Problem 91 - Chemistry The Central Science In Si Units

Question

A sample of \(500 \mathrm{mg}\) of an unknown monoprotic acid was dissolved in \(50.0 \mathrm{~mL}\) of water and titrated with \(0.200 \mathrm{M}\) KOH. The acid required \(20.60 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(10.30 \mathrm{~mL}\) of base had been added in the titration, the pH was found to be \(4.20 .\) What is the \(p K_{a}\) for the unknown acid

Step-by-Step Solution

Answer

(a) Molar mass of the acid is 121.36 g/mol. (b) \( pK_a = 4.20 \).

Step 1: Calculate Moles of KOH used

The moles of KOH used at the equivalence point can be calculated using its concentration and the volume of solution used. Use the formula: \( \text{moles of KOH} = M \times V \), where \( M = 0.200 \text{ M} \) and \( V = 20.60 \text{ mL} = 0.02060 \text{ L} \). The calculation is: \( \text{moles of KOH} = 0.200 \times 0.02060 = 0.00412 \text{ mol} \).

Step 2: Equivalence point indicates moles of acid

Since it is a monoprotic acid, the moles of acid present in the solution are equal to the moles of KOH used at the equivalence point, which is 0.00412 mol.

Step 3: Calculate Molar Mass of the Acid

To find the molar mass of the acid, use the formula: \( \text{Molar Mass} = \frac{\text{mass of acid}}{\text{moles of acid}} \). Here, the mass of the acid is 500 mg = 0.500 g. Thus, \( \text{Molar Mass} = \frac{0.500}{0.00412} \approx 121.36 \text{ g/mol} \).

Step 4: Determine pK_a from pH given at half-equivalence point

At the half-equivalence point, the concentration of the acid equals the concentration of its conjugate base, meaning \( \text{pH} = \text{p}K_a \). The volume at the half-equivalence point is half of \(20.60 \text{ mL} = 10.30 \text{ mL} \), and the pH is given as 4.20. Therefore, \( \text{p}K_a = 4.20 \).

Key Concepts

Monoprotic Acid

A monoprotic acid is an acid that can donate only one proton (hydrogen ion) per molecule to a base in a chemical reaction. This is in contrast to diprotic or triprotic acids, which can donate two or three protons, respectively. In simpler terms, each molecule of a monoprotic acid, when it dissociates, yields just one hydrogen ion. Common examples of monoprotic acids include hydrochloric acid (HCl) and acetic acid (CH₃COOH). Understanding that an unknown acid is monoprotic is crucial in titration processes, as it simplifies the calculations related to the equivalence point. For the titration of a monoprotic acid, the amount of base added to reach the equivalence point is a direct measure of the number of moles of hydrogen ions present.

Molar Mass Calculation

Molar mass is a key property of a compound, defining the mass of one mole of its molecules. Calculating the molar mass of an unknown substance using titration data involves understanding the relationship between mass, moles, and molar mass. The formula used is: \[\text{Molar Mass} = \frac{\text{mass of acid}}{\text{moles of acid}}\]Here, the mass of the monoprotic acid was known (500 mg). By using titration data, we determined the moles of the acid, which in this case equaled the moles of KOH due to the 1:1 reaction ratio. This direct relationship allows us to solve for the molar mass of the unknown acid. In the provided exercise, it is calculated to be approximately 121.36 g/mol.This value plays a significant role in identifying the acid or determining its purity.

Equivalence Point

In a titration, the equivalence point is a critical moment where the amount of titrant added is stoichiometrically equal to the amount of substance present in the sample. For a monoprotic acid, this is the point where all available protons have reacted with the base. The equivalence point is distinct from the endpoint, which might be indicated by a color change due to an indicator. At this stage for our exercise, 20.60 mL of 0.200 M KOH was required to fully neutralize the acid. Understanding the equivalence point is vital in determining the exact quantity of acid initially present. Since the moles of acid equal the moles of KOH at this point, it allows for precise calculations of concentration and reactivity of the components.

pKa Determination

The pKa value of an acid is a measure of its acid strength and is a log-transformed form of the acid dissociation constant, Ka. It tells you how readily the acid donates its proton. During a titration of a monoprotic acid, determining the pKa becomes simple at the half-equivalence point, where the concentrations of the acid and its conjugate base are equal. At this juncture, the pH of the solution directly equates to the pKa of the acid. In our exercise, after 10.30 mL of KOH was added, the solution's pH was measured as 4.20. Hence, the pKa of the unknown monoprotic acid was 4.20. Accurately determining pKa is crucial for comparing acid strengths and making predictions about reactions in different pH environments.