration is key to understanding how much of the acid dissociates in a solution. Initially, when you mix benzoic acid in water, it doesn't completely break apart into ions. This is because it's a weak acid and partially dissociates.
The equilibrium concentration of each species involved is determined by how much the acid dissociates and what remains in the solution.
To find these concentrations, we start by considering the initial concentration of benzoic acid, which we calculated using its molar mass. As benzoic acid partially dissociates into \( ext{C}_6 ext{H}_5 ext{CO}_2^-\) and \( ext{H}^+\), we introduce the variable \(x\) to denote the amount that dissociates at equilibrium.
At this stage, you might have learned about the ICE table, which helps immensely in organizing this process.Understanding equilibrium concentrations helps:

• Predict how the acid and its ions behave at equilibrium.
• Establish what is left of the initial acid concentration.
• Determine the amount of ions formed in the solution.

ICE Table Analysis

The ICE table is a systematic tool used by chemists to track the changes in concentrations of each species in a chemical reaction. ICE stands for Initial, Change, and Equilibrium, and it provides a clear framework to approach equilibrium problems.
Here's how we use it for weak acid dissociation:
Initial: Record the initial concentrations of reactants and products. For our benzoic acid solution, we started with a substantial amount of the acid and zero concentration of the ions it dissociates into.

• Initial concentration of benzoic acid was calculated to be approximately \(4.58 \times 10^{-3} \text{ mol/L}\).
• The concentrations of \( ext{C}_6 ext{H}_5 ext{CO}_2^-\) and \( ext{H}^+\) were both 0 initially.

Change: Represent the change in concentration as the reaction proceeds to equilibrium. This is where we introduce \(x\) for the change:

• Benzoic acid decreases by \(x\), while the ions each increase by \(x\).

Equilibrium: Sum the Initial and Change to find equilibrium concentrations.
Utilizing the given \( K_a \), the dissociation constant, we plug these values into the equilibrium expression to solve for \(x\). This tells us the concentrations at equilibrium. Using an ICE table helps:

• Visually organize and simplify complex equilibrium calculations.
• Track how the concentration of each species changes.
• Use mathematical relations to find unknowns associated with equilibrium.

pH Calculation

The pH of a solution is a measure of its acidity or basicity, specifically how many hydrogen ions (\( ext{H}^+\)) are present. It's an essential part of understanding the chemistry of any solution. The process of calculating the pH of a benzoic acid solution, which partially dissociates, requires a series of logical steps.
Once you've found the concentration of \( ext{H}^+\) ions using the ICE table analysis (denoted as \(x\) in the equilibrium concentration), you can use the formula for pH:\[pH = -\log_{10} [H^+]\]Given that \([H^+]\) was solved to be approximately \(1.22 \times 10^{-4} \text{ mol/L}\), the pH calculation becomes straightforward.
Some points to keep in mind:

• A lower pH indicates a higher acidity.
• In our calculation, the solution's pH was approximately 3.91, indicating it's acidic, but not strongly.
• The assumption that x is small compared to the initial concentration is valid, reflected in the minor correction when compared to initial concentration.

Understanding pH provides insights into:

• How acidic or basic a solution is.
• The potential behavior of a solution in chemical reactions.
• The biological significance, as pH can affect the activity of enzymes.