![]() 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.ģ) Let us calculate the phosphate ion concentration when Ca 3(PO 4) 2 first begins to precipitate:Ĥ) Let us calculate the concentration of the aluminum ion when the phosphate is the value just above:Įxample #6: A dilute solution of AgNO 3 is added slowly and continuously to a second solution containing both Cl¯ and CrO 4 2¯. 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. I have to do the above calculations because AlPO 4 and Ca 3(PO 4) 2 ionize differently in solution. We go off to the Internet and find:Ģ) However AlPO 4, with the larger K sp, is not the more soluble: What is the concentration of this ion when the second ion begins to precipitate?ġ) We need to know the K sp of more soluble product. ![]() The concentration of the first ion to precipitate (either Al 3+ or Ca 2+) decreases as its precipitate forms. (2.63 x 10¯ 4 / 0.207) times 100 = 0.127%Įxample #5: Sodium phosphate is added to a solution that contains 0.0099 M aluminum nitrate and 0.021 M calcium chloride. The K sp of AgCl is 1.77 x 10¯ 10 of AgBr it is 5.35 x 10¯ 13Ģ) Calculate the silver ion concentration when AgCl (the more soluble) just begins to precipitate:ģ) Calculate the bromide concentration at the above silver ion concentration:Ĥ) Calculate the percent bromide remaining in solution: ![]() Calculate the % of bromide ion present when the choride starts to precipitate. This is the molarity of the carbonate when NiCO 3 (the more soluble) is ready to precipitate.Ĥ) Let us solve for the copper(II) ion still in solution at the above carbonate concentration:ĥ) What percent of the copper(II) ion is still in solution when NiCO 3 begins to precipitate? The less soluble substance will have already preciptated when the NiCO 3 begins to precipitate: After a bit of Internet sleuthing, we find this:ģ) We begin with the more soluble substance (the larger K sp). 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?Ģ) We need to know the K sp values for NiCO 3 and CuCO 3. Let it dissolve to the maximum:ģ) What is the when = 7.937 x 10¯ 4 M?Įxample #3: A solution of 0.10 M (each) Ni 2+ and Cu 2+ are separated using selective precipitation by the addition of solid Na 2CO 3. What are the values of, , and in a solution at equilibrium with both substances?Ģ) PbSO 4 is the more soluble. ![]() (2.54 x 10¯ 5 M / 1.0 x 10¯ 4 M) x 100 = 25% (this is the amount remaining in solution)Įxample #2: The solubility products of PbSO 4 and SrSO 4 are 6.3 x 10¯ 7 and 3.2 x 10¯ 7, respectively. What is when BaSO 4 begins to precipitate?ģ) At the above, what is the ?Ĥ) What percentage of the SeO 4 2¯ has precipitated? This is not the case if the two substances ionize differently (see Example #5).Ģ) BaSO 4 is the more soluble, so it precipitates last. Since they both ionized in EXACTLY the same way, we can compare K sp values to see which is more soluble. 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.)ġ) Which is the more soluble? We need to look at how the two substances ionize: 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 2SO 4 and 1.0 x 10¯ 4 M in sodium selenate, Na 2SeO 4. 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. ![]()
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