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.
<u>Answer:</u> The volume of stock solution needed is 90 mL
<u>Explanation:</u>
To calculate the molarity of the diluted solution, we use the equation:
where,
are the molarity and volume of the stock sulfuric acid solution
are the molarity and volume of diluted sulfuric acid solution
We are given:
Putting values in above equation, we get:
Hence, the volume of stock solution needed is 90 mL
Answer:
Explanation:
Your nuclear equation is
The main point to remember in balancing nuclear equations is that
- the sum of the superscripts and must be the same on each side of the equation.
- the sum of the subscripts must be the same on each side of the equation.
Then
85 = 0 + y, so y = 85 - 0 = 0
35 = -1 + x, so x = 35 + 1 = 36
The nucleus with atomic number 36 and atomic mass 85 is krypton-85.
The nuclear equation becomes
Answer:
May be the instrument is incorrect or may be error in it.
Explanation:
The copper have not been detected by this test because the test may be not for the detection of copper, may be it is used for identification of another minerals. If there is copper in the lake sample but can't be detected in the test so it means that the instrument which is used for detection is not the right one or having error in that instrument. Every mineral has a specific type of instrument that detect its presence, if we use incorrect instrument for the mineral then we can't detect the presence of that specific mineral.
Answer:
The volume you need to transfer from the stock solution is 0.145 l
Explanation:
Since the number of moles of lactose in the volume of stock solution that you transfer will be the same as the number of moles of lactose in the final solution, you can use this expression:
number of moles in volume to transfer = number of moles in the final solution
Since number of moles = concentration * volume (if the concentration is expressed in molarity), then:
Ci * Vi = Cf * Vf
where:
Ci = concentration of the stock solution.
Vi = volume of the stock solution to be transferred.
Cf = concentration of the final solution
Vf = volume of the final solution
Then, replacing with the data:
518 mM * Vi = 16.7 mM * 4.5 l
Vi = 16.7 mM * 4.5 l / 518 mM
<u>Vi = 0.145 l or 145 ml</u>
Notice that any concentration unit can be used, as long as the units of the concentration of the stock and final solution are the same.
