Selective Precipitation

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Example #1: The Ksp for BaSO4 is 1.1 x 10¯10 and that for BaSeO4 is 2.8 x 10¯11. A 1.0 M solution of BaCl2 is added slowly to a solution that is 1.0 x 10¯4 M in sodium sulfate, Na2SO4 and 1.0 x 10¯4 M in sodium selenate, Na2SeO4. What is the approximate percentage of one anion has precipitated at the point which the second anion just begins to precipitate? (Assume the addition of the BaCl2 solution does not change the overall solution volume to any significant degree.)

Solution:

1) Which is the more soluble? We need to look at how the two substances ionize:

BaSO4 ---> Ba2+ + SO42¯
BaSeO4 ---> Ba2+ + SeO42¯

Since they both ionized in EXACTLY the same way, we can compare Ksp values to see which is more soluble. This is not the case if the two substances ionize differently (see Example #5).

2) BaSO4 is the more soluble, so it precipitates last. What is [Ba2+] when BaSO4 begins to precipitate?

1.1 x 10¯10 = [Ba2+] [1.0 x 10¯4]

[Ba2+] = 1.1 x 10¯6 M

3) At the above [Ba2+], what is the [SeO42¯]?

2.8 x 10¯11 = [1.1 x 10¯6] [SeO42¯]

[SeO42¯] = 2.54 x 10¯5 M

4) What percentage of the SeO42¯ has precipitated?

(2.54 x 10¯5 M / 1.0 x 10¯4 M) x 100 = 25% (this is the amount remaining in solution)

75% of the selenate has precipitated


Example #2: The solubility products of PbSO4 and SrSO4 are 6.3 x 10¯7 and 3.2 x 10¯7, respectively. What are the values of [SO42¯], [Pb2+], and [Sr2+] in a solution at equilibrium with both substances?

Solution:

1) Which is the more soluble? We need to look at how the two substances ionize:

PbSO4 ---> Pb2+ + SO42¯
SrSO4 ---> Sr2+ + SO42¯

Since they both ionized in EXACTLY the same way, we can compare Ksp values to see which is more soluble. This is not the case if the two substances ionize differently (see Example #5).

2) PbSO4 is the more soluble. Let it dissolve to the maximum:

6.3 x 10¯7 = [Pb2+] [SO42¯]

[Pb2+] = [SO42¯] = 7.937 x 10¯4 M

3) What is the [Sr2+] when [SO42¯] = 7.937 x 10¯4 M?

3.2 x 10¯7 = [Sr2+] (7.937 x 10¯4)

[Sr2+] = 4.0 x 10¯4 M


Example #3: A solution of 0.10 M (each) Ni2+ and Cu2+ are separated using selective precipitation by the addition of solid Na2CO3. Assuming no volume change upon this addition, how much of the first precipitated ion (in %) remains at the point where the second ion begins to precipitate?

Solution:

1) Which is the more soluble? We need to look at how the two substances ionize:

NiCO3 ---> Ni2+ + CO32¯
CuCO4 ---> Cu2+ + CO32¯

Since they both ionized in EXACTLY the same way, we can compare Ksp values to see which is more soluble. This is not the case if the two substances ionize differently (see Example #5).

2) We need to know the Ksp values for NiCO3 and CuCO3. After a bit of Internet sleuthing, we find this:

NiCO3 ⇒ 1.4 x 10¯7
CuCO3 ⇒ 2.5 x 10¯10

3) We begin with the more soluble substance (the larger Ksp). The less soluble substance will have already preciptated when the NiCO3 begins to precipitate:

Ksp = [Ni2+] [CO32¯]

1.4 x 10¯7 = (0.10) (x)

x = 1.4 x 10¯6 M

This is the molarity of the carbonate when NiCO3 (the more soluble) is ready to precipitate.

4) Let us solve for the copper(II) ion still in solution at the above carbonate concentration:

Ksp = [Cu2+] [CO32¯]

2.5 x 10¯10 = (x) (1.4 x 10¯6)

x = 1.786 x 10¯4 M

5) What percent of the copper(II) ion is still in solution when NiCO3 begins to precipitate?

(1.786 x 10¯4 / 0.10) times 100 = 0.1786%

To two sig figs, this is 0.18%.


Example #4: A 0.207 M NaBr and 0.0870 M NaCl mixture is going to be treated with AgNO3. Calculate the % of bromide ion present when the choride starts to precipitate. The Ksp of AgCl is 1.77 x 10¯10; of AgBr it is 5.35 x 10¯13

Solution:

1) Which is the more soluble? We need to look at how the two substances ionize:

AgCl ---> Ag+ + Cl¯
AgBr ---> Ag+ + Br¯

Since they both ionized in EXACTLY the same way, we can compare Ksp values to see which is more soluble. This is not the case if the two substances ionize differently (see Example #5).

