Question

A 0.1276 -g sample of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with a \(0.0633 \mathrm{M} \mathrm{NaOH}\) solution. The volume of base required to bring the solution to the equivalence point was \(18.4 \mathrm{~mL}\). (a) Calculate the molar mass of the acid. (b) After \(10.0 \mathrm{~mL}\) of base had been added during the titration, the \(\mathrm{pH}\) was determined to be 5.87. What is the \(K_{\mathrm{a}}\) of the unknown acid?

Short answer

The molar mass of the acid is approximately 109.43 g/mol. The \(K_a\) is approximately \(1.74 \times 10^{-6}\).

Step-by-step solution

Calculate Moles of NaOH

First, calculate the moles of NaOH used at the equivalence point using the formula \( n = M \cdot V \), where \( M \) is the molarity and \( V \) is the volume in liters. Substitute the known values: \( n = 0.0633 \: \text{M} \times 0.0184 \: \text{L} \). This gives \( n = 1.16552 \times 10^{-3} \: \text{moles} \).

Determine Moles of Acid

Since the acid is monoprotic, it donates one proton per molecule, meaning it reacts in a 1:1 molar ratio with NaOH. Thus, the moles of acid are equal to the moles of NaOH at the equivalence point, which is \( 1.16552 \times 10^{-3} \: \text{moles} \).

Calculate Molar Mass of the Acid

Use the formula for molar mass: \( \text{Molar Mass} = \frac{\text{mass of acid}}{\text{moles of acid}} \). Substitute the values: \( \text{Molar Mass} = \frac{0.1276 \: \text{g}}{1.16552 \times 10^{-3} \: \text{moles}} \), which results in \( \text{Molar Mass} \approx 109.43 \: \text{g/mol} \).

Determine Moles and Concentration of HA at 10 mL of NaOH

10 mL of NaOH corresponds to \( 0.0633 \: \text{M} \times 0.0100 \: \text{L} = 6.33 \times 10^{-4} \: \text{moles} \) of NaOH. At this point, the moles of remaining HA are \( 1.16552 \times 10^{-3} \: \text{moles} - 6.33 \times 10^{-4} \: \text{moles} = 5.3252 \times 10^{-4} \: \text{moles} \). The concentration of HA is \( \frac{5.3252 \times 10^{-4} \: \text{moles}}{0.035 \: \text{L}} \approx 0.0152 \: \text{M} \).

Calculate \([A^-]\) at 10 mL of NaOH

The moles of \( A^- \) formed are equal to the moles of NaOH added, which is \( 6.33 \times 10^{-4} \: \text{moles} \). Its concentration is \( \frac{6.33 \times 10^{-4} \: \text{moles}}{0.035 \: \text{L}} \approx 0.01809 \: \text{M} \).

Use Henderson-Hasselbalch Equation to find \( pK_a \)

Apply the Henderson-Hasselbalch equation: \( pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \). Substitute \( pH = 5.87 \), \([A^-] = 0.01809 \: \text{M} \), and \([HA] = 0.0152 \: \text{M}\). Solve for \( pK_a \): \( 5.87 = pK_a + \log\left(\frac{0.01809}{0.0152}\right) \). This leads to \( pK_a \approx 5.76 \).

Calculate \( K_a \)

Convert \( pK_a \) to \( K_a \) using \( K_a = 10^{-pK_a} \). Substitute \( pK_a = 5.76 \): \( K_a = 10^{-5.76} \approx 1.74 \times 10^{-6} \).

Key concepts

Molar Mass Calculation

To determine the molar mass of an unknown acid, we need to know both its mass and the amount of substance in moles. In this exercise, we are given a sample mass of 0.1276 grams of the monoprotic acid. Now, let's focus on how we calculate the moles of the acid. Since the titration process involves a 1:1 molar reaction between the acid and the base (NaOH), the moles of acid at the equivalence point are equal to the moles of NaOH required to reach that point.

• First, calculate the moles of NaOH. Use the formula: \[ n = M \cdot V \] where \( M \) is the molarity and \( V \) is the volume in liters.
• Substitute the given values: \( n = 0.0633 \: \text{M} \times 0.0184 \: \text{L} = 1.16552 \times 10^{-3} \: \text{moles} \).

Since the moles of the acid equal the moles of NaOH, you have 1.16552 x 10^-3 moles of the acid. Now, calculate the molar mass using the molar mass formula:

• \[ \text{Molar Mass} = \frac{\text{mass of acid}}{\text{moles of acid}} \]
• Insert the values: \[ \text{Molar Mass} = \frac{0.1276 \: \text{g}}{1.16552 \times 10^{-3} \: \text{moles}} \approx 109.43 \: \text{g/mol} \].

In conclusion, the molar mass of the unknown acid is 109.43 g/mol. Understanding this calculation allows you to identify acids based not just on their formula but also on their molar mass.

Henderson-Hasselbalch Equation

This equation is a valuable tool in acid-base chemistry. Used predominantly to find the pH of buffer solutions, it relates the pH to the pKa and the concentrations of the acid \([HA]\) and its conjugate base \([A^-]\). In our exercise, after 10 ml of NaOH had been added, the pH was recorded as 5.87. Using the Henderson-Hasselbalch equation enables us to find the pKa of the unknown acid.
Let's break it down:

• The equation: \[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
• With the given pH being 5.87, substitute to find pK\_a: \[ 5.87 = pK_a + \log\left(\frac{0.01809}{0.0152}\right) \]
• The ratio of \([A^-]/[HA]\) reveals how much of the acid has converted into its conjugate base.

Solving the equation for pKa, you find that pKa is approximately 5.76. This means the acid has moderate strength, not too strong, and not too weak. Applying these methods helps deepening your understanding of buffer systems and equilibrium.

Acid Dissociation Constant

The acid dissociation constant, denoted as \(K_a\), reflects the extent to which an acid releases protons in solution. It is essential for understanding an acid's strength. In chemical terms, \(K_a\) quantifies the equilibrium concentration of hydrogen ions, conjugate base, and un-ionized acid. From our exercise, we first determined pKa through the Henderson-Hasselbalch equation, and we will now convert that into \(K_a\).
To calculate \(K_a\):

• Remember that \(pK_a\) is the negative logarithm of \(K_a\), given by \[ pK_a = -\log_{10}(K_a) \]
• Thus, you can derive \(K_a\) as follows: \[ K_a = 10^{-pK_a} \]
• Use the value found in the previous section: \[ K_a = 10^{-5.76} \approx 1.74 \times 10^{-6} \]

This \(K_a\) value suggests the unknown acid is weak, as stronger acids have larger \(K_a\) values. By knowing \(K_a\), you can predict how the acid will behave in different pH environments, fundamental to understanding chemical equilibria in solutions.