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

01

Determine the total number of valence electrons

Trifluoroacetic acid has the chemical formula \(\mathrm{CF}_{3}\mathrm{CO}_{2}\mathrm{H}\). The total number of valence electrons for each atom are, C = 4, F = 7 (3 times), O = 6 (2 times) and H = 1, which leads to a total of 4 + (3*7) + (2*6) + 1 = 32 valence electrons.

02

Draw the central atom and connect outer atoms

Place the C atom in the center and connect it to the F and O atoms. Remember that C can create four covalent bonds. Place the three F atoms around the central C to form a \(\mathrm{CF}_{3}\) unit and bond the other O atoms creating 3 single bonds. Attach the H atom to the second O atom, which has a single bond with the C atom.

03

Complete the octets for outer atoms

Complete the octet for the outer atoms (F and O) by adding lone pair electrons. Three F atoms have one bond, so add three lone pairs (6 electrons) to each F atom. One O atom has a single bond with C, so add three lone pairs (6 electrons) to that O atom. The second O atom has double bond with C, so add two lone pairs (4 electrons) to that O atom. The H atom has one bond, so leave it as is.

04

Check if the central atom has a complete octet

After Steps 2 and 3, count the electron pairs around the central C atom. There should be four pairs. Two single bonds, with the F atom and the O atom; one double bond with the second O atom. The central C atom now has a complete octet. b) Write the balanced chemical equation for the reaction of trifluoroacetic acid with NaOH:

05

Write the unbalanced equation

In this reaction, trifluoroacetic acid (\(\mathrm{CF}_{3}\mathrm{CO}_{2}\mathrm{H}\)) reacts with sodium hydroxide (NaOH) to produce the trifluoroacetate ion (\(\mathrm{CF}_{3}\mathrm{COO}^{-}\)) and water (H\(_{2}\)O): \( \mathrm{CF}_{3}\mathrm{CO}_{2}\mathrm{H} + \mathrm{NaOH} \rightarrow \mathrm{CF}_{3}\mathrm{COO}^{-} + \mathrm{H}_{2}\mathrm{O}\)

06

Balance the equation

Since there is one of each element on both sides, the equation is already balanced: \( \mathrm{CF}_{3}\mathrm{CO}_{2}\mathrm{H} + \mathrm{NaOH} \rightarrow \mathrm{CF}_{3}\mathrm{COO}^{-} + \mathrm{H}_{2}\mathrm{O}\) c) Draw the Lewis structure of the trifluoroacetate ion, showing resonance if present: The process is similar to drawing the Lewis structure for trifluoroacetic acid. Just replace the H atom in the carboxyl group with an O atom to form the trifluoroacetate ion. The octets would complete, but there will be a negatively charged oxygen atom. d) Calculate the milliliters of a 0.500 M solution of NaOH needed to neutralize 10.5 mL of trifluoroacetic acid:

07

Calculate moles of trifluoroacetic acid

First, calculate the moles of trifluoroacetic acid present in 10.5 mL. To do this, use the density and convert mL to grams: 10.5 mL * 1.489 g/mL = 15.6345 g Trifluoroacetic acid The molar mass of trifluoroacetic acid is \(\mathrm{C}=12.01 (\mathrm{g/mol})+3\mathrm{F}=3\times19 (\mathrm{g/mol})+\mathrm{C}=12.01 (\mathrm{g/mol})+2\mathrm{O}=2\times16 (\mathrm{g/mol})+\mathrm{H}=1 (\mathrm{g/mol})\Rightarrow m=2\times12.01+3\times19+2\times16+1=114.02 \mathrm{g/mol}\) By dividing the grams of trifluoroacetic acid by its molar mass, the number of moles is obtained: 15.6345 g / 114.02 g/mol = 0.137 moles Trifluoroacetic acid

08

Calculate moles of NaOH required for neutralization

From the balanced equation, it is known that 1 mole of trifluoroacetic acid reacts with 1 mole of NaOH. Therefore, 0.137 moles of NaOH are needed to neutralize the given amount of trifluoroacetic acid.

09

Calculate the volume of 0.500 M NaOH solution needed

Now, calculate the volume of the 0.500 M NaOH solution required to neutralize 0.137 moles of trifluoroacetic acid: Volume = moles / concentration Volume = 0.137 moles / 0.500 M = 0.274 L Convert the volume to mL: 0.274 L * 1000 mL/L = 274 mL So, 274 mL of a 0.500 M solution of NaOH is needed to neutralize 10.5 mL of trifluoroacetic acid.

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