Solubility PDF

Solubility (mol dm – 3 or g dm – 3) Molar solubility Mass solubility (mol dm – 3) (g dm – 3)  Definitions: Solubility: Ø maximum amount of solute (in moles/grams) that saturates 1 dm3 solvent at a given temperature;...

Solubility (mol dm – 3 or g dm – 3) Molar solubility Mass solubility (mol dm – 3) (g dm – 3)  Definitions: Solubility: Ø maximum amount of solute (in moles/grams) that saturates 1 dm3 solvent at a given temperature; Molar solubility: Ø number of moles of solute in 1 dm3 of saturated solution at a given temperature; OR number of moles of a compound that dissolves to give 1 dm3 of a saturated solution. mass solubility: Ø number of grams of solute in 1 dm3 of a saturated solution at a specified temperature; Solubility of a salt: Ø amount of the salt that will saturate a given amount of solvent at a stated temperature.  Definitions: Unsaturated solution: Ø Solution that can dissolve more solute at a given temperature; Ø Solution that has less than maximum amount of solute dissolved at specified temperature. Saturated Solution: Ø solution whose undissolved solid solute is in equilibrium with its ions in solution at a stated temperature; Ø Solution that has dissolved the maximum amount of solute at a given temperature; Ø rate of solution equals rate of crystallization; Ø at a given temperature, concentrations of solute and its ions/molecules are constant; Ø Any added solute dissolves at the same rate as its ions/molecules in solution crystallize out.  Supersaturated solution: Ø Solution that contains more solute than it can normally hold at specified temperature; Ø Solution that dissolved more than the maximum amount of solute (reached by heating, then cooling); Ø not a true saturated solution; Ø quickly forms crystals when stirred or small crystals of the solute or dust particles get into it to create centres of nucleation; Ø Example: Na2S2O3 and Na2SO4.10H2O readily form supersaturated solutions. Solubility of a substance is a physical property like melting and boiling points.  Solubility curves/graphs: Ø graphs showing a plot of solubility on the y – axis against temperature on the x – axis; Ø graphs which show the variation of solubility of substances with temperature; Ø graphs obtained by plotting solubility of a solute against temperature; Ø graphs that shows number of grams of solute that will dissolve in 100 g of water over a temperature range of 00C to 100 0C.  Interpreting solubility curves/graphs: For comparing the solubilities of different solutes in a given solvent, several curves can be plotted on the same axes; 1. Solubilities of KNO3 and NaNO3 increase rapidly with increasing temperature; 2. Solubility of NaCl is independent of temperature; 3. Na2SO4 shows a sharp break at 35 0C. Ø the salt exists in solution as Na2SO4.10H2O at temperatures below 35 0C; but gives up its water of crystallization above 35 0C; Ø 35 0C is the transition point; Ø when it cools, saturated solution of Na2SO4 will be deposited; (i) Hydrated crystals Na2SO4.10H2O exists at temperatures below 35 0C; (ii) Anhydrous salt Na2SO4 exists at temperatures above 35 0C.  Examination style questions: 1. How much KNO3 will dissolve in 20 g of water at 25 0C? Solution: 100 g dissolves 42 g !" 20 g dissolves x 42 = 8.4 g #"" 2. At 40 0C, how much potassium nitrate can be dissolved in 300 g of H2O? Solution: 100 g H2O dissolves 65 g KNO3 $"" 300 g H2O dissolves x 65 = 192 g KNO3 #"" 3. How many grams of NaCl can be dissolved in 200 g of water at 70 0C? Ans: 84 g NaCl 4. A saturated solution of NaCl is formed from 100 g of water. If the solution cools from 80 0C to 50 0C, how many grams ppt forms? curve Solution: Trace from graph 80 0C, say solubility = x Trace from graph 50 0C, say solubility = y Find difference: x – y = z g of ppt forms. 