Question

1) The average human body contains 5.70 L of blood with a Fe2+ concentration of 2.60×10−5 M . If a person ingests 12.0 mL of 16.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion? (When species combine to produce a coordination complex, the equilibrium constant for the reaction is called is the formation constant, Kf.

For example, the iron(II) ion, Fe2+, can combine with the cyanide ion, CN−, to form the complex [Fe(CN)6]4− according to the equation

Fe2+(aq)+6CN−(aq)⇌[Fe(CN)6]4−(aq)

where Kf=4.21×1045.)

2) Consider a solution that is 1.5×10−2 M in Ba2+and 1.9×10−2 M in Ca2+.

Ksp(BaSO4)=1.07×10−10
Ksp(CaSO4)=7.10×10−5
What minimum concentration of Na2SO4 is required to cause the precipitation of the cation that precipitates first?

3) Calculate the solubility (in grams per 1.00×102mL of solution) of magnesium hydroxide in a solution buffered at pH = 12.
Calculate the solubility (in grams per 1.00×102mL of solution) of magnesium hydroxide in pure water.
How does the solubility of Mg(OH)2 in a buffered solution compare to the solubility of Mg(OH)2 in pure water?

4) Use the appropriate values of Ksp and Kf to find the equilibrium constant for the following reaction:
FeS(s)+6CN−(aq)⇌Fe(CN)4−6(aq)+S2−(aq)

5) A 115.0 −mL sample of a solution that is 3.0×10−3 M  in AgNO3  is mixed with a 225.0 −mL sample of a solution that is 0.11 M in NaCN. A complex ion forms.After the solution reaches equilibrium, what concentration of Ag+(aq) remains?   

Answer 1

2)   

Ksp(BaSO4)=1.07×10^−10 , Ksp(CaSO4)=7.10×10^−5

Ksp of BaSO4 lesser than CaSO4 . so BaSO4 will precipitate first

BaSO4 ------------------> Ba+2 + SO4-2

Ksp = [Ba+2] [SO4-2]

1.07×10^−10 = (1.5×10^−2) [SO4-2]

[SO4-2] = 1.5 x 10^-2 M

Na2SO4 concentration = 1.5 x 10^-2 M

3) magnesium hydroxide in a solution buffered at pH = 12.

A) volume = 1.00 x 10^2 mL = 100 mL

pH = 12

pOH = 14 -12 = 2

[OH-] = 10^-2 = 0.01 M

Mg (OH)2 <-----------------> Mg+2 + 2OH-

S 0.01

Ksp = [Mg+2] [OH-]^2

5.61 x 10^-12 = S x (0.01)^2

S = 5.61 x 10^-8 M

molar solubility = 5.61 x 10^-8 mol / L

= 5.61 x 10^-8 x (58.32) g / L

= 3.27 x 10^-6 g / L

= 3.27 x 10^-6 g x 1000 / 100

= 3.27 x 10^-5 g / 100 mL

B ) in pure water.

Ksp = [Mg+2] [OH-]^2

5.61 x 10^-12 = S x (2S)^2

S = 1.12 x 10^-4 mol / L

S = 1.12 x 10^-4 x 58.32 x 1000 / 100

S = 0.0653 g / 100 mL

solubility = 0.0653 g / 100 mL

C)

due to common ion effect the solubility of Mg(OH)2 in a buffered solution is lesser than to the solubility of Mg(OH)2 in pure water.