# Solubility of Salts - Ksp. Ksp Solubility

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1 Solubility of Salts - Ksp We now focus on another aqueous equilibrium system, slightly soluble salts. These salts have a Solubility Product Constant, K sp. (We saw this in 1B with the sodium tetraborate solubility lab.) For example, K sp is defined for Cu(OH) 2 (s) as follows: Cu(OH) 2 (s) Cu 2+ (aq) + 2 OH (aq) K sp =[Cu 2+ ][OH ] 2 = 4.8x10 20 at 25 C For a list of Ksp values at 25 C, refer to Appendix D-Table D.3. These slightly soluble salts dissolve until K sp is satisfied. At this point, we say the solution is saturated, and no more salt will dissolve. In terms of Q sp we have the following possibilities: 1. Q sp < K sp 2. Q sp = K sp 3. Q sp > K sp Other Aspects of Ionic Equilibria 1 Ksp Solubility Write the chemical equilibrium and the K sp equilibrium-constant expression for the solubility of Ca 3 (PO 4 ) 2. Solubility and Ksp The solubility of salts (saturated solution) are often expressed in one or more of the following units with the temperature also specified: mol/l (molar solubility) g/l, g/ml, g/100 ml or mg/l (Note that these are NOT densities.) ppm or ppb (parts per million or parts per billion) for very insoluble salts. The above units for solubility are directly related to K sp through equilibrium stoichiometry. Let s do some examples. Other Aspects of Ionic Equilibria 2

2 Solubility and K sp 1. What is the molar solubility of Ca 3 (PO 4 ) 2 at 25 C given a K sp of 2.0x10 29? 1.1. What is the concentration of calcium ions? 1.2. What is this solubility in g/l solution? 1.3. What is this solubility in parts per million (ppm) and parts per billion (ppb)? (Assume a soln density of 1.0 g/ml) mass of component in soln ppm = x10 6 total mass of soln mass of component in soln ppb = x10 9 total mass of soln Other Aspects of Ionic Equilibria 3 K sp from Experimental Data A saturated solution of magnesium hydroxide in water has a ph of Calculate the K sp for magnesium hydroxide. Other Aspects of Ionic Equilibria 4

3 Relative Molar Solubility of Salts The relative MOLAR solubility of salts (saturated solution) can be determined by comparing K sp values. The greater the K sp the more ions are in solution, hence the greater the molar solubility. However, you can only directly compare salts that give equivalent numbers of ions in solution. For example, you can compare the K sp s of all salts with a 1:1 ion ratio like: AgBr, BaSO 4 etc. Or, you could compare K sp s of all salts with 1:2 and 2:1 ion ratios like: BaCl 2, Ag 2 SO 4, etc. But you can t compare the Ksp AgBr directly to the Ksp Ag 2 SO 4 to determine which is more soluble. Which salt has a greater molar solubility soluble: AgBr or BaSO 4? Other Aspects of Ionic Equilibria 5 Ksp Values and Solubility Calcs. LIMITATIONS EXIST! Unfortunately, solubilities calculated using K sp values sometimes deviate appreciably from the experimentally measured solubilities! Why? Reason 1 of 2: Some salts do not completely dissociate 100% into their respective ions. For example, PbCl 2 exists in three different forms in water, each with its own individual K value. 1. PbCl 2 (s) PbCl 2 (aq) (no ions) K = 1.1x10-3 ; S = PbCl 2 (s) PbCl + (aq) + Cl (aq) (two ions) K = 6.9x10-4 ; S = PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) (all ions) K = 1.7x10-5 ; S = The experimental solubility of PbCl 2 is M. Over twice the value predicted by K sp alone. Other Aspects of Ionic Equilibria 6