5. A 200 cm3 of NaNO3 solution was prepared at a temperature of 70 0C. What mass of the salt crystals form if the solution was cooled to 20 0C? Solution: Trace from graph 80 0C, say solubility = x Trace from graph 50 0C, say solubility = y Find difference: x – y = z g. But volume prepared is 200, total mass = 2z g of crystals formed. 6. At 40 0C, 6.9 g of Na2SO4 dissolves in 30 g of water, what is its solubility in g/100 g H2O? Solution:  Let x = solubility of salt in g/100 g H2O 6.9 g of salt dissolves in 30 g H2O X g of salt dissolves in 100 g H2O ⇒ 30x = 6.9 𝑥 100 '.) * #"" x= = 23 g $" 7. If 10.2 g of a salt Q dissolves in 15.4 cm3 of distilled water at 40 0C, calculate the solubility of Q in (i) mol dm – 3; (ii) g dm – 3. [Mr of Q = 331] Solution: + #".! (i) n(Q) = = = 0.0308 mol , $$# - "."$"/ "."$"/ * #""" "."$"/ c(Q) = = !" = = = 2.00 mod dm – 3.. #0.1 *#" #0.1 "."#01  2 (ii) c = ⇒ ρ = cM = 2.00 x 331 = 662 g dm – 3. , Y TR 8. If 12.20 g of Pb(NO3)2 were dissolved in 21 cm3 of distilled water at 28 0C, determine the solubility of the solute in (i) mol dm – 3; (ii) g dm – 3. [O = 16; N = 14; Pb = 207; Ans: (i) = 1.76; (ii) = 583] 9. Water was added to 120 g of a salt MCl2 to produce 60 cm3 of saturated solution at 25 0C. Its solubility at 25 0C is 8.0 mol dm – 3. Find the mass of the salt that remains undissolved. [M = 24; Cl = 35.5]  10. In the experiment to determine the solubility of KClO3 at different temperatures, the following results were obtained: Temperature/0C 0 10 20 30 40 50 60 Solubility in g dm-3 14 17 20 24 29 34 40 (i) Draw a graph of temperature on the x – axis against solubility on the y – axis; (ii) Use your graph to determine: (𝛼) the solubility of KClO3 at 55 0C; (𝛽) how much of the KClO3 will crystalize out when a saturated solution is cooled from 55 0C to 25 0C? Solution:  Solubility curve of KClO3 45 40 35 Solubility in mol dm - 3 30 25 20 15 10 5 0 0 10 20 30 40 50 60 70 Temperature/0C  Uses of solubility curves/graphs: 1. Find how the solubility of given substance varies with temperature; 2. Find how to purify crystals; 3. Determine the temperature at which a hot solution of a given salt will start to crystalize on cooling; 4. Find how substances are separated; 5. Determine differences in solubilities of two or more solutes/substances. 6. Determine whether a solution of known concentration is saturated or not; 7. Maximum solubility of substance at a range of temperatures can be obtained from the curves.  Solubility rules: Generalization on the solubility of common ionic compounds/salts in water at 25 0C/room temperature. It provides information on the species that will form a ppt. In general, a molar solubility of 0.0100 mol dm – 3 or greater for a substance is considered soluble. Soluble compounds: Ø Compounds containing alkali metal ions and ammonium ion: Li+, Na+, K+, NH19 ; Ø Nitrates (NO; $ ); Ø Halides (Cl – , Br –, I –,); Ø Sulphates (SO!; 1 ); Ø Bicarbonates (HCO; $ ). Insoluble compounds: Ø Carbonates of Ca2+, Pb2+, Ba2+, Zn2+; Ø Hydroxide of Al3+, Zn2+, Pb2+; Ø Halides of Ag+, Pb2+; Ø Sulphates of Ba2+, Pb2+, Ag+, Ca2+; Ø Sulphides, S2 –; Ø Oxides, O 2 –. Sparingly soluble compounds: Ø Ca(OH)2; Ø CaS  Factors that affect solubility of substances: 1. Nature of solvent and solute; 2. Temperature; 3. Charge density of ions; 4. pH or common ion effect; 5. Complex ion formation; 6. Pressure(gases); Nature of solvent and solute: Ø When the molecules of the solute are similar in structure and electrical properties with the solvent, solubility will be high; Ø Polar solvents like water can readily dissolve polar and ionic solutes like NaCl but can not dissolve non – polar solutes like naphthalene; Ø Non – polar solvents like carbon tetrachloride have strong dissolving action on non – polar solutes (like dissolves like); Ø Solubility of gases in liquids is governed by Henry’s law which states that: amount of gas that would be dissolved in a given amount of solvent liquid would directly be proportional to the pressure of the gas above the solution at a given temperature.  