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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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g When 2.50 g of methane (CH4) burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion (in kJ) per mole
Anna [14]

Answer:

-800 kJ/mol

Explanation:

To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).

First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:

Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol

moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄

Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:

ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol

Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).

3 0
2 years ago
1. What is oxidation? * O a. it is loss of electrons O b. it is gain of electrons O c. it is loss of protons O d. none of the ab
Lemur [1.5K]

Answer:

yes

Explanation:

who wants free points

7 0
2 years ago
What is the partial pressure of argon, par, in the flask? express your answer to three significant figures and include the appro
monitta
Given which are missing in your question:
the flask is filled with 1.45 g of argon at 25 C° 
So according to this formula (Partial pressure):
PV= nRT
first, we need n, and we can get by substitution by:
n =  1.45/mass weight of argon
   = 1.45 / 39.948 = 0.0363 mol of Ar
we have R constant = 0.0821
and T in kelvin = 25 + 273 = 298
and V = 1 L
∴ P * 1 = 0.0363* 0.0821 * 298 = 0.888 atm

3 0
2 years ago
2. 27.8 mL of an unknown were added to a 50.0-mL flask that weighs 464.7 g. The total mass of the flask and the liquid is 552.4
agasfer [191]

Answer:

d=4.24x10^{-4}\frac{lb}{in^3}

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:

d=\frac{m}{V}

Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:

m=552.4g-464.7g=87.7g

So that we are now able to calculate the density in g/mL first:

d=\frac{87.7g}{27.8mL}=3.15g/mL

Now, we proceed to the conversion to lb/in³ by using the following setup:

d=3.15\frac{g}{mL}*\frac{1lb}{453.6g}*\frac{1in^3}{16.3871mL}\\\\d=4.24x10^{-4}\frac{lb}{in^3}

Regards!

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2 years ago
A balance is used to measure mass
FrozenT [24]

if this is true or false its true hope i helped

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