Problem 108 - Chemistry The Central Science In Si Units

Question

Trifluoroacetic acid has the chemical formula \(\mathrm{CF}_{3} \mathrm{CO}_{2} \mathrm{H}\). It is a colorless liquid that has a density of \(1.489 \mathrm{~g} / \mathrm{mL}\). (a) Trifluoroacetic acid contains one \(\mathrm{CF}_{3}\) unit and is connected to the other \(\mathrm{C}\) atom which bonds with both O's. Draw the Lewis structure for trifluoroacetic acid. (b) Trifluoroacetic acid can react with \(\mathrm{NaOH}\) in aqueous solution to produce the trifluoroacetate ion, \(\mathrm{CF}_{3} \mathrm{COO}^{-}\). Write the balanced chemical equation for this reaction. (c) Draw the Lewis structure of the trifluoroacetate ion, showing resonance if present. (d) How many milliliters of a \(0.500 \mathrm{M}\) solution of \(\mathrm{NaOH}\) would it take to neutralize \(10.5 \mathrm{~mL}\) of trifluoroacetic acid?

Step-by-Step Solution

Answer

275 mL of 0.500 M NaOH is needed.

Step 1: Lewis Structure of Trifluoroacetic Acid

To draw the Lewis structure for \( \text{CF}_3\text{CO}_2\text{H} \), first identify the central atoms and the bonding arrangement. The \( \text{CF}_3 \) group is attached to a \( \text{C} \) atom which is bonded to two \( \text{O} \) atoms and one \( \text{H} \) atom. Each \( \text{F} \) is singly bonded to the first \( \text{C} \) (\( \text{F}-\text{C} \)), while the second \( \text{C} \) atom is doubly bonded to one \( \text{O} \) and singly bonded to the other \( \text{O} \) with an \( \text{O}-\text{H} \) bond, forming a carboxylic acid group. The \( \text{C}-\text{O} \) single bond allows for the \( \text{O}-\text{H} \) linkage.

Step 2: Balanced Chemical Equation

The balanced chemical equation for the reaction of trifluoroacetic acid with sodium hydroxide (\( \text{NaOH} \)) is:\[ \text{CF}_3\text{COOH} + \text{NaOH} \rightarrow \text{CF}_3\text{COO}^- \text{Na}^+ + \text{H}_2\text{O} \] In this reaction, the hydrogen ion from the carboxylic acid is replaced by a sodium ion, forming water as a byproduct.

Step 3: Lewis Structure of Trifluoroacetate Ion With Resonance

The trifluoroacetate ion \( \text{CF}_3\text{COO}^- \) has resonance structures. Upon losing the \( \text{H}^+ \), the negative charge is delocalized over the two oxygen atoms. The primary resonance structures have one \( \text{C}-\text{O} \) as a double bond and the other \( \text{C}-\text{O} \) with a single bond carrying the negative charge; these structures alternate showing the delocalization of electrons.

Step 4: Calculate Volume of NaOH Solution Needed

First, find the moles of trifluoroacetic acid in \( 10.5 \text{ mL} \) using its density:\[ 10.5 \text{ mL} \times 1.489 \text{ g/mL} = 15.6345 \text{ g} \]Find moles of trifluoroacetic acid (molar mass \( \approx 114.02 \text{ g/mol} \)):\[ \frac{15.6345 \text{ g}}{114.02 \text{ g/mol}} \approx 0.1372 \text{ mol} \]Since the reaction equation shows a 1:1 molar ratio, moles of \( \text{NaOH} \) needed = moles of trifluoroacetic acid = \( 0.1372 \text{ mol} \).The volume of \( 0.500 \text{ M} \) \( \text{NaOH} \) solution required is:\[ V = \frac{0.1372 \text{ mol}}{0.500 \text{ mol/L}} = 0.2744 \text{ L} = 274.4 \text{ mL} \]

Key Concepts

Lewis Structure

Understanding the Lewis structure of a molecule is a fundamental step in visualizing its bonding and molecular geometry. For trifluoroacetic acid, \( \text{CF}_3\text{CO}_2\text{H} \), we start by recognizing the central structure. It consists of a \( \text{CF}_3 \) group connected to a central carbon atom, which is further bonded to two oxygen atoms and one hydrogen atom.
This forms the characteristic carboxylic acid group. In this structure:

• Every fluorine (\( \text{F} \)) is singly bonded to the first carbon (\( \text{C} \)).
• The central carbon atom forms a double bond with one oxygen (\( \text{C}=\text{O} \)).
• The other oxygen is connected through a single bond and is attached to a hydrogen (\( \text{O}-\text{H} \)).

Visualizing this can help in predicting properties and possible interactions in chemical reactions.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products through breaking and forming bonds. When trifluoroacetic acid reacts with sodium hydroxide (\( \text{NaOH} \)), a classic acid-base reaction occurs. Here’s what happens:

• The hydrogen ion \( \text{H}^+ \) from the carboxylic group is replaced by sodium \( \text{Na}^+ \).
• This produces trifluoroacetate ion \( \text{CF}_3\text{COO}^- \) and water \( \text{H}_2\text{O} \) as the byproduct.

The balanced equation for this reaction is:\[ \text{CF}_3\text{COOH} + \text{NaOH} \rightarrow \text{CF}_3\text{COO}^- \text{Na}^+ + \text{H}_2\text{O} \]This equation highlights the exchange of ions that characterizes many common neutralization reactions.

Resonance

Resonance is an essential concept in chemistry that describes the delocalization of electrons in a molecule or ion. For the trifluoroacetate ion (\( \text{CF}_3\text{COO}^- \)), resonance plays a crucial role.
Due to the loss of a hydrogen ion, a negative charge is introduced.

• This negative charge is shared between two oxygen atoms, not fixed to one.
• Multiple resonance structures can be drawn, illustrating that the double bond can shift between the two oxygen atoms.

This electron delocalization provides additional stability to the trifluoroacetate ion. Understanding resonance helps appreciate how molecules stabilize through electron sharing, rather than forming a singular rigid structure.

Stoichiometry

Stoichiometry is the area of chemistry that deals with the quantitative relationships in chemical reactions. In this exercise, we calculate the volume of \( \text{NaOH} \) needed to neutralize a given volume of trifluoroacetic acid.
First, we calculate the mass of the acid using its density, and then find the number of moles. The conversion is as follows:

• Using the density \( 1.489 \text{ g/mL} \), calculate the mass of \( 10.5 \text{ mL} \) trifluoroacetic acid.
• Convert this mass into moles using its molar mass (\( 114.02 \text{ g/mol} \)).

For a 1:1 stoichiometric ratio between the acid and \( \text{NaOH} \), the moles of acid equate the moles of base required.
Therefore, divide the moles by the molarity (\( 0.500 \text{ M} \) \( \text{NaOH} \)) to find the volume of base needed: \[ \text{Volume} = \frac{\text{moles of acid}}{\text{molarity of } \text{NaOH}} \]
This calculation leads us to find that \( 274.4 \text{ mL} \) of sodium hydroxide is required for neutralization.