Problem 5 - Chemistry and Chemical Reactivity

Question

What is the pH of the solution that results from adding \(30.0 \mathrm{mL}\) of \(0.015 \mathrm{M} \mathrm{KOH}\) to \(50.0 \mathrm{mL}\) of \(0.015 \mathrm{M}\) benzoic acid?

Step-by-Step Solution

Answer

The pH of the solution is approximately 3.31.

Step 1: Calculate Moles of KOH and Benzoic Acid

Calculate the moles of each component in the solutions. For KOH, use \( n = C \times V \), where \( C = 0.015 \ \mathrm{mol/L} \) and \( V = 0.0300 \ \mathrm{L} \). This gives \( 4.50 \times 10^{-4} \ \mathrm{mol} \) of KOH. Similarly, calculate moles of benzoic acid using \( V = 0.0500 \ \mathrm{L} \), giving \( 7.50 \times 10^{-4} \ \mathrm{mol} \) of benzoic acid.

Step 2: Determine the Excess Reagent

Identify the limiting reagent by comparing the moles of KOH and benzoic acid. Since moles of KOH \( (4.50 \times 10^{-4}) \) are less than moles of benzoic acid \( (7.50 \times 10^{-4}) \), KOH is the limiting reagent.

Step 3: Calculate Moles of Remaining Benzoic Acid

Subtract the moles of KOH from the moles of benzoic acid to find the moles of benzoic acid that did not react: \( 7.50 \times 10^{-4} - 4.50 \times 10^{-4} = 3.00 \times 10^{-4} \ \mathrm{mol} \).

Step 4: Find the Volume of the Solution

The total volume of the mixed solution is the sum of the volumes of the KOH solution and benzoic acid solution: \( 30.0 \ \mathrm{mL} + 50.0 \ \mathrm{mL} = 80.0 \ \mathrm{mL} = 0.0800 \ \mathrm{L} \).

Step 5: Calculate the Concentration of Remaining Benzoic Acid

Divide the moles of excess benzoic acid by the total volume of the solution: \( \frac{3.00 \times 10^{-4} \ \mathrm{mol}}{0.0800 \ \mathrm{L}} = 3.75 \times 10^{-3} \ \mathrm{M} \).

Step 6: Calculate the pH of the Solution

Use the expression for the weak acid dissociation: \( K_a = \frac{[H^+][A^-]}{[HA]} \). For benzoic acid, assume \( [H^+] = [A^-] \), so \( K_a = \frac{[H^+]^2}{[HA]} \). Solve for \( [H^+] = \sqrt{K_a \times [HA]} \). With \( K_a = 6.3 \times 10^{-5} \) and \( [HA] = 3.75 \times 10^{-3} \), \( [H^+] = \sqrt{6.3 \times 10^{-5} \times 3.75 \times 10^{-3}} \) which gives \( [H^+] = 4.86 \times 10^{-4} \). Finally, calculate pH: \( \text{pH} = -\log_{10}([H^+]) = -\log_{10}(4.86 \times 10^{-4}) \approx 3.31 \).

Key Concepts

Acid-Base Reactions

In chemistry, understanding acid-base reactions is essential, especially when dealing with solutions like KOH and benzoic acid. An acid-base reaction involves an acid donating protons (H+) to a base.
When you mix KOH, a strong base, with benzoic acid, a weak acid, the hydroxide ions (OH-) from KOH will react with the hydrogen ions (H+) from benzoic acid to produce water (H2O). This is a classic acid-base neutralization reaction.

• The reacting species are KOH (base) and benzoic acid (acid).
• The product of the reaction is water and benzoate ion.
• The process continues until one of the reactants is fully consumed.

To determine which reactant is entirely used, we identify the limiting reagent, which in this situation, happens to be KOH.

Weak Acid Dissociation

Weak acid dissociation is a vital point in understanding pH calculations of weak acids like benzoic acid. Unlike strong acids, weak acids don't dissociate completely in water. Instead, they establish an equilibrium between the acid and its ions.
The dissociation of a weak acid HA into its ions can be represented as:
\[ HA \leftrightharpoons H^+ + A^- \]
The equilibrium constant for this reaction, called the acid dissociation constant \(K_a\), determines the extent to which the acid breaks into ions. For benzoic acid:

• \(K_a = 6.3 \times 10^{-5} \).
• The concentration of hydrogen ions \([H^+]\) can be found using: \([H^+] = \sqrt{K_a \times [HA]}\).

Understanding this concept helps in calculating the concentrations of reactants and products at equilibrium, allowing us to determine the final pH of the solution.

Moles and Molarity

Calculating moles and molarity is frequently the first step in solving problems involving solutions.
Moles measure the quantity of a substance, and molarity (M) is the number of moles per liter of solution. To find the moles of a solute, use the formula:\[ n = C \times V \]where:

• \(n\) is the number of moles
• \(C\) is molarity in moles/liter
• \(V\) is volume in liters

For example, in our problem, we found:

• Moles of KOH = \(0.015 \, \mathrm{M} \times 0.0300 \, \mathrm{L} = 4.50 \times 10^{-4} \, \text{mol} \)
• Moles of benzoic acid = \(0.015 \, \mathrm{M} \times 0.0500 \, \mathrm{L} = 7.50 \times 10^{-4} \, \text{mol} \)

This straightforward calculation sets the stage for further analysis, such as determining the limiting reagent.

Limiting Reagent

In chemical reactions, especially those involving mixed solutions, it's crucial to identify the limiting reagent. This is because the limiting reagent determines how much product can be formed before it runs out.
To find the limiting reagent in our equation, compare the initial moles of the reactants. The limiting reagent is the one that will be consumed first, stopping the reaction:

• KOH: \(4.50 \times 10^{-4} \, \text{mol}\)
• Benzoic Acid: \(7.50 \times 10^{-4} \, \text{mol}\)
• Thus, KOH is the limiting reagent because it is present in fewer moles.

Recognizing the limiting reagent is key as it helps predict the amount of products formed and excess reactants left behind, which influence the final pH calculation.