Problem 89 - Chemistry The Central Science

Question

A sample of \(0.1687 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.1150 \mathrm{M} \mathrm{NaOH}\). The acid required \(15.5 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molecular weight of the acid? (b) After \(7.25 \mathrm{~mL}\) of base had been added in the titration, the \(\mathrm{pH}\) was found to be 2.85 . What is the \(K_{a}\) for the unknown acid?

Step-by-Step Solution

Answer

The molecular weight of the unknown monoprotic acid is approximately \(94.6 \mathrm{~g/mol}\), and the \(K_a\) of the acid is approximately \(1.95 \times 10^{-3}\).

Step 1: 1. Calculate moles of NaOH at the equivalence point

At the equivalence point, the number of moles of NaOH is equal to the number of moles of the unknown acid. We can find the moles of NaOH using the volume of NaOH titrated and its molarity. Moles of NaOH = Molarity × Volume Moles of NaOH = 0.1150 M × 15.5 mL × (1 L / 1000 mL) Moles of NaOH = 0.0017825 mol

Step 2: 2. Calculate moles of the acid

Since the moles of NaOH and the acid are equal at the equivalence point, the moles of the unknown acid are also 0.0017825 mol.

Step 3: 3. Calculate molecular weight of the acid

We can use the mass and moles of the acid to calculate the molecular weight. Molecular weight = mass / moles Molecular weight = 0.1687 g / 0.0017825 mol Molecular weight ≈ 94.6 g/mol So, the molecular weight of the acid is approximately 94.6 g/mol.

Step 4: 4. Calculate molarity of the acid

Now we can calculate the molarity of the acid using the volume of water it is dissolved in: Molarity of the acid = moles / volume Molarity of the acid = 0.0017825 mol / 0.025 L Molarity of the acid = 0.0713 M

Step 5: 5. Calculate moles of NaOH after 7.25 mL of base

We'll use the same molarity formula to calculate the moles of NaOH present after 7.25 mL of base has been added: Moles of NaOH = Molarity × Volume Moles of NaOH = 0.1150 M × 7.25 mL × (1 L / 1000 mL) Moles of NaOH = 0.00083375 mol

Step 6: 6. Calculate moles of the acid after 7.25 mL of base

To calculate the moles of the acid after 7.25 mL of base has been added, we subtract the moles of NaOH from the initial moles of the acid: Moles of the acid = Initial moles - moles of NaOH Moles of the acid = 0.0017825 mol - 0.00083375 mol Moles of the acid = 0.00094875 mol

Step 7: 7. Calculate the Ka constant

We have the pH and the moles of the acid and base after 7.25 mL of base has been added. We can use the Henderson-Hasselbalch equation to find the Ka: \[pH = pKa + \log \frac{[A^-]}{[HA]}\] Rearrange the equation to find pKa: \[pKa = pH - \log \frac{[A^-]}{[HA]}\] Since we know the pH, we need to find the concentrations of [A^-] and [HA]. The total volume of the solution at this point is the sum of the initial volume of the acid (25 mL) and the volume of NaOH added (7.25 mL): Total volume = 25 mL + 7.25 mL = 32.25 mL = 0.03225 L [A^-] = moles of NaOH / total volume [A^-] = 0.00083375 mol / 0.03225 L [A^-] = 0.02585 M [HA] = moles of the acid / total volume [HA] = 0.00094875 mol / 0.03225 L [HA] = 0.02942 M Now we can find the pKa: \[pKa = 2.85 - \log \frac{0.02585}{0.02942}\] \[pKa ≈ 2.71\] Finally, to find the Ka, we use: \[Ka = 10^{-pKa}\] \[Ka ≈ 1.95 × 10^{-3}\] So, the Ka of the unknown acid is approximately 1.95 × 10^{-3}.

Key Concepts

Molecular Weight Calculation

Understanding the molecular weight of a compound is crucial in chemistry because it gives insight into the compound's composition and how it behaves in reactions. To calculate the molecular weight of an unknown acid during titration, we use its mass and the amount in moles it reacts with.
To start, determine the moles of the titrant, in this case, NaOH, used at the equivalence point. At the equivalence point, it is known that the moles of titrant and the moles of unknown acid are equal. Therefore:

• Calculate moles of NaOH: Moles = Molarity × Volume
• In our scenario, that is: \(0.1150\, \text{M} \times 15.5\, \text{mL} \times \frac{1\, \text{L}}{1000\, \text{mL}} = 0.0017825\, \text{mol}\)

Knowing the moles of the acid lets us calculate the molecular weight, using its given mass.
The molecular weight formula is:

\[\text{Molecular Weight} = \frac{\text{Mass}}{\text{Moles of Acid}}\]

Here, \(0.1687\, \text{g} \div 0.0017825\, \text{mol} \approx 94.6\, \text{g/mol}\).
In summary, finding the molecular weight provides a vital step in identifying the unknown acid's chemical characteristics.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is an essential tool for understanding the properties of a buffer solution, providing a direct relationship between pH, pKa (the negative base-10 logarithm of the acid dissociation constant), and the concentrations of an acid and its conjugate base.
The equation is given as:

\[pH = pKa + \log \frac{[A^-]}{[HA]}\]

In the context of a titration experiment, it helps calculate the pKa by considering the amount of base added. This comes into play particularly after a portion of the base is added but before reaching the equivalence point.

• First, determine the changes in concentration of the acid and its conjugate base once some NaOH is added.
• Use these concentrations in the equation to find the pKa.

From the given values:

• Total Volume after base addition = Initial acid volume + Volume NaOH = \(25\,\text{mL} + 7.25\,\text{mL} = 32.25\,\text{mL}\)
• Concentration of conjugate base \([A^-]\) = Moles of NaOH/Total Volume = \(0.02585\, \text{M}\)
• Concentration of acid \([HA]\) = Moles of Acid/Total Volume = \(0.02942\, \text{M}\)

Plugging into the Henderson-Hasselbalch equation provides:

\[pKa = 2.85 - \log \frac{0.02585}{0.02942} \approx 2.71\]

This simple step-by-step calculation helps determine the pKa, crucial for further calculating \(Ka\), the acid dissociation constant.

Ka Determination

The acid dissociation constant, \(K_a\), is a measure of an acid's strength. It reflects the extent of ionization of an acid in solution. For monoprotic acids, determining \(K_a\) helps understand how readily the acid donates protons.
From the problem, the pKa is computed using the Henderson-Hasselbalch equation. Once the pKa is known, it can be converted to \(K_a\) using the formula:

\[Ka = 10^{-pKa}\]

In our case, the calculated pKa is approximately 2.71. Therefore:

\[Ka \approx 10^{-2.71} \approx 1.95 \times 10^{-3}\]

This calculation shows that the unknown monoprotic acid partially dissociates in the solution, providing a quantitative measure of its acid strength.
Understanding \(K_a\) allows chemists to better predict the behavior of the acid in various chemical reactions and also compare it with other acids. Knowledge of \(K_a\) is essential in titration studies and crucial in understanding acids' practical applications.