Temperature: Ø Solubility of most substances increases with increasing temperature; Ø Solubility of few ones (CaO, Ca(OH)2, CaSO4 decreases with temperature rise. Ø In most cases, solubility of solids in liquid increases with temperature; Ø Some solutes become less soluble at higher temperatures; Ø Temperature effect depends on heat of solution of solute; Ø For gases, solubility decreases as temperature increases; Ø the kinetic energy of gas molecules increases; Ø they have higher ability to escape the attraction of solvent molecules and return to the gas phase.  Charge density: Ø Solubility of ionic compounds depends on the strength of its ionic bond; Ø Stronger bond results in lower solubility; > Ø Strength of bond depends on charge density ( ) of the cation and anion; ? Ø Ion has high charge density if its charge is large and its radius is small and vice versa; Ø Ion with low charge density will form weak ionic bonds than that with high charge density; Ø Compound with weak ionic bonds will be more soluble in water than one with strong ionic bonds; Ø Polyatomic ions such as NH19 , NO; !; $ or SO1 are larger than and have lower charge density than monoatomic ions with the same charge; Weak ionic bonds (more soluble) Strong ionic bonds (less soluble) NH4Cl LiCl LiNO3 LiF Al2(SO4)3 Al2S3  pH Ø Solubility of insoluble acids and bases are affected by pH; Ø Insoluble acids tend to dissolve more in basic solutions; Ø Insoluble bases tend to dissolve more in acidic solutions; Ø Example: solubility of Mg(OH)2: Ø Mg(OH)2(s) ⇌ Mg !9 ; (BC) + 2OH(BC) Ø Adding OH – ions from a base (increases pH) shifts equilibrium from right to left; Ø Solubility of Mg(OH)2 decreases (common ion effect); Ø Adding H+ ions from an acid (decreasing pH) shifts equilibrium to the product side (H+ removes OH – to form H2O; Ø Solubility of Mg(OH)2 increases. Ø pH also influences solubility of salts containing basic anion and anion from weak acid; Ø Example: solubility equilibrium for BaF2: BaF2 ⇌ Ba2+ + 2F – ; Ø In acidic medium, H+ ions pick up F – ions to form HF; Ø Equilibrium shifts to the product side; Ø BaF2 solubility increases; BaF2(s) ⇌ Ba!9 ; (BC) +2F(BC) 9 2H(BC) ; + 2F(BC) ⇌ 2HF(aq) 9 Overall: BaF2(s) + 2H(BC) ⇌ Ba!9 (BC) + 2HF(aq) Solubilities of salts that contain anions such as Cl –, Br –, I – and NO; $ (anions from strong acid) that do not hydrolyze are not affected by pH.  Complex ion formation: Ø Transition metals have particular tendency to form complex ions; Ø Have more than one oxidation number or variable oxidation states; Ø Act as Lewis acids in reactions with many molecules/ions that serve as electron pair donors/Lewis bases; Ø Example: CuSO4 dissolves in water to produce blue solution; Ø Hydrated Cu2+ are responsible for this colour; Ø Addition of a few drops of NH3(aq) forms light/pale blue ppt (Cu(OH)2) Ø Cu!9 BC + OH ; BC ⟶ Cu(OH)2(s) Ø OH – ions are supplied by the ammonia solution; Ø If excess NH3(aq) is added, the blue ppt dissolves (soluble) to produce a dark blue solution due to the formation of the complex ion [Cu(NH3)4]2+ Ø Cu(OH)2(s) + 4NH3(aq) ⟶ [Cu(NH3)4]!9 ; (BC) + 2OH BC Ø Formation of the complex ion increases the solubility of Cu(OH)2  Solubility and solubility product of sparingly soluble ionic compounds: Solubility product: product of the molar concentration of ions