4 Ksp Values and Solubility Calcs. LIMITATIONS Reason 2: Some anions in salts are strongly basic. They react with water in a K b equilibrium. This reduces the anion concentration available to satisfy K sp. This in turn increases the solubility of the salt. The three common ions that are basic enough to have considerable reaction with water are: S 2-, CO 3 2-, and PO For example consider Ca 3 (PO 4 ) 2 : The Ksp equilibrium is: Ca 3 (PO 4 ) 2 3 Ca + (aq) + 2 PO 4 3 (aq) K sp = 2.0x10 29 Write the chemical reaction for the Kb equilibrium (hydrolysis) for PO 4 3 (aq): K b =? (How do you determine the value?) The experimental solubility of Ca 3 (PO 4 ) 2 is somewhat higher than predicted by K sp alone because some phosphate ion is removed from solution through the K b reaction. K sp values are still useful for estimating solubilities, predicting trends and predicting relative solubilities. We just need to keep in mind that they have limitations! We previously determined the molar solubility of Ca 3 (PO 4 ) 2 (s) using Ksp alone. Now let s determine its solubility including Kb with Ksp to find Knet. Other Aspects of Ionic Equilibria 7 The K sp of Sulfide Salts INCLUDES Kb Since the sulfide ion is very basic, Kb significantly increases the solubility of sulfide salts. For the metal sulfides, the K sp values are actually Knet for the overall equilibrium that includes two processes: 1) the dissolution of the metal sulfide 2) the Kb hydrolysis reaction of the the sulfide ion. (See the footnote at the bottom of Table D.3 in our text book) For CuS write the chemical reaction for each of the above equilibria, and for the net equilibrium: 1. K 1 2. K b >> 1 Net: Ksp = K1*Kb = 6x10-37 (Table D.3) Determine the ph of a saturated CuS solution. Other Aspects of Ionic Equilibria 8

5 Solubility of Salts and the Common Ion Effect Like the percent ionization of weak acids and bases, the solubility of salts is influenced by the presence of a common ion. These salts are less soluble when a common ion is present just like a weak acid s ionization is limited in the presence of significant conjugate base. Play Movie 1. The Ksp of silver iodide in water is 8.3x10 17 M. Calculate the molar solubility of silver iodide in: 1. Pure water M NaI (Common Ion Effect) Other Aspects of Ionic Equilibria 9 Solubility of Salts and the Common Ion Effect 1. The barium ion, Ba 2+ (aq), is poisonous when ingested. The lethal does in mice is about 12 mg Ba 2+ per kg of body mass. Despite this fact, BaSO 4 is widely used in medicine to obtain X-ray images of the gastrointestinal tract since Ba is a very good x-ray absorbing element. a) Explain why BaSO 4 (s) is safe to ingest, even though Ba 2+ (aq) is poisonous. b) Calculate the concentration of Ba 2+, in milligrams per liter, in saturated BaSO 4 (aq) at 25 C. c) Soluble MgSO 4 is often mixed with BaSO 4 when ingested. What function does the MgSO 4 serve? Other Aspects of Ionic Equilibria 10

6 Solubility of Salts with ph Adjustments We have already seen that basic salts containing S 2-, CO 3 2-, and PO 4 3- ions are more soluble in water than expected because of the reaction of the basic anion with water through K b. All salts that contain a basic anion will have their solubility increased in an acidic solution. In acidic solutions the basic anion reacts with the acid, forcing more of the salt to dissolve to reach equilibrium. For example, PbF2 and Mg(OH) 2+ are practically insoluble in water. However, they are very soluble in dilute acids. Play Movie Other Aspects of Ionic Equilibria 11 Solubility of Salts with ph Adjustments 1. Which of the following slightly soluble salts will be more soluble in acidic solution than pure water? For those where the solubility increases, write the net ionic chemical reaction that occurs when a strong acid is present in solution: a) Al(OH) 3 (s) b) BaSO4(s) c) BaC2O4(s) d) PbCl2 Other Aspects of Ionic Equilibria 12

7 Solubility and ph 1. Calculate the molar solubility of iron (II) hydroxide (Ksp = 7.9x10-16 ) in a) A solution buffered at ph=7.00: b) What could you say about the solubility of iron (II) hydroxide in a solution buffered at ph=10.00 Other Aspects of Ionic Equilibria 13 Solubility and ph 1. A chief component in marble is calcium carbonate. Marble has been widely used for statues and ornamental work on buildings, but marble is readily attacked by acids. Assume that the overall reaction that occurs in a dilute acid is CaCO 3 (s) + H 3 O + (aq) Ca 2+ (aq) + HCO 3 (aq) + H 2 O(l) Determine the equilibrium constant for this reaction* and then determine the solubility of calcium carbonate in a buffer with ph = 5.6 (ph stays constant). *Hint: What combination of reactions yields the overall reaction? Other Aspects of Ionic Equilibria 14

