Problem 90 - Chemistry The Central Science In Si Units

Question

A sample of \(0.2140 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.0950 \mathrm{M}\) \(\mathrm{NaOH}\). The acid required \(30.0 \mathrm{~mL}\). of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(15.0 \mathrm{~mL}\) of base had been added in the titration, the \(\mathrm{pH}\) was found to be \(6.50 .\) What is the \(K_{a}\) for the unknown acid?

Step-by-Step Solution

Answer

The molar mass is 75.09 g/mol and \( K_a \) is \( 3.16 \times 10^{-7} \).

Step 1: Determine Moles of NaOH Used

First, calculate the moles of NaOH required to reach the equivalence point. Use the formula for moles: \[ \text{Moles of NaOH} = M \times V = 0.0950 \, \text{mol/L} \times 0.0300 \, \text{L} = 0.00285 \, \text{mol} \]

Step 2: Determine Moles of Acid

Since the acid is monoprotic, it donates one proton per molecule, meaning the moles of acid equal moles of NaOH at the equivalence point: \[ 0.00285 \text{ mol of acid} = 0.00285 \text{ mol of NaOH} \]

Step 3: Calculate Molar Mass of the Acid

Using the given mass and the moles of acid, calculate the molar mass: \[ \text{Molar mass} = \frac{\text{Mass of acid}}{\text{Moles of acid}} = \frac{0.2140 \, \text{g}}{0.00285 \, \text{mol}} = 75.09 \, \text{g/mol} \]

Step 4: Determine Concentration of Acid at Halfway Point

When 15.0 mL of NaOH is added, the volume of the solution at this point is \(25.0 \, \text{mL} + 15.0 \, \text{mL} = 40.0 \, \text{mL}\). The concentration of the acid at this point is half of the initial acid concentration because it's at the halfway point to equivalence.

Step 5: Calculate Initial Concentration of Acid

The initial concentration of acid can be calculated as: \[ [\text{HA}]_0 = \frac{\text{moles of acid}}{\text{volume of solution in liters}} = \frac{0.00285 \, \text{mol}}{0.025 \, \text{L}} = 0.114 \, \text{M} \]

Step 6: Use Henderson-Hasselbalch Equation to Find \(K_a\)

At the halfway point, we use the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pK}_a + \log\left(\frac{[A^-]}{[HA]}right) \] Since the concentrations of \( [A^-] \) and \( [HA] \) are equal, \[ \text{pH} = \text{pK}_a = 6.50 \] Now find \( K_a \): \[ K_a = 10^{-\text{pK}_a} = 10^{-6.50} = 3.16 \times 10^{-7} \]

Step 7: Conclusion

The molar mass of the acid is 75.09 g/mol and the \( K_a \) of the acid is \( 3.16 \times 10^{-7} \).

Key Concepts

Molar Mass Calculation

Calculating the molar mass of a substance is fundamental in chemistry, especially during titrations. In this scenario, we are given a sample of a monoprotic acid with a known mass and discuss how to use this information to find its molar mass. The molar mass is calculated by using the formula:

• Molar Mass (g/mol) = Mass of the substance (g) / Moles of the substance (mol)

Knowing the amount of acid and having determined the moles at the equivalence point, where moles of the acid equal moles of NaOH, simplifies this process. By dividing the given mass (0.2140 g) by the moles of acid (0.00285 mol), we find the molar mass to be 75.09 g/mol. This value represents the mass of one mole of the unknown acid.

Equivalence Point

The equivalence point in a titration is a crucial concept where the amount of titrant added exactly neutralizes the analyte solution. In our specific case, the equivalence point occurs when all the acid is neutralized by the baseline, meaning that every mole of the titrant (NaOH) corresponds to a mole of acid.

At this juncture, the moles of NaOH used equate precisely to the moles of the unknown monoprotic acid. This relationship helps us determine various unknown properties of the acid, such as its molar mass, because the stoichiometry tells us the exact amount of acid present initially.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a fantastically useful tool in acid-base chemistry. It gives a relationship between the pH of a solution and the concentrations of an acid and its conjugate base. It is represented as follows:

• \[\text{pH} = \text{pK}_a + \log\left(\frac{[A^-]}{[HA]}\right)\]

In our problem, halfway to the equivalence point (when 15 mL of NaOH is added), the concentrations of the acid ([HA]) and its conjugate base ([A⁻]) are equal, simplifying the equation to:

• pH = pKₐ

This allows us to directly equate the pH to the pKₐ, and subsequently compute the acid dissociation constant \( K_a \) using the formula \( K_a = 10^{-\text{pK}_a} \).

Acid Dissociation Constant (Ka)

The acid dissociation constant, symbolized as \( K_a \), provides insight into the strength of an acid within a solution. It effectively measures how well an acid dissociates into its ions in water. A lower \( K_a \) value signifies a weaker acid, as it indicates lesser dissociation.

In our titration exercise, after noting the pH at the halfway point, the Henderson-Hasselbalch application revealed the pKₐ to be 6.50, which can be used to find \( K_a \) as follows:

• \[K_a = 10^{-\text{pK}_a} = 10^{-6.50} = 3.16 \times 10^{-7}\]

This small \( K_a \) value confirms that the unknown monoprotic acid is relatively weak, as it does not dissociate significantly in water.