2) Calculate the silver ion concentration when AgCl (the more soluble) just begins to precipitate:

1.77 x 10¯10 = (x) (0.0870)

x = 2.0345 x 10¯9 M

3) Calculate the bromide concentration at the above silver ion concentration:

5.35 x 10¯13 = (2.0345 x 10¯9) (x)

x = 2.63 x 10¯4 M

4) Calculate the percent bromide remaining in solution:

(2.63 x 10¯4 / 0.207) times 100 = 0.127%

Example #5: Sodium phosphate is added to a solution that contains 0.0099 M aluminum nitrate and 0.021 M calcium chloride. The concentration of the first ion to precipitate (either Al3+ or Ca2+) decreases as its precipitate forms. What is the concentration of this ion when the second ion begins to precipitate?

Solution:

1) We need to know the Ksp of more soluble product. We go off to the Internet and find:

AlPO4 ⇒ 9.8 x 10¯22
Ca3(PO4)2 ⇒ 2.0 x 10¯29

2) However AlPO4, with the larger Ksp, is not the more soluble:

Ksp = [Al3+] [PO43¯]

9.8 x 10¯22 = (x)(x)

x = 3.13 x 10¯11 mol/L

Ksp = [Ca2+]3 [PO43¯]2

2.0 x 10¯29 = (3x)3(2x)2

x = 7.137 x 10¯7 mol/L

Ca3(PO4)2 is the more soluble.

I have to do the above calculations because AlPO4 and Ca3(PO4)2 ionize differently in solution. AlPO4 gives 2 ions per formula unit (just like in Examples #1-4) while Ca3(PO4)2 gives five per formula unit. Because of the difference in production of ions per formula unit, I must determine which substance is more soluble by calculation rather than by examining the Ksp values.

3) Let us calculate the phosphate ion concentration when Ca3(PO4)2 first begins to precipitate:

Ksp = [Ca2+]3 [PO43¯]2

2.0 x 10¯29 = (0.021)3 (x)2

x = 1.47 x 10¯12 mol/L

4) Let us calculate the concentration of the aluminum ion when the phosphate is the value just above:

Ksp = [Al3+] [PO43¯]

9.8 x 10¯22 = (x) (1.47 x 10¯12)

x = 6.7 x 10¯10 mol/L


Example #6: A dilute solution of AgNO3 is added slowly and continuously to a second solution containing both Cl¯ and CrO42¯. If the CrO42¯ concentration is 0.010 M when the first trace of Ag2CrO4 precipitate appears, what is the concentration of the Cl¯ at that point? (Ksp of AgCl is 1.77 x 10¯10; Ksp of Ag2CrO4 is 1.12 x 10¯12)

Solution:

1) Calculate the silver ion concentration required to just precipitate silver chromate:

Ksp = [Ag+]2 [CrO42¯]

1.12 x 10¯12 = (x)2 (0.01)

x = 1.0583 x 10¯5 M

2) Calculate the chloride ion concentration at the above silver ion concentration:

Ksp = [Ag+] [Cl¯]

1.77 x 10¯10 = (1.0583 x 10¯5) (x)

x = 1.67 x 10¯5 M


Example #7: A 0.050 M solution of hydrogen sulfide (H2S) contains 0.15 M nickel chloride NiCl2, and 0.35 M mercury(II) nitrate Hg(NO3)2. What pH is required to precipitate the maximum amount of HgS but none of the NiS?

Solution:

1) We need some values, so off we go to the Internet:

Ksp for NiS = 4 x 10-20
Ksp for HgS = 2 x 10-53

Ka1 for H2S = 1.0 x 10-7
Ka2 for H2S = 1.3 x 10-13

2) We do not want any NiS to precipitate. What concentration of NiS will not precipitate?

NiS(s) <==> Ni2+(aq) + S2-(aq)

Ksp = [Ni2+] [S2-]

4 x 10-20 = (0.15) (x)

x = 5.164 x 10-10 M (I'll keep some guard digits)

3) The [S2-] cannot exceed 5.164 x 10-10 M. What pH creates [S2-] = 5.164 x 10-10 M?

Please refer to equation 9.13 at this location.

The x refers to the hydrogen ion concentration.

5.164 x 10-10 = [(1.3 x 10-13) (0.050)] / [x2 + (1.0 x 10-7)(x) + 1.3 x 10-13]

(5.164 x 10-10) (x2) + 5.164 x 10-17) (x) + 6.7132 x 10-23 = 6.5 x 10-15

(5.164 x 10-10) (x2) + 5.164 x 10-17) (x) - 6.5 x 10-15 = 0

I used a quadratic equation solver.

x = 0.0035477862 M (I kept a few extra digits)

pH = - log 0.0035477862 = 2.45

This question was asked on Yahoo Answers in 2007.


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