Chemistry 115 Quiz 2

April 19, 2001 Name ___________________________________

  1. 200 mL of 0.150 M Al2(SO4)3 solution is added to 300 mL of 0.100 M NaOH. Calculate (4)
  1. the number of moles of Al(OH)3 formed. SO42- and Na+ are spectator ions

    2. Al3+ +

    3 OH- à

    Al(OH)3

    2 x 200mL x 0.150M = 60.0mmol

    300mL x 0.100M = 30.0 mmol

    0

    -10 mmol = 50.0mmol

    Limiting reagent -30.0 mmol = 0

    = 10.0 mmol formed.

    200 + 300 = 500 mL total volume

    Moles = |_0.010 moles__

  2. the final molar concentration of each ion.

Spectator ions are changed only by the new combined volume.

[SO42-] = 3 x 30.0 mmol / 500 mL = 90.0 mmol / 500 mL = 0.180 M

[Na+] = 30.0 mmol / 500 mL = 0.0600 M

Excess Al3+ will determine its concentration, except for very small portion from solubility of Al(OH)3

[Al3+] = 50.0 mmol / 500 mL = 0.10 M

Ksp = [Al3+][OH-]3 = 1.3 x 10-33 = [x + 0.10][3x]3 ~ 0.10 x 27x3

x ~ (1.3 x 10-33 / 2.7)1/3 ~ (4.81 x 10-34)1/3 = 7.8 x 10-12

[OH_] = 3x = 3(7.8 x 10-9) = 2.35 x 10-11

[Al3+] = |_0.100 M____

[SO42-] = |_0.180 M_____

[Na+] = |_0.0600 M____

[OH-] = |_2.4 x 10-11 M_

 

 
  1. What concentration of SO42- is necessary to start the precipitation of BaSO4 from a saturated solution of BaF2? Ksp BaSO4 = 1.1 x 10-10 = [Ba2+][SO42-]; Ksp BaF2 = 1.0 x 10-6 = [Ba2+][F-]2 (3)

    2. First determine the concentration of Ba2+ in a saturated solution of BaF2

    Ksp BaF2 = 1.0 x 10-6 = [x][2x]2 = 4 x3; x = (1.0 x 10-6 / 4)1/3 = (2.5 x 10-7)1/3 = 6.3 x 10-3 M Ba2+

    Then use this number with Ksp for BaSO4 to determine the [SO42-] concentration that just satisfies it

    Ksp BaSO4 = 1.1 x 10-10 = [Ba2+][SO42-] = [6.3 x 10-3][x]; x = 1.1 x 10-10/6.3 x 10-3 = 1.7 x 10-8 M

    Note that this is a very small concentration that will start the precipitation.

    [SO42-] = |_1.7 x 10-8 M _

     

     

  2. Determine the molar solubility of lead azide, Pb(N3)2 in a strongly buffered solution with pH = 3.00, given that [H+] = 1.0 x 10-3 M (4)

    3. Pb(N3)2(s) = Pb2+ + 2 N3- Ksp = 2.5 x 10-9 = [Pb2+][N3-]2

    HN3 + H2O = H3O+ + N3- Ka = 1.9 x 10-5 = [H+][N3-] / [HN3]

    Because the H+ will react with the N3- ion, and more solid will dissolve so that the Ksp is fulfilled.

    [Pb2+] = 1/2 { [N3-] + [HN3] } = solubility because no more reaction of Pb2+

    [HN3] = [H+][N3-] / Ka = 1.0 x 10-3 [N3-] / 1.9 x 10-5 = 52.63 [N3-]

    [Pb2+] = 1/2 { [N3-] + 52.63 [N3-] } = 1/2 {53.63 [N3-]} = 26.82 [N3-]

    Ksp = 2.5 x 10-9 = [Pb2+][N3-]2 = [26.82 [N3-]][N3-]2 = 26.82 [N3-]3

    [N3-] = (2.5 x 10-9 / 26.82)1/3 = (9.32 x 10-11)1/3 = 4.53 x 10-4 M

    [Pb2+] = 26.82 x [N3-] = 26.82 x 4.53 x 10-4 = 1.22x 10-2

    M = |_1.2 x 10-2 M______

  3. A solution that is 0.20 M in Ni2+ and 0.20 M in Cd2+ is to be saturated with H2S to separate the two ions.

Ksp CdS = 3 x 10-28; Kspa CdS = 3 x 10-7; Ksp NiS = 4 x 10-20; Kspa = 4 x 101

  1. Which cation is expected to precipitate at the lowest pH? Cd2+ because it has smaller Ksp (4)

    2. Cation = |_Cd2+______

  2. What H+ concentration will permit the maximum precipitation of this cation without precipitation of the second cation? Work with more soluble salt to keep it in solution.

