Problem 90 - Chemistry The Central Science

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 pH was found to be \(6.50 .\) What is the \(K_{a}\) for the unknown acid?

Step-by-Step Solution

Answer

The molar mass of the unknown acid is calculated as follows: \(molar\ mass = \frac{0.2140 g}{(0.0950 M \times 0.0300 L)}\). After finding the moles of the unknown acid, we determine the Ka using the formula \(Ka = 10^{-6.50}\).

Step 1: Calculate moles of NaOH used to reach the equivalence point

At the equivalence point, the moles of the unknown acid equal the moles of NaOH. We are given the volume and concentration of NaOH: Volume of NaOH = 30.0 mL (convert to L by dividing by 1000) Concentration of NaOH = 0.0950 M We can find the moles of NaOH using the formula: moles = concentration × volume: moles of NaOH = 0.0950 M × 0.0300 L

Step 2:Find the moles of the unknown acid

Since the moles of the unknown acid equal the moles of NaOH at the equivalence point, we can say: moles of unknown acid = moles of NaOH

Step 3: Calculate the molar mass of the unknown acid

To find the molar mass of the acid, we will use the formula: molar mass = mass / moles We are given the mass of the unknown acid (0.2140g), and we have just calculated the moles. Now we can solve for molar mass: molar mass = 0.2140 g / moles of unknown acid

Step 4: Use the titration halfway point to find the Ka of the acid

At the halfway point of the titration, the pH equals the pKa of the acid. In this case, the pH was found to be 6.50, so pKa = 6.50. Ka is the antilogarithm of the pKa, so to find the Ka, use the formula: Ka = 10^(-pKa) Ka = 10^(-6.50)

Key Concepts

Molar Mass Calculation

When determining the molar mass of an unknown substance, especially during an acid-base titration, specific steps need to be followed. The molar mass is obtained by dividing the mass of the substance by the number of moles present.
Here’s how it's done in the given scenario:

• First, calculate the moles of NaOH used by multiplying its concentration by the volume in liters. In our example, that's 0.0950 M multiplied by 0.0300 L.
• Understand that at the equivalence point, the moles of acid equal the moles of NaOH. So, once you have the moles of NaOH, you also know the moles of acid.
• Finally, divide the given mass of the acid, 0.2140 g, by the moles of acid that you've just found to get the molar mass.

This calculation is crucial for identifying unknown compounds or verifying purity.

Equivalence Point

In a titration, the equivalence point is when the exact amount of titrant required to neutralize the analyte has been added. At this point, moles of acid equal the moles of base.
Reaching the equivalence point allows for the accurate calculation of unknown concentrations and characteristics of the sample.

• In the monoprotic acid example, 30.0 mL of 0.0950 M NaOH was the point at which the moles matched those of the acid perfectly.
• This is detectable by a color change if an indicator is used, or by a pH meter showing a sudden change in pH level.

Knowing the equivalence point is key to accurate molarity and mass determinations in titration.

pKa and Ka Relation

The pKa and Ka values describe the strength of an acid. pKa is simply the negative logarithm of the Ka value, which expresses the acid's dissociation in water.
The smaller the pKa, the stronger the acid. Here’s how to find it during a titration:

• In the example, the pH at half the equivalence point was 6.50, thus indicating pKa is also 6.50.
• The Ka value itself is derived by taking the antilog of the negative pKa: Ka = 10^(-pKa).
• This conversion helps chemists understand how completely an acid can lose a proton, impacting reaction dynamics.

Understanding pKa and Ka is crucial when predicting how a particular acid will behave in solution.

Monoprotic Acid

A monoprotic acid is an acid that donates one proton, or hydrogen ion, per molecule during the dissociation process. When dissolving in water, it releases one hydrogen ion.
This property affects how an acid's reaction is computed in a titration:

• Monoprotic acids include common examples like hydrochloric acid (HCl) and acetic acid (CH₃COOH).
• In calculations, this simplifies the process as it's a 1:1 mole ratio with the base in titrations, like our scenario with NaOH.

Knowing whether an acid is monoprotic or not is essential for understanding its behavior in chemical reactions and titrations.