8 Solubility of Salts and ph-applications A principle cause of sinkhole formation: dissolution of limestone (CaCO 3 ) by naturally acidic rain water as it percolates through the bedrock. Voids in the bed rock can cause sudden collapse of the overlying ground. Dissolved CaCO 3 Sinkhole forming Sinkhole collapses Other Aspects of Ionic Equilibria 15 Sinkhole Formation The large sinkhole shown here has destroyed several buildings and part of a highway. Other Aspects of Ionic Equilibria 16

9 Solubility of Salts and ph-applications 1. Two of the main crystalline components of kidney stones are calcium phosphate and calcium oxalate. What happens to the solubility of these compounds as ph is increased? 2. Tooth enamel consists mainly of a mineral called hydroxyapatite, Ca 5 (PO 4 ) 3 (OH)(s), K sp = 6.8x10 37, that is insoluble in pure water. When acids dissolve tooth enamel, the result is tooth decay. Write the net ionic equation that occurs between hydroxyapatite and H 3O + (aq). Note, the phosphate ions are also pronated along with the hydroxide. 3. Fluoridation of water and the use of fluoride toothpaste causes the OH ion in hydroxyapatite to be replaced with F forming Ca 5 (PO 4 ) 3 F (K sp =6.8x10 60 ). Suggest a reason why fluoride helps prevent tooth decay. Other Aspects of Ionic Equilibria 17 Complex Ion Formation, Kf Complex ions are ions that usually contain a transition metal cation and one or more ligands. A ligand is either a neutral molecule or an anion that bonds to the metal cation through a Lewis acid-base reaction: Metal Ion: Lewis acid - LIMITING REACTANT Ligand: Lewis Base - EXCESS REACTANT Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) 2+ (aq) K f = 1.7x10 7 =? (Write the expression) What can you conclude about the favorability of complex ion formation? What can you conclude about the free (uncomplexed) metal cation concentration in solutions where a complex ion can form and the ligand is in excess? Other Aspects of Ionic Equilibria 18

10 Complex Ion Formation, Kf Determine the concentration of free (uncomplexed) silver ions in solution when 5.0 ml of a 6.0 M ammonia solution (ligand) is added to 45.0 ml of a 0.10 M silver nitrate solution (metal cation). Other Aspects of Ionic Equilibria 19 Complex Ion Formation and Solubility Formation of complex ions is one way to dissolve insoluble salts. Consider the following sequence of additions to a solution containing Ag + ions. Precipitation of AgCl Complex ion formation Precipitation of AgBr Complex ion formation AgCl is precipitated by adding NaCl(aq) to AgNO3(aq) The AgCl is dissolved by adding excess aqueous NH3 The silver-ammonia complex ion is changed to insoluble AgBr on addition of NaBr(aq). The AgBr is dissolved by adding excess Na2S2O3(aq). Other Aspects of Ionic Equilibria 20

11 Complex Ion Formation and Solubility Using equilibrium constants (K sp and K f ) we can show why each process in the previous illustration can be made favorable. Step (a): Precipitation of AgCl (Ksp = 1.8x10-10 ) Ag + (aq) + Cl (aq) AgCl(s) Kppt =? What concentration of Cl (aq) is required to precipitate 99.9% of the Ag + (aq) ions from a 0.10 M AgNO3(aq) solution. What mass of NaCl (molar mass = g/mol) must be added to ml of the solution to accomplish this? (Assume that the volume change of the solution is negligible upon addition of the NaCl.) Other Aspects of Ionic Equilibria 21 Complex Ion Formation and Solubility Step (b): AgCl(s) dissolves when NH 3 (aq) is added: AgCl(s) Ag + (aq) + Cl - (aq) K sp = 1.8x10 10 Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) 2+ (aq) K f = 1.7x10 7 AgCl(s) + 2 NH 3 (aq) Ag(NH 3 ) 2+ (aq) + Cl (aq) K net =? A high concentration of the ligand, NH3(aq), in solution ensures that almost all the AgCl(s) will dissolve. This is an application of Le Chatelier s Principle. This is a key point for dissolving insoluble salts by formation of a complex ion: Salt + Ligand > Complex ion + Anion K net = Ksp*Kf Other Aspects of Ionic Equilibria 22

