Problem 91 - 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 pH was found to be \(2.85\). What is the \(K_{a}\) for the unknown acid?

Step-by-Step Solution

Answer

Moles of \(\mathrm{OH^-}\) = \((0.0155\,\mathrm{L})\times(0.1150\,\mathrm{M})=0.0017825\,\mathrm{mol}\) #tag_title# Step 2: Calculate the molecular weight of the acid#tag_content# Knowing the moles of the acid, we can now calculate its molecular weight. We have the mass of the sample (0.1687 g) and the moles of the acid (0.0017825 mol). The molecular weight can be calculated using the following formula: Molecular weight = \(\frac{\text{Mass of acid}}{\text{Moles of acid}}\) Molecular weight = \(\frac{0.1687\,\mathrm{g}}{0.0017825\,\mathrm{mol}} = 94.61\,\mathrm{g/mol}\) #tag_title# Step 3: Find the concentration of ions at pH 2.85 after 7.25 mL of base is added#tag_content# From the given pH of 2.85, we can calculate the concentration of \(\mathrm{H^+}\) ions using the formula: \(\mathrm{pH}=-\log{[\mathrm{H^+}]}\) \[ [\mathrm{H^+}] = 10^{-\mathrm{pH}} = 10^{-2.85} = 1.4125\times 10^{-3}\,\mathrm{M}\] #tag_title# Step 4: Determine the \(K_{a}\) of the unknown acid#tag_content# Now that we have the concentration of \(\mathrm{H^+}\) ions, we can find the concentration of acid (\([\mathrm{HA}]\)) and its conjugate base (\([\mathrm{A^-}]\)). The concentration of \(\mathrm{A^-}\) ions is equal to the moles of \(\mathrm{OH^-}\) ions added (7.25 mL of 0.1150 M \(\mathrm{NaOH}\)). Moles of \(\mathrm{OH^-}\) = \((7.25\times 10^{-3}\,\mathrm{L})\times(0.1150\,\mathrm{M}) = 8.3375\times 10^{-4}\,\mathrm{mol}\) \[\text{Concentration of }\mathrm{A^-} = \frac{8.3375\times 10^{-4}\,\mathrm{mol}}{25.0\times 10^{-3}\,\mathrm{L} + 7.25\times 10^{-3}\,\mathrm{L}} = 2.703 \times 10^{-2}\,\mathrm{M}\] To find the concentration of \(\mathrm{HA}\), we can assume that the initial concentration of acid (before adding any base) remains the same: \[[\mathrm{HA}] = \frac{0.0017825\,\mathrm{mol}}{25.0\times 10^{-3}\,\mathrm{L}} - [\mathrm{A^-}] = 7.13\times 10^{-2}\,\mathrm{M}\] Now we can calculate the \(K_{a}\) value using the following equation: \[K_{a} = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}\] \[K_{a} = \frac{(1.4125\times 10^{-3}\,\mathrm{M})(2.703\times 10^{-2}\,\mathrm{M})}{(7.13\times 10^{-2}\,\mathrm{M})} = 9.064 \times 10^{-5}\] The molecular weight of the acid is approximately 94.61 g/mol and the \(K_{a}\) is \(9.064 \times 10^{-5}\).

Step 1: Calculate the moles of \(\mathrm{OH^-}\) ions required to reach equivalence

To reach the equivalence point, there must be equal moles of \(\mathrm{OH^-}\) ions and acid. We are given the volume of \(\mathrm{0.1150 M}\) \(\mathrm{NaOH}\) needed to reach the equivalence point (15.5 mL). To calculate the moles of \(\mathrm{OH^-}\) ions, we can use the following equation: Moles of \(\mathrm{OH^-}\) = (Volume of \(\mathrm{NaOH}\) in L) * (Concentration of \(\mathrm{NaOH}\) in M)

Key Concepts

Equivalence Point

Understanding the equivalence point is central to mastering titrations. It's the moment in a titration when the amount of added titrant is stoichiometrically equivalent to the amount of substance in the sample. For example, in a monoprotic acid titration, the equivalence point occurs when the moles of hydroxide ions (OH^-) added from the base (like NaOH) equal the moles of hydrogen ions (H^+) present in the acid.

At this juncture, all the acid has been neutralized, and it can be mathematically expressed by the moles of acid originally in solution. In practical terms, it's usually detected by a sudden change in pH or indicated by a color change if a pH indicator is used. Importantly, the equivalence point allows us to deduce valuable information such as the molecular weight of the acid in question by relating the moles of base to the mass of the acid.

Molecular Weight Calculation

The molecular weight of a compound is a critical piece of information, particularly when dealing with unknown substances. In our titration scenario, the molecular weight calculation starts by identifying the moles of base used to neutralize the acid. In this case, the problem stipulates a specific volume and molarity of NaOH used to reach the equivalence point.

Once we have the number of moles of base, stoichiometry tells us that, due to the 1:1 ratio in the reaction of a monoprotic acid with a base, this will also be the number of moles of acid present. To find the molecular weight (MW), we take the mass of the acid sample (given) and divide it by the moles of acid (calculated). Mathematically, it's MW = (mass of acid) / (moles of acid). This simple step provides the formula weight of the unknown acid.

Acid Dissociation Constant (Ka)

The acid dissociation constant, or Ka, is a quantitative measure of an acid's strength – the higher the Ka, the stronger the acid. It's determined by the ratio of the concentration of the dissociated form (ions) to the undissociated form (the intact acid).

To calculate the Ka value during a titration, we measure the pH at a point before the equivalence point – typically at half the equivalence volume. The pH value at this midway stage helps us infer the concentration of H^+ ions, and using the Henderson-Hasselbalch equation, we can compute the Ka. It's a bit like detective work, where we use the clues given at a particular point in the reaction to reveal the inherent properties of the acid being studied.

Stoichiometry

Stoichiometry is all about the quantitative relationships between the substances as they react together according to a balanced chemical equation. It forms the bedrock of chemistry calculation – converting between moles, grams, and molecules to predict the outcomes of reactions.

In the context of a titration, stoichiometry is used to determine how much titrant is needed to react completely with the analyte. For a monoprotic acid reacting with a base like NaOH, the ratio is often 1:1, meaning one mole of acid reacts with one mole of base. Understanding stoichiometry enables us to make accurate predictions and calculations about the amount of substances involved, and it's why we can confidently use the data from a titration to find other properties like molecular weight or the acid dissociation constant.