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Organic Chem Precipitation

Essay by   •  February 24, 2018  •  Exam  •  1,604 Words (7 Pages)  •  827 Views

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Will a precipitate form when 1.0 ppm fluoride is added to drinking water that contains 0.010 M Ca2+? What is the maximum concentration of Ca2+ to which 1.0 ppm fluoride can be added without a precipitate forming? 
Unbalance reaction: CaF2 (s) [pic 1] Ca2+(aq) + F-(aq)         Ksp = 4.9x10-11.

1. The pre-equilibrium concentrations are 1.0 ppm for fluoride, and 0.010 M for Ca2+. ppm is parts-per-million, and is the same as mg/L. For this problem we want the [F-] in molarity:

              1.0 mg/L      1 g

[F-] in M =  ---------- x -------  =  5.2x10-5 M

             19.0 g/mol   1000 mg

2. The balanced reaction is: CaF2 (s) [pic 2] Ca2+(aq) + 2 F-(aq)

and the equilibrium constant expression for this reaction: Ksp = [Ca2+][F-]2

3. Q = [Ca2+][F-]2

Q = (0.010)(5.2x10-5)2

Q = 2.7x10-11

Since Q < Ksp, the reaction will proceed in the forward direction and no precipitate will form.

4. Step 4. is not necessary in this problem.

5. The second part of this problem asks for the maximum concentration of Ca2+ to which 1.0 ppm fluoride can be added without a precipitate forming. We can calculate the concentration of Ca2+ at which a precipitate forms. Use the equilibrium constant expression and a fluoride concentration of 1.0 ppm (5.2x10-5 M):

Ksp = [Ca2+][F-]2

4.9x10-11 = [Ca2+](5.2x10-5 M)2

[Ca2+] = 1.8x10-2 M

[Ca2+] < 1.8x10-2 M

Note that since precipitation depends on [F-]2, increasing the fluoride to 10.0 ppm would lower the maximum Ca2+ concentration to 1.8x10-4 M.

Will a precipitate form in a sample of rain runoff from an old coal mine if it is neutralized to a pH of 7.0 with 1 M NaOH? Assume that the water sample contains approximately 0.001 M Fe3+and 0.010 M H2SO4. If a precipitate forms, what is the resulting Fe3+(aq) concentration? What is the Fe3+ concentration if enough NaOH is added to raise the pH of the solution to 10.0?
Unbalance reaction: Fe(OH)3 (s) [pic 3] Fe3+(aq) + OH-(aq)         Ksp = 4x10-38.

1. Note that exact volumes and concentrations are not given for the original solution, but that NaOH is added to the sample until it reaches a pH of 7. At this point the solution will be at equilibrium, with or without a precipitate, and the equilibrium OH-(aq) concentration is therefore 1.0x10-7 M. Since NaOH was added to the original sample, the [Fe3+] will be less than the original concentration, but it probalby won't be a lot less. To determine if a precipitate forms, we can try a Q calculation assuming that the solution was diluted no more than a factor of 10, so take [Fe3+] to be 0.0001 M.

2. The balanced reaction is: Fe(OH)3 (s) [pic 4] Fe3+(aq) + 3 OH-(aq)

and the equilibrium constant expression for this reaction is: Ksp = [Fe3+][OH-]3

3. Q = [Fe3+][OH-]3

Assuming [Fe3+] = 0.0001 M.

Q = (0.0001 M Fe3+)(1.0x10-7 M OH-)3

Q = 1.0x10-25

Since Q > Ksp, the reaction will proceed in the reverse direction and a precipitate will form. Note that even assuming a dilution factor of ten for [Fe3+], Q is still > > Ksp.

4. Step 4. is not necessary in this problem.

5. We can now calculate the equilibrium concentration of Fe3+ using the equilibrium constant expression:

Ksp = [Fe3+][OH-]3

4x10-38 = [Fe3+](1.0x10-7)3

[Fe3+] = 4x10-17 M

Check results: Q = (4x10-17)(1.0x10-7)3 =4x10-38. Q = Keq, so the system is at equilibrium.

The second part of this problem is solved in the same way. First find [OH-] for a pH of 10:

pOH = 14 - 10 = 4
[OH
-] = 10-4

Now solve for [Fe3+]: 4x10-38 = [Fe3+](1.0x10-4)3

[Fe3+] = 4x10-26 M

Check results: Q = (4x10-26)(1.0x10-4)3 =4x10-38. Q = Keq, so the system is at equilibrium.

Note that the [Fe3+] is much lower at pH=10 than at pH=7. This decrease in the solubility of the solid due to the presence of one of the ions from another source is called the common-ion effect. When the [OH-] went up, the [Fe3+] had to decrease to maintain equilibrium.

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