NiS(s) +

2 H+ =

Ni2+ +

H2S

 

2x

0.20 M

0.10 (sat'd)

Kspa = 40 = [Ni2+][H2S] / [H+]2 = [0.20][0.10] / [H+]2

[H+] = (2.0 x 10-2/ 40 )1/2 = (5.0 x 10-4 )1/2 = 2.24 x 10-2 M

[H+] = |_2.2 x 10-2 M ___
  1. The equilibrium constant for the formation of the TlCl4- complex ion from Tl3+ and Cl- is 1 x 1018. If 0.080 mol of Tl(NO3)3 is dissolved in 1.00 L of 0.50 M NaCl, what are the concentrations of TlCl4- and Tl3+ in solution? Kf is large; mainly products. (3)

Tl3+ +

4 Cl- =

[TlCl4-]

0.080 - 0.080 = 0 limiting reag

0.50 - 4(.080) = 0.18 mol/L

0 + 0.080 = 0.080 mol/L

x (at equilibrium)

0.18 + 4x

0.080-x

To reach equilibrium, complex ion formed in the reaction must dissociate.

Kf = 1 x 1018 = [TlCl4-] / [Tl3+][Cl-]4 = [0.080 - x] / [x][0.18 + 4x]4 ~ [0.080] / [x][0.18]4 since K is large

X ~ 0.080 / (0.18)4 (1 x 1018) = 7.62 x 10-17M

[TlCl4-] = |_0.080 M ___

[Tl3+] = |_7.6 x 10-17M _

6. Using the Ksp for Cu(OH)2 (1.6 x 10-19) and the overall formation constant for Cu(NH3)42+ (1.0 x 1013)

(3)

  1. Calculate a value for the equilibrium constant for the reaction:

    2. Cu(OH)2(s) + 4 NH3(aq) = Cu(NH3)42+ + 2 OH-

    K = Ksp x Kf = 1.6 x 10-19 x 1.0 x 1013 = 1.6 x 10-6

    K = |_1.6 x 10-6_____

  2. Calculate the molar solubility of Cu(OH)2 in 5.0 M NH3. In 5.0 M NH3, the concentration of OH- is about 0.010 M.

Cu(OH)2 (s) +

4 NH3 =

Cu(NH3)42+ +

2 OH-

 

5.0 M - 4x

x

0.010 + 2x

K = 1.6 x 10-6 = [Cu(NH3)42+][OH-]2 / [NH3]4 = [x][0.010+2x]2 / [5.0 -4x]4 ~ [x][0.010]2 / [5.0]4

x ~ 1.6 x 10-6 x [5.0]4 / [0.010]2 ~ 10 M > 5M so approx not good; reasonable answer if 5.0 M NH3 is maintained and OH- doesn't build up. Calculation suggests Cu(OH)2 is "very soluble"

Cu(OH)2 (s) +

4 NH3 =

Cu(NH3)42+ +

2 OH-

 

4x

1.25 M - x

2.50 M - 2x

K = 1.6 x 10-6 = [Cu(NH3)42+][OH-]2 / [NH3]4 = [1.25 -x][2.50 -2x]2 / [4x]4 ~ [1.25][2.50]2 / [4x]4

x ~ [(1.25)(2.50)2 / 256(1.6 x 10-6)]1/4 ~ (1.907 x 104)1/4 = 11.8 M > 1.25 M; not completely reacted with NH3 since x is unreasonably large. Approximate to find the concentrations of all species that fit the equilibrium constant.

Try solubility = 0.5 M; K = (0.5)(1.0)2 / (3.0)4 = 6.2 x 10-3 still too large

Try solubility = 0.25 M; K = (0.25)(0.51)2 / (4.0)4 = 2.5 x 10-4 still too large

Try solubility = 0.125 M; K = (0.125)(0.26)2 / (4.0)4 = 6.6 x 10-5 getting closer

Try solubility = 0.10 M; K = (0.10)(0.21)2 / (4.6)4 = 9.8 x 10-6 still a bit smaller

Try solubility = 0.075 M; K = (0.075)(0.16)2 / (4.7)4 = 3.9 x 10-6

Try solubility = 0.06 M; K = (0.06)(0.13)2 / (4.76)4 = 2.0 x 10-6

Try solubility = 0.058 M; K = (0.058)(0.126)2 / (4.768)4 = 1.8 x 10-6

Try solubility = 0.056 M; K = (0.056)(0.122)2 / (4.776)4 = 1.6 x 10-6

M = |_0.056 M____