**Example #1:** The K_{sp} for BaSO_{4} is 1.1 x 10¯^{10} and that for BaSeO_{4} is 2.8 x 10¯^{11}. A 1.0 M solution of BaCl_{2} is added slowly to a solution that is 1.0 x 10¯^{4} M in sodium sulfate, Na_{2}SO_{4} and 1.0 x 10¯^{4} M in sodium selenate, Na_{2}SeO_{4}. 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 BaCl_{2} 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:

BaSO_{4}---> Ba^{2+}+ SO_{4}^{2}¯

BaSeO_{4}---> Ba^{2+}+ SeO_{4}^{2}¯Since they both ionized in EXACTLY the same way, we can compare K

_{sp}values to see which is more soluble. This is not the case if the two substances ionize differently (see Example #5).

2) BaSO_{4} is the more soluble, so it precipitates last. What is [Ba^{2+}] when BaSO_{4} begins to precipitate?

1.1 x 10¯^{10}= [Ba^{2+}] [1.0 x 10¯^{4}][Ba

^{2+}] = 1.1 x 10¯^{6}M

3) At the above [Ba^{2+}], what is the [SeO_{4}^{2}¯]?

2.8 x 10¯^{11}= [1.1 x 10¯^{6}] [SeO_{4}^{2}¯][SeO

_{4}^{2}¯] = 2.54 x 10¯^{5}M

4) What percentage of the SeO_{4}^{2}¯ 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 PbSO_{4} and SrSO_{4} are 6.3 x 10¯^{7} and 3.2 x 10¯^{7}, respectively. What are the values of [SO_{4}^{2}¯], [Pb^{2+}], and [Sr^{2+}] 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:

PbSO_{4}---> Pb^{2+}+ SO_{4}^{2}¯

SrSO_{4}---> Sr^{2+}+ SO_{4}^{2}¯Since they both ionized in EXACTLY the same way, we can compare K

_{sp}values to see which is more soluble. This is not the case if the two substances ionize differently (see Example #5).

2) PbSO_{4} is the more soluble. Let it dissolve to the maximum:

6.3 x 10¯^{7}= [Pb^{2+}] [SO_{4}^{2}¯][Pb

^{2+}] = [SO4^{2}¯] = 7.937 x 10¯^{4}M

3) What is the [Sr^{2+}] when [SO_{4}^{2}¯] = 7.937 x 10¯^{4} M?

3.2 x 10¯^{7}= [Sr^{2+}] (7.937 x 10¯^{4})[Sr

^{2+}] = 4.0 x 10¯^{4}M

**Example #3:** A solution of 0.10 M (each) Ni^{2+} and Cu^{2+} are separated using selective precipitation by the addition of solid Na_{2}CO_{3}. 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:

NiCO_{3}---> Ni^{2+}+ CO_{3}^{2}¯

CuCO_{4}---> Cu^{2+}+ CO_{3}^{2}¯Since they both ionized in EXACTLY the same way, we can compare K

_{sp}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 K_{sp} values for NiCO_{3} and CuCO_{3}. After a bit of Internet sleuthing, we find this:

NiCO_{3}⇒ 1.4 x 10¯^{7}

CuCO_{3}⇒ 2.5 x 10¯^{10}

3) We begin with the more soluble substance (the larger K_{sp}). The less soluble substance will have already preciptated when the NiCO_{3} begins to precipitate:

K_{sp}= [Ni^{2+}] [CO_{3}^{2}¯]1.4 x 10¯

^{7}= (0.10) (x)x = 1.4 x 10¯

^{6}MThis is the molarity of the carbonate when NiCO

_{3}(the more soluble) is ready to precipitate.

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

K_{sp}= [Cu^{2+}] [CO_{3}^{2}¯]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 NiCO_{3} 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 AgNO_{3}. Calculate the % of bromide ion present when the choride starts to precipitate. The K_{sp} 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¯_{sp}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 Al^{3+} or Ca^{2+}) 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 K_{sp} of more soluble product. We go off to the Internet and find:

AlPO_{4}⇒ 9.8 x 10¯^{22}

Ca_{3}(PO_{4})_{2}⇒ 2.0 x 10¯^{29}

2) However AlPO_{4}, with the larger K_{sp}, is not the more soluble:

K_{sp}= [Al^{3+}] [PO_{4}^{3}¯]9.8 x 10¯

^{22}= (x)(x)x = 3.13 x 10¯

^{11}mol/LK

_{sp}= [Ca^{2+}]^{3}[PO_{4}^{3}¯]^{2}2.0 x 10¯

^{29}= (3x)^{3}(2x)^{2}x = 7.137 x 10¯

^{7}mol/LCa

_{3}(PO_{4})_{2}is the more soluble.

I have to do the above calculations because AlPO_{4} and Ca_{3}(PO_{4})_{2} ionize differently in solution. AlPO_{4} gives 2 ions per formula unit (just like in Examples #1-4) while Ca_{3}(PO_{4})_{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 K_{sp} values.

3) Let us calculate the phosphate ion concentration when Ca_{3}(PO_{4})_{2} first begins to precipitate:

K_{sp}= [Ca^{2+}]^{3}[PO_{4}^{3}¯]^{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:

K_{sp}= [Al^{3+}] [PO_{4}^{3}¯]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 AgNO_{3} is added slowly and continuously to a second solution containing both Cl¯ and CrO_{4}^{2}¯. If the CrO_{4}^{2}¯ concentration is 0.010 M when the first trace of Ag_{2}CrO_{4} precipitate appears, what is the concentration of the Cl¯ at that point? (K_{sp} of AgCl is 1.77 x 10¯^{10}; K_{sp} of Ag_{2}CrO_{4} is 1.12 x 10¯^{12})

**Solution:**

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

K_{sp}= [Ag^{+}]^{2}[CrO_{4}^{2}¯]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:

K_{sp}= [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 (H_{2}S) contains 0.15 M nickel chloride NiCl_{2}, and 0.35 M mercury(II) nitrate Hg(NO_{3})_{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:

K_{sp}for NiS = 4 x 10^{-20}

K_{sp}for HgS = 2 x 10^{-53}K

_{a1}for H_{2}S = 1.0 x 10^{-7}

K_{a2}for H_{2}S = 1.3 x 10^{-13}

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

NiS(s) <==> Ni^{2+}(aq) + S^{2-}(aq)K

_{sp}= [Ni^{2+}] [S^{2-}]4 x 10

^{-20}= (0.15) (x)x = 5.164 x 10

^{-10}M (I'll keep some guard digits)

3) The [S^{2-}] cannot exceed 5.164 x 10^{-10} M. What pH creates [S^{2-}] = 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)] / [x^{2}+ (1.0 x 10^{-7})(x) + 1.3 x 10^{-13}](5.164 x 10

^{-10}) (x^{2}) + 5.164 x 10^{-17}) (x) + 6.7132 x 10^{-23}= 6.5 x 10^{-15}(5.164 x 10

^{-10}) (x^{2}) + 5.164 x 10^{-17}) (x) - 6.5 x 10^{-15}= 0I 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.