12 Calculations: Complex Ion Formation and Solubility Step (b): What is the molar solubility of AgCl(s) in a) pure H 2 O(l): b) a solution where the concentration of NH3(aq) begins at 0.10 M: Other Aspects of Ionic Equilibria 23 Calculations: Complex Ion Formation and Solubility Step (c): Ag(NH 3 ) 2+ (aq) precipitates as AgBr(s) when NaBr(aq) is added. Find Knet for this system. Determine the concentration of complexed ion still in solution where the Ag(NH 3 ) 2+ (aq) and NaBr(aq) concentrations both begin at 0.10 M. Other Aspects of Ionic Equilibria 24

13 Step (d): AgBr(s) dissolves as Na2S 2 O 3 (aq) is added. (Write the net ionic equation for the reaction below.) Complex Ion Formation and Solubility Crystals of AgBr can be removed from black and white photographic film by reacting the AgBr(s) with aqueous sodium thiosulfate. In order to dissolve 2.5 g of AgBr in 1.00 L of solution, what concentration of thiosulfate ion is needed in solution at equilibrium? Now calculate how many grams of solid Na 2 S 2 O 3 (molar mass = g/mol ) must be added to dissolve the AgBr. Other Aspects of Ionic Equilibria 25 Amphoterism of Metal Oxides and Hydroxides Definition of amphoteric: An amphoteric substance is slightly soluble in water, but soluble in either acidic or basic solutions. Example Al 3+ : (waters of hydration omitted for clarity) 1. Al(OH)3(s) + 3 H + (aq) Al 3+ (aq) + 3 H2O(l) Acidic solution, ph < 7 } 2. Al 3+ (aq) + OH (aq) AlOH 2+ (aq) Weakly basic solution, 7 < ph < 9 3. AlOH 2+ (aq) + OH (aq) Al(OH)2 + (aq) 4. Al(OH)2 + (aq) + OH (aq) Al(OH)3(s) Basic solution, ph Al(OH)3(s) + OH (aq) Al(OH)4 (aq) Strongly basic solution, ph > 10 Other amphoteric oxides and hydroxides: Cr 3+, Zn 2+, and Sn 2+ Each will complex a different # of hydroxide ions - you will have to check a table of K f values. ph << 10 ppt ph = 10 ph > 10 Other Aspects of Ionic Equilibria 26

14 Zinc hydroxide is insoluble in water (K sp = 3.0x10 16 ), but amphoteric. Write net ionic equations to show why Zn(OH) 2 is readily soluble in the following solutions: (a) CH 3 COOH(aq) Amphoterism of Metal Oxides and Hydroxides (b) NH 3 (aq): Zn 2+ ions form the following complex ion: [Zn(NH 3 )4] 2+ with a Kf = 4.1x10 8 (c) OH (aq) Zn 2+ ions form the following complex ion: [Zn(OH)4] 2- with a Kf = 4.6x10 17 (c.1) What is the minimum ph required to dissolve 1.5 g of Zn(OH)2 in a liter of solution? Other Aspects of Ionic Equilibria 27 Precipitation Reactions and Separation of Ions Recall from previous slides about solubility: Q sp < K sp : unsaturated soln. - no precipitation. Q sp = K sp : saturated soln. - equilibrium. Q sp > K sp : supersaturated soln. - a ppt. should form. Using K sp values: 1. We can determine if a ppt will form when two solutions are mixed that contain a cation and an anion that can combine to form a slightly soluble salt. 2. We can determine if different cations in solution can be quantitatively separated from each other by selective precipitation. In other words, we can determine if one cation can be removed from the solution as a precipitated salt before a second cation in solution is also precipitated. Other Aspects of Ionic Equilibria 28

