Problem 27 - Chemical Principles

Question

Suppose that \(4.25 \mathrm{~g}\) of an unknown monoprotic weak acid, \(\mathrm{HA}\), is dissolved in water. Titration of the solution with \(0.350 \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq})\) required \(52.0 \mathrm{~mL}\) to reach the stoichiometric point. After the addition of \(26.0 \mathrm{~mL}\), the \(\mathrm{pH}\) of the solution was found to be \(3.82\). (a) What is the molar mass of the acid? (b) What is the value of \(\mathrm{p} K_{\mathrm{a}}\) for the acid?

Step-by-Step Solution

Answer

The molar mass of the acid is found after calculating the moles of HA at the stoichiometric point and equating it to the moles of NaOH used. pKa is calculated based on the pH at half-titration and the molar concentration of HA remaining at that point.

Step 1: Calculate moles of NaOH used at the stoichiometric point

To determine the moles of NaOH used at the stoichiometric point, use the molarity and the volume of the NaOH solution. The formula to convert from molarity (M) and volume (L) to moles is moles = Molarity × Volume. First, convert the volume from mL to L by dividing by 1000. Moles of NaOH = 0.350 M × (52.0 mL / 1000).

Step 2: Determine moles of HA at the stoichiometric point

Since HA is a monoprotic acid, it reacts with NaOH in a 1:1 molar ratio, this means the moles of HA is equal to the moles of NaOH at the stoichiometric point. Use the moles calculated in Step 1 for the moles of HA.

Step 3: Calculate the molar mass of the acid, HA

The molar mass of HA can be calculated using the mass of HA and the moles of HA at the stoichiometric point. Molar mass (M) of HA = Mass of HA / Moles of HA.

Step 4: Calculate the initial moles of HA before adding NaOH

To find the initial moles of HA, use the molar mass calculated in Step 3 and the given mass of HA. Initial moles of HA = 4.25 g / Molar mass of HA.

Step 5: Calculate the moles of HA remaining after adding 26.0 mL of NaOH

Calculate the moles of NaOH added when 26.0 mL was introduced using the molarity of NaOH and converting the volume to liters. Moles of NaOH at 26.0 mL = 0.350 M × (26.0 mL / 1000). The moles of HA that have reacted = Moles of NaOH at 26.0 mL. The moles of HA remaining = Initial moles of HA - Moles of HA that have reacted.

Step 6: Calculate the concentration of HA remaining

The concentration of HA ([HA]) is equal to the moles of HA remaining divided by the total volume of the solution in liters. The total volume is the initial volume of the HA solution plus 26.0 mL of NaOH added. [HA] = Moles of HA remaining / Total Volume in L.

Step 7: Convert pH to concentration of H+

The pH is the negative logarithm of the hydrogen ion concentration. To find the concentration of H+, use the formula [H+] = 10^(-pH).

Step 8: Calculate the value of Ka for the acid

Using the expression for the equilibrium constant, Ka, for the weak acid HA (Ka = [H+] [A-] / [HA]) and knowing that [H+] = [A-] at half-titration (which is the point where pH was measured), we can rearrange the formula to solve for Ka. Since [H+] = [A-], the formula simplifies to Ka = [H+]^2 / [HA].

Step 9: Calculate the pKa of the acid

The pKa is the negative logarithm of the Ka value. pKa = -log(Ka).

Key Concepts

Molar Mass Calculation

Understanding how to calculate molar mass is fundamental in chemistry, particularly in titration problems like the one presented. In essence, the molar mass is the mass of one mole of a substance, which is the molecular weight in grams per mole (g/mol). To calculate it, you need the mass of the substance and the number of moles.

Let's talk about how we calculated the molar mass of the unknown acid HA in the exercise. We started by finding the moles of NaOH at the stoichiometric point, which is where the acid and base completely neutralize each other. Since the reaction is 1:1, this also gave us the moles of HA. We then took the initial mass of the acid (4.25 g) and divided it by the moles of HA calculated from the titration with NaOH to determine the molar mass of the acid. This step is crucial because the molar mass ties together mass and mole quantities, two of the core foundations of stoichiometry.

pH and pKa Relationship

The pH and pKa values are quintessential parameters in acid-base chemistry. The pH quantifies the acidity or basicity of a solution by representing the concentration of hydrogen ions (H+). On the other hand, pKa is the acid dissociation constant and it's a characteristic property of an acid that quantifies its strength.

The relationship between pH and pKa comes into play when dealing with buffer solutions or during titration of weak acids, as in our exercise. At halfway to the equivalence point, called the half-neutralization or half-titration point, the pH of the solution equals the pKa of the weak acid because the amount of the acid equals the amount of its conjugate base. In this scenario, the Henderson-Hasselbalch equation (pH = pKa + log([A-]/[HA])), usually used to find the pH of a buffer, simplifies to pH = pKa. This is exactly what we utilized in the problem to determine the acid's pKa after figuring out the pH at the half-titration point.

Weak Acid Titration

Titration is an analytical technique used to determine the concentration of an unknown solution. When a weak acid is titrated, as in our example, it's important to realize that the weak acid doesn't fully dissociate in water. As NaOH is added during the titration, it reacts with the HA to form water and the conjugate base A-.

At the halfway point of the titration, the amounts of HA and A- are equal. This is important for calculating the pKa. As you add more NaOH, you decrease the concentration of HA while increasing the concentration of A-. Once you reach the stoichiometric point (equivalence point), all the original HA has reacted and you can use the amount of NaOH added to determine the original concentration of HA. This relationship is fundamental to characterizing the acid involved and understanding the titration curve.

Stoichiometry

Stoichiometry is the portion of chemistry that speaks to the quantification of reactants and products in chemical reactions. It allows chemists to make predictions about the outcomes of those reactions. Essential to stoichiometry is the mole concept, which offers a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure.

In the problem at hand, stoichiometry was the link between the mass of the unknown acid and the volume of NaOH solution used. By understanding the stoichiometry of the neutralization reaction, which in this case is a 1:1 molar ratio of HA to NaOH, we could calculate the molar mass of HA and use the midpoint titration data to calculate the pKa. With stoichiometry, we weren't just mixing chemicals; we were quantitatively analyzing the acid to reveal its identity.