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vaieri [72.5K]
3 years ago
12

Consider the pka (3.75) of formic acid, h-cooh as a reference. with appropriate examples, show how inductive, dipole, and resona

nce effects influence the strength of carboxylic acids.
Chemistry
1 answer:
Luden [163]3 years ago
3 0
Formic acid is the simplest carboxylic acid with a structure of HCOOH and has a pka of 3.75. The pka refers to the acidity of the molecule, which in this example refers to the molecules ability to give up the proton of the O-H. A decrease in the pka value corresponds to an increase in acidity, or an increase in the ability to give up a proton. When an acid gives up a proton, the remaining anionic species (in this case HCOO-) is called the conjugate base, and an increase in the stability of the conjugate base corresponds to an increase in acidity.

The pka of a carboxylic can be affected greatly by the presence of various functional groups within its structure. An example of an inductive effect changing the pka can be shown with trichloroacetic acid, Cl3CCOOH. This molecule has a pka of 0.7. The decrease in pka relative to formic acid is due to the presence of the Cl3C- group, and more specifically the presence of the chlorine atoms. The electronegative chlorine atoms are able to withdraw the electron density away from the oxygen atoms and towards themselves, thus helping to stabilize the negative charge and stabilize the conjugate base. This results in an increase in acidity and decrease in pka.

The same Cl3CCOOH example can be used to explain how dipoles can effect the acidity of carboxylic acids. Compared to standard acetic acid, H3CCOOH with a pka of 4.76, trichloroacetic acid is much more acidic. The difference between these structures is the presence of C-Cl bonds in place of C-H bonds. A C-Cl bond is much more polar than a C-H bond, due the large electronegativity of the chlorine atom. This results in a carbon with a partial positive charge and a chlorine with a partial negative charge. In the conjugate base of the acid, where the molecule has a negative charge localized on the oxygen atoms, the dipole moment of the C-Cl bond is oriented such that the partial positive charge is on the carbon that is adjacent to the oxygen atoms containing the negative charge. Therefore, the electrostatic attraction between the positive end of the C-Cl dipole and the negative charge of the anionic oxygen helps to stabilize the entire species. This level of stabilization is not present in acetic acid where there are C-H bonds instead of C-Cl bonds since the C-H bonds do not have a large dipole moment.

To understand how resonance can affect the pka of a species, we can simply compare the pka of a simple alcohol such as methanol, CH3OH, and formic acid, HCOOH. The pka of methanol is 16, suggesting that is is a very weak acid. Once methanol gives up that proton to become the conjugate base CH3O-, the charge cannot be stabilized in any way and is simply localized on the oxygen atom. However, with a carboxylic acid, the conjugate base, HCOO-, can stabilize the negative charge. The lone pair electrons containing the charge on the oxygen atom are able to migrate to the other oxygen atom of the carboxylic acid. The negative charge can now be shared between the two electronegative oxygen atoms, thus stabilizing the charge and decreasing the pka.
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How to determine how many valence electrons an element has?
In-s [12.5K]
Usually (ignoring transition metals, as they kinda get trickier), the element's valency can be found out by its group (column) number. Usually, we ignore the transition metal block while counting these columns, so Aluminium is in group 3, for example. Since Aluminium is in group 3, it has 3 valence electrons.
8 0
3 years ago
A hydrocarbon sample was burned in a bomb calorimeter. The temperature of the calorimeter and the 1.00 kg of water rose from 20.
fomenos

Answer:

The heat released by the combustion is 20,47 kJ

Explanation:

Bomb calorimeter is an instrument used to measure the heat of a reaction. The formula is:

Q = C×m×ΔT + Cc×ΔT

Where:

Q is the heat released

C is specific heat of water (4,186kJ/kg°C)

m is mass of water (1,00kg)

ΔT is temperature change (23,65°C - 20,45°C)

And Cc is heat capacity of the calorimeter (2,21kJ/°C)

Replacing these values the heat released by the combustion is:

<em>Q = 20,47 kJ</em>

6 0
3 years ago
Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.
maria [59]
The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
5 0
3 years ago
A solution is made by adding 35.5 mL of concentrated hydrochloric acid ( 37.3 wt% , density 1.19 g/mL1.19 g/mL ) to some water i
erastova [34]

Answer:

1.73 M

Explanation:

We must first obtain the concentration of the concentrated acid from the formula;

Co= 10pd/M

Where

Co= concentration of concentrated acid = (the unknown)

p= percentage concentration of concentrated acid= 37.3%

d= density of concentrated acid = 1.19 g/ml

M= Molar mass of the anhydrous acid

Molar mass of anhydrous HCl= 1 +35.5= 36.5 gmol-1

Substituting values;

Co= 10 × 37.3 × 1.19/36.5

Co= 443.87/36.6

Co= 12.16 M

We can now use the dilution formula

CoVo= CdVd

Where;

Co= concentration of concentrated acid= 12.16 M

Vo= volume of concentrated acid = 35.5 ml

Cd= concentration of dilute acid =(the unknown)

Vd= volume of dilute acid = 250ml

Substituting values and making Cd the subject of the formula;

Cd= CoVo/Vd

Cd= 12.16 × 35.5/250

Cd= 1.73 M

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3 years ago
Explain the relationship between the following terms: eutrophication, nutrients, dissolved oxygen leads, and algae.
makvit [3.9K]

Answer:

Eutrophication is the enrichment of a body of water with excessive nutrients (nitrogen and phosphorus), which causes algal growth and subsequent decline of dissolved oxygen after decomposition.

6 0
3 years ago
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