15 Single Precipitation Reactions Calcium ion triggers clotting of blood. Therefore, when blood is donated the receiving bag contains sodium oxalate to precipitate the Ca 2+, and thus prevent clotting. Typically, blood contains 9.7x10-5 g Ca 2+ /ml. To remove the calcium ions, a medical technologist treats a 104-mL blood sample with ml of M Na 2 C 2 O 4. Calculate the [Ca 2+ ] left in solution after the treatment. (K sp for calcium oxalate = 2.6x10-9 ) Other Aspects of Ionic Equilibria 29 Single Precipitation Reactions 1. Show that a precipitate of Mg(OH) 2 will form in an aqueous solution that is M MgCl 2 and M NaOH. 2. Show that a precipitate of Mg(OH) 2 will also form in an aqueous solution that is M MgCl 2 and M NH 3. Hint: Find the [OH - ] produced by the hydrolysis of NH3. Is it enough to ppt the Mg 2+ ion? 2.1. Explain why the Mg(OH) 2 precipitate can be prevented from forming in (2) by adding NH 4 Cl to the solution containing ammonia. Other Aspects of Ionic Equilibria 30

16 Single Precipitation Reactions Calculate the minimum ph needed to precipitate Mn(OH) 2 so completely that the concentration of Mn 2+ is less than 1 µg per liter, that is 1 part per billion (ppb). Other Aspects of Ionic Equilibria 31 Selective Precipitation of Cations Differences in molar solubilities between compounds containing a common ion can be used to selectively precipitate one ion from solution leaving the other ion in solution. We can calculate the amount of the common ion needed to reach saturation for the most soluble ionic salt. Addition of just a little less of the common ion will insure the most complete separation possible. Quantitative, or complete, separation is considered possible if 99.9% of the least soluble salt precipitates before any of the most soluble salt starts to precipitate. Typically, the added reagent (containing the common ion) is quite concentrated so that its addition does not appreciable change the volume of the solution and dilute the solution containing the cations to be separated. Solution of cations Add ppt agent Least soluble cation ppts first Centrifuge and decant Selective ppt complete A ml solution has [Ba 2+ ] = M and [Sr 2+ ] = M. A M Na 2 SO 4 (aq) solution is added drop-wise. Which ion, Ba 2+ or Sr 2+, will precipitate first? Ksp for BaSO4 is 1.1x10-10, Ksp for SrSO4 is 3.4x10-7 Can the two cations be quantitatively (99.9%) separated by selective precipitation? Other Aspects of Ionic Equilibria 32

17 Precipitation Reactions: Applications Precipitation of Mg 2+ as as Mg(OH)2 from sea water is the source of Mg metal. Titration of solutions containing Cl (aq) with AgNO3 to quantitatively determine [Cl ]. For this to work,we must be sure that almost all of the Cl precipitates as we add the AgNO3. How can we be sure of this? Determination of the amount of SO4 2 in solution by precipitating as BaSO4(s). The BaSO4 formed is filtered off, dried and weighed. For this to work, the precipitation of SO4 2 must be complete (99.9%). How can we be sure of this? Complex Ion Formation: Complex ion formation can be used to extract gold from low-grade gold containing rock. The formation constant of Au(CN)2 is very large. A very small concentration of Au + ions are formed through oxidation of Au. This oxidation takes place in the presence of CN ions. Complex ion formation removes the Au + ions from solution, so more are formed. As a result, even though Au is not oxidized normally by air, bubbling air through a suspension of Au containing ore in the presence of CN leads to formation of a solution of the complex ion: 4Au(s) + 8CN (aq) + O2(g) +2H2O(l) 4Au(CN)2 (aq) + 4OH (aq) The resulting solution is filtered and the Au + reduced to Au(s). Aluminum ore contains Fe2O3 impurities along with the desired aluminum hydroxide, Al2O3. A strong base is added to dissolve the Al2O3 as the Al(OH)4 complex ion (Kf = 3x10 33 ). The iron (III) ion does not form a complex ion with hydroxide, therefore the Fe2O3 does not dissolve. The resulting solution is filtered and acid is added to the filtrate to precipitate Al(OH)3. The filtrate must not be made too acidic or the Al(OH)3 will redissolve! Other Aspects of Ionic Equilibria 33

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