Acetic acid: ka = 1.75x105 Ammonia: ko = 1.8x105. Hydrofluoric acid: 7.2 x104 Hydrogen sulfide: kal = 5.7x108 Ka2 = 1.2x10-15 Solubility Product Constants, kp Nickel hydroxide 2.0x1014 Nickel sulfide: 1.4x10 24 Manganous 4.5 x1014 Manganous sulfide: 1.4 x1015 hydroxide: Aluminum chromium 1.0x10 30 hydroxide: 1.9x10 33 hydroxide: 1. The pka value of calcium fluoride is 10.41. Calculate the solubility in g/L at 25 C in: a. pure water b. 0.0200 M solution of calcium nitrate c. 0.0.30M hydrofluoric acid acid 2. A 200 ml of 0.05 F hydrogen sulfide solution is added to 25 ml 1.5 Facetic acid solution. Compute the sulfide ion concentration of the solutiow 3. A 50.0 ml of 0.02 F ammonia solution is added to a solution containing 0.004 g Ni2", 0.003 g Mn2+, and 0.002 g Cr2* per 750 ml a. calculate whether or not a precipitate will form b. if answer to (a) is yes, formulas of precipitates c. which ion will precipitate first d. what pH is necessary to prevent precipitation of all cations 4. Calculate the change in pH that results from adding 25 ml of 0.010F HCI to 500ml buffer solution which is 0.25M ammonia and 0.15M ammonium chloride.
Basics in Organic Reactions Mechanisms
In organic chemistry, the mechanism of an organic reaction is defined as a complete step-by-step explanation of how a reaction of organic compounds happens. A completely detailed mechanism would relate the first structure of the reactants with the last structure of the products and would represent changes in structure and energy all through the reaction step.
Heterolytic Bond Breaking
Heterolytic bond breaking is also known as heterolysis or heterolytic fission or ionic fission. It is defined as breaking of a covalent bond between two different atoms in which one atom gains both of the shared pair of electrons. The atom that gains both electrons is more electronegative than the other atom in covalent bond. The energy needed for heterolytic fission is called as heterolytic bond dissociation energy.
Polar Aprotic Solvent
Solvents that are chemically polar in nature and are not capable of hydrogen bonding (implying that a hydrogen atom directly linked with an electronegative atom is not found) are referred to as polar aprotic solvents. Some commonly used polar aprotic solvents are acetone, DMF, acetonitrile, DMSO, etc.
Oxygen Nucleophiles
Oxygen being an electron rich species with a lone pair electron, can act as a good nucleophile. Typically, oxygen nucleophiles can be found in these compounds- water, hydroxides and alcohols.
Carbon Nucleophiles
We are aware that carbon belongs to group IV and hence does not possess any lone pair of electrons. Implying that neutral carbon is not a nucleophile then how is carbon going to be nucleophilic? The answer to this is that when a carbon atom is attached to a metal (can be seen in the case of organometallic compounds), the metal atom develops a partial positive charge and carbon develops a partial negative charge, hence making carbon nucleophilic.
pls answer the thirs item
![Acetic acid: ka = 1.75x105
Ammonia: ko = 1.8x105. Hydrofluoric acid: 7.2 x104
Hydrogen sulfide: kal = 5.7x108
kaz = 1.2x10-15
Solubility Product Constants, kap
Nickel hydroxide 2.0x10-14
Nickel sulfide:
1.4x10-24
Manganous
4.5 x1014
Manganous sulfide: 1.4 x1015
hydroxide:
Aluminum
chromium
1.0x10 30
hydroxide:
1.9x10 33
hydroxide:
1. The pksa value of calcium fluoride is 10.41. Calculate the solubility in g/L at 25 oC in:
a. pure water
b. 0.0200 M solution of calcium nitrate
c. 0.0.30M hydrofluoric acid acid
2. A 200 ml of 0.05 F hydrogen sulfide solution is added to 25 mL 1.5 Facetic acid solution. Compute the sulfide
ion concentration of the solution
3. A 50.0 ml of 0.02 F ammonia solution is added to a solution containing 0.004 g Ni?, 0.003 g Mn2+, and 0.002
g Cr2 per 750 mla
a. calculate whether or not a precipitate will form
b. if answer to (a) is yes, formulas of precipitates
c. which ion will precipitate first
d. what pH is necessary to prevent precipitation of all cations
4. Calculate the change in pH that results from adding 25 ml of 0.010F HCI to 500ml buffer solution which is
0.25M ammonia and 0.15M ammonium chloride.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F65995b94-2dd7-4a67-a870-8a89877dce6c%2F335c8036-a284-46c9-a67f-e1e435a49019%2F9t8mtde_processed.png&w=3840&q=75)
![](/static/compass_v2/shared-icons/check-mark.png)
Step by step
Solved in 5 steps with 4 images
![Blurred answer](/static/compass_v2/solution-images/blurred-answer.jpg)
![Chemistry](https://www.bartleby.com/isbn_cover_images/9781305957404/9781305957404_smallCoverImage.gif)
![Chemistry](https://www.bartleby.com/isbn_cover_images/9781259911156/9781259911156_smallCoverImage.gif)
![Principles of Instrumental Analysis](https://www.bartleby.com/isbn_cover_images/9781305577213/9781305577213_smallCoverImage.gif)
![Chemistry](https://www.bartleby.com/isbn_cover_images/9781305957404/9781305957404_smallCoverImage.gif)
![Chemistry](https://www.bartleby.com/isbn_cover_images/9781259911156/9781259911156_smallCoverImage.gif)
![Principles of Instrumental Analysis](https://www.bartleby.com/isbn_cover_images/9781305577213/9781305577213_smallCoverImage.gif)
![Organic Chemistry](https://www.bartleby.com/isbn_cover_images/9780078021558/9780078021558_smallCoverImage.gif)
![Chemistry: Principles and Reactions](https://www.bartleby.com/isbn_cover_images/9781305079373/9781305079373_smallCoverImage.gif)
![Elementary Principles of Chemical Processes, Bind…](https://www.bartleby.com/isbn_cover_images/9781118431221/9781118431221_smallCoverImage.gif)