Question

(a) A 0.1044-g sample of an unknown monoprotic acid requires \(22.10 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{NaOH}\) to reach the end point. What is the molar mass of the unknown? (b) As the acid is titrated, the \(\mathrm{pH}\) of the solution after the addition of \(11.05 \mathrm{~mL}\) of the base is \(4.89 .\) What is the \(K_{a}\) for the acid? (c) Using Appendix D, suggest the identity of the acid.

Short answer

The molar mass of the unknown monoprotic acid is approximately 94.43 g/mol. The Ka of the unknown acid is \(6.69 \times 10^{-6}\). Comparing these values with Appendix D, the unknown acid is most likely benzoic acid, which has a molar mass of approximately 122.12 g/mol and a Ka value of \(6.5 \times 10^{-5}\).

Step-by-step solution

Calculate the moles of NaOH used in the titration

We will need to find moles of NaOH by using the volume of NaOH required to reach the endpoint of the titration and the molarity of the solution. Moles of NaOH = (Volume of NaOH in L) x (Molarity of NaOH) Moles of NaOH = (22.10 mL / 1000) x 0.0500 M = \(1.105 \times 10^{-3}\) moles **Step 2: Calculate moles of the acid**

Determine the moles of the unknown monoprotic acid

As the acid is monoprotic, it will react with NaOH in a 1:1 ratio. Moles of monoprotic acid = Moles of NaOH = \(1.105 \times 10^{-3}\) moles **Step 3: Calculate the molar mass of the unknown acid**

Determine the molar mass of the unknown acid

Using the moles of monoprotic acid calculated and the given mass of the acid, we can calculate the molar mass of the unknown acid. Molar mass = (Mass of the acid) / (Moles of acid) Molar mass = 0.1044 g / \(1.105 \times 10^{-3}\) moles = 94.43 g/mol **Step 4: Calculate the concentration of the acid and the base at pH 4.89**

Determine the concentrations of the acid and the base

To calculate the Ka, we will need to know the concentration of the acid, its conjugate base, and the hydrogen ion concentration. We can calculate these values using the pH of the solution (4.89) after adding 11.05 mL of 0.0500 M NaOH: \(H^{+}\) concentration = \(10^{-pH}\) = \(10^{-4.89}\) = \(1.29 \times 10^{-5}\) M At that point in the titration, the amounts of both the acid and its conjugate base are identical, as half of the titration is complete. So, we have: [H+] = [A-] = \(1.29 \times 10^{-5}\) M Initial moles of acid = \(1.105 \times 10^{-3}\) moles found in step 2. At half titration, moles of acid = 1/2 \(\times\) \(1.105 \times 10^{-3}\) moles (half of initial moles). Total volume at half titration = volume of acid + 11.05 mL of NaOH = 22.1 mL (assuming the acid had negligible initial volume). Concentration of acid, [HA] = moles of acid / volume at half titration = (1/2 \(\times\) \(1.105 \times 10^{-3}\) moles) / (22.1 mL / 1000 L) = \(2.50 \times 10^{-5}\) M **Step 5: Calculate the Ka of the unknown monoprotic acid**

Determine the Ka of the unknown acid

Now that we have the concentrations of the acid, its conjugate base, and the hydrogen ion concentration, we can calculate the Ka of the unknown acid using the expression: Ka = \([H^{+}] [A^{-}]/[HA]\) Since [H+] = [A-]: Ka = \(\frac{(1.29 \times 10^{-5})^2}{2.50 \times 10^{-5}}\) = \(6.69 \times 10^{-6}\) **Step 6: Identify the unknown acid**

Compare the Ka value with Appendix D to identify the acid

Using the Ka value of \(6.69 \times 10^{-6}\) and Appendix D, find an acid with a molar mass close to 94.43 g/mol and a Ka value close to \(6.69 \times 10^{-6}\). The acid most closely matching these values is benzoic acid, with a molar mass of approximately 122.12 g/mol and a Ka value of \(6.5 \times 10^{-5}\).

Key concepts

Molar Mass Calculation

To find the molar mass of an unknown acid during titration, you need to know both the mass of the acid sample and the moles of acid that reacted. Here's how this process works:

1. **Identify Moles of Reactant**: When you titrate, you typically react your acid with a base. In this case, we use NaOH. You can find the moles of NaOH by multiplying its volume (converted to liters) by its molarity. For instance, if you use 22.10 mL of 0.0500 M NaOH, you have:

• Moles of NaOH = (22.10 mL / 1000) × 0.0500 M = 1.105 × 10^{-3} moles

2. **Determine Moles of Acid**: Since the acid is monoprotic, there is a 1:1 reaction with NaOH, meaning you also have 1.105 × 10^{-3} moles of the acid.

3. **Calculate Molar Mass**: Finally, divide the mass of your acid sample by the moles of acid to find the molar mass. In our example:

• Molar Mass = 0.1044 g / 1.105 × 10^{-3} moles = 94.43 g/mol

This step is crucial to determine the identity or purity of the acid being analyzed.

pH Calculation

Calculating the pH in a titration involves understanding the concentration of hydrogen ions in the solution. Here's a simplified explanation:

1. **Find Hydrogen Ion Concentration**: Given a pH value, you can find the concentration of hydrogen ions using the formula:

• \[ [H^+] = 10^{-pH} \]
• For a pH of 4.89, the \[ [H^+] = 10^{-4.89} = 1.29 × 10^{-5} \, \text{M} \]

2. **Understand Equal Concentrations**: During a titration when you've added half the amount of base, the concentrations of the acid (HA) and its conjugate base (A-) are equal. At this point:

• \[ [H^+] = [A^-] \]

3. **Concentration Calculations**: With known volumes and moles, calculate the concentration of the acid and its conjugate base.

pH calculations help in understanding the strength and behavior of the acid during titration, providing essential data to find the dissociation constant, \( K_a \).

Acid Dissociation Constant (Ka)

The acid dissociation constant, \( K_a \), measures the strength of an acid in solution. To calculate it, follow these guidelines:

1. **Gather Necessary Concentrations**: You need the concentrations of the hydrogen ions, the conjugate base, and the remaining acid at a given point in the titration.

2. **Use the Ka Expression**: The expression for the dissociation constant is:

• \[ K_a = \frac{[H^+][A^-]}{[HA]} \]

Insert values to compute \( K_a \). For example, with given concentrations during the titration:

• \[ K_a = \frac{(1.29 × 10^{-5})^2}{2.50 × 10^{-5}} = 6.69 × 10^{-6} \]

3. **Identify the Acid**: Compare \( K_a \) values from reference tables to identify the likely acid. Knowing \( K_a \) explains how easily the acid gives up protons in solution and helps determine its chemical identity and purity.