in saturated solution of sparingly solution salt raised to their appropriate powers OR Ionic product of a saturated solution of a sparingly soluble salt at standard temperature (of 25 0C or 298 K); Sparingly/slightly soluble solute is one that dissolves until a saturated solution is formed; Precipitation reactions Example: for saturated solution of Ca(OH)2, the following equilibrium is established between the undissolved Ca(OH)2 and its ions in solution: dissolution Ca(OH)2(s) ⇌ precipitation Ca!9 ; (BC) + 2OH(BC) heterogeneous equilibrium; involves solid reactant in equilibrium with ions in aqueous solution: [MB&' ][NO( ]& Ksp = , Ksp = solubility product constant [MB NO &] Omitting [Ca OH. ] which is an intensive property and does not depend on how much of it is present in equilibrium, Ksp = [Ca.0][OH 1]. AgNO3(aq) + HCl(aq) ⇋ AgCl(s) + HNO3(aq) Ag 0 1 (67) + Cl(67) ⇋ AgCl(s) : Keq = [;$ ] AgCl(s) ⇋ Ag 0 (67) + Cl1 (67) : [Ag+][Cl –] = = Ksp ?%& Ksp is determined experimentally Ksp is directly proportional to solubility When Ksp is known solubility expression allows the calculation of the concentrations of ions in equilibrium with the solid solute in aqueous solution at 25 0 C Determination of Ksp for solids 1. The solid is put in water at 25 0 C. 2. Stir the mixture until a saturated solution forms. 3. The molar solubility of the solid is measured and the Ksp is calculated.  Uses of Ksp values 1. Calculation of concentrations of ions in saturated solutions; 2. Determine of whether a precipitate will form or not; 3. For separating precipitate (fractional precipitation); 4. Determine the degree of solubility of substances. Factors that affect Ksp Ø Temperature Ø Pressure Ø Composition Ksp is defined in terms of activity rather than concentration due to its dependent on the conditions above (it is influenced by the surroundings). Relationship between Qsp and Ksp Qsp = Ksp: equilibrium system Ions are forming a ppt just as fast as the solid is dissolving Qsp< Ksp: Unsaturated solution No ppt forms Solution will dissolve more solute Qsp = Ksp: Supersaturated solution Solubility is exceeded Ppt forms and will continue until the concentration of ions in the solution decrease to such a point that Qsp = Ksp, when the system is at equilibrium  Fractional precipitation Ø Process whereby some ions are removed from solution while leaving other ions with similar properties still in solution. Ø Example: Ø If solid NaCl is added slowly to a solution that is 0.010 mol dm – 3 each in Cu+, Ag+, Au+ which salt precipitates fast? Cu+ Ksp = [Cu+ ][Cl–] = 1.9 x 10 – 7 Ag+ Ksp = [Ag+ ][Cl–] = 1.8 x 10 – 10 Au+ Ksp = [Au+ ][Cl–] = 2.0 x 10 – 13 Cl– ions are being added, which system does Qsp > Ksp fast? System with the smallest Ksp (will be most insoluble salt) Au+ precipitates out fast as AuCl, Ag+ as AgCl and Cu+ as CuCl  Examination style questions: 1. The solubility of CaSO4 is found to be 0.65 g dm – 3 at 25 0C. Calculate the value of Ksp for CaSO4 at this temperature. [Ca = 40; S = 32; O = 16] Solution CaSO4(s) ⇌ Ca.0.1 (67) + SOA(67) Let s = molar solubility (in mold dm – 3 of CaSO4) Parameter CaSO4 Ca2+ 𝐒𝐎𝟐; 𝟒 Initial conc. c 0 0 Change in conc. -s +s +s Equilibrium conc. c–s s s Writing the solubility product constant, Ksp for CaSO4 ⟹ Ksp = [Ca2+][𝐒𝐎𝟐% 𝟒 ] = (s)(s) = s 2 M(CaSO4) = 40 + 32 + 64 = 136 g mol – 1 2 ".'0 From c = ⟹c= = 4.78 x 10 – 3 = 0.00478 mol dm – 3 = s , #$' Ksp = (0.00478)2 = 2.28 x 10 – 5  2. Calculate the solubility of Cu(OH)2 at 20 0C in (i) mol dm – 3; (ii) g dm – 3. [Cu = 64; O = 16; H = 1; Ksp = 2.2 x 10 – 20] Solution: Cu HO 2(s) ⇌ Cu.0 1 (67) + 2OH(67) Parameter Cu(HO)2 Cu2+ 2𝐎𝐇 ; Initial conc. c 0 0 Change in conc. -s +s + 2s Equilibrium conc. c–s s 2s (i) Ksp = [Cu2+][OH –]2 = s(2s)2 = s(4s2) = 4s3 !.! * #"(&D ⟹ 2.2 x 10 – 20 = 4s3 ⟹ S3 = = 5.5 x 10 – 21 1 ! s = ∛ 5.5 x 10 – 21 = (5.5 x 10 – 21 ) = 1.77 x 10 – 7 mol dm – 3 " M(Cu(OH)2) = 64 + 34 = 98 g mol – 1 G (ii) c = H ⟹ ρ = c M = 1.77 x 10 – 7 x 98 = 1.73 x 10 – 5 g dm – 3 3. Calculate the molar solubility for Sr3(PO4)2 [Ksp = 1.0 x 10 – 31]  Common ion effect 3. The molar solubility of PbCl2 in a 0.01 mol dm – 3 NaCl solution is 1.7 x 10 – 3 mol dm – 3 at 28 0C. Determine the Ksp for PbCl2. [Pb = 207; Cl = 35.5] Solution: PbCl2(s) ⇌ Pb.0 1 (67) + 2Cl(67) Parameter PbCl2 Pb2+ 2𝐂𝐥; Initial conc. c 0 0.01 Change in conc. -s + 1.7 x 10 – 3 + 2(1.7 x 10 – 3 ) Equilibrium conc. c–s 1.7 x 10 – 3 0.0134 Ksp = [Pb2+][Cl –]2 = (1.7 x 10 – 3)(0.0134)2 = 3.05 x 10 - 7 Ø PbCl2 is not dissolved in pure water; but in 0.01 mol dm – 3 NaCl solution which contains 0.01 mol dm – 3 Cl – and Na+. Ø Na+ does not affect the equilibrium and so does not appear in the Ksp expression. 1. The molar solubility of PbF2 in a 0.10 mol dm – 3 Pb(NO3)2 solution at 25 0C is 3.1 x 10 - 4 mol dm – 3. Calculate the value of Ksp for PbF2. [Pb = 207; F = 19; N = 14; O = 16] 2. At 30 0C, the molar solubility of CoCO3 in a 0.02 mol dm – 3 Na2CO3 solution is 1.0 x 10 – 9 mol dm – 3. Find the value of Ksp for CoCO3. [Co = 59; C = 12; O = 16] 3. The solubility of Ag2SO4 in a solution containing 28.4 g Na2SO4 per dm3 is 4.3 x 10 – 3 mol dm – 3. What is the Ksp for Ag2SO4? [Ag = 108; S = 32; O = 16; Na = 23] 1. Calculate the moles of MgF2 that will dissolve in 1.00 dm3 of (a) Pure water; (b) 0.010 mol dm – 3 Mg(OH)2; (c) 0.010 mol dm – 3 NaF. [Ksp (MgF2) = 6.4 x 10 - 9] (Ans: (a) 1.17 x 10 – 3, (b) 4.0 x 10 – 4, (c) 6.4 x 10 – 5) In a solution of common ion, the solubility of the compound decreases drastically 4. Estimate the solubility of Ag2CrO4 in pure water at 25 0C if the solubility product constant for silver chromate is 1.1 x 10-12. (Ans: 1.1 x 10-12 = [2x]2[x] x = 6.50 x 10-5 mol dm – 3 ) 5. Calculate the solubility product constant for lead(II) chloride, if 50.0 cm3 of a saturated solution of lead(II) chloride was found to contain 0.2207 g of lead(II) chloride dissolved in it at 30 0C (Ans: Ksp = [0.0159][0.0318]2 = 1.61 x 10-5) 6. Estimate the solubility of barium sulphate in a 0.020 mol dm – 3 sodium sulphate solution. The solubility product constant for barium sulphate is 1.1 x 10-10 at 20 0C. (Ans: 1.1 x 10-10 = [x][0.020 + x] = [x][0.020] x = 5.5 x 10-9 mol dm – 3 )  Precipitation or no precipitation 7. 25.0 cm3 of 0.0020 mol dm – 3 potassium chromate are mixed with 75.0 cm3 of 0.000125 mol dm – 3 lead(II) nitrate. Will a precipitate of lead(II) chromate form? [Ksp of lead(II) chromate is 1.8 x 10-14]. 8. If equal volumes of 0.010 mol dm – 3 K2SO4 and 0.10 mol dm – 3 Pb(NO3)2 solutions are mixed, will a ppt form? [Ksp(PbSO4) = 1.8 x 10 – 8, Qsp = 2.4 x 10 – 4] Since equal volumes of the solutions are mixed, each concentration is halved in the final solution 9. If equal volumes of 0.004 mol dm – 3 Na2SO4 and 0.004 mol dm – 3 Pb(NO3)2 solutions are mixed, will a ppt form? [Ksp(PbF2) = 1.8 x 10 – 8, Qsp = 4.0 x 10 – 6] Since equal volumes of the solutions are mixed, each concentration is halved in the final solution 10. If equal volumes of 0.004 mol dm – 3 KF and 0.004 mol dm – 3 Pb(NO3)2 solutions are mixed, will a ppt form? [Ksp(PbF2) = 3.7 x 10 – 8, Qsp = 8.0 x 10 – 9] Since equal volumes of the solutions are mixed, each concentration is halved in the final solution 11. Equal volumes of the following solutions are mixed. Determine whether a ppt will form or not (a) 3.0 x 10 – 6 mol dm – 3 NiCl2 and 6.0 x 10 – 4 Na2Co3 mol dm – 3 [Ksp NiCO3 = 6.6 x 10 - 9] (b) 2.0 x 10 – 2 mol dm – 3 NaF and 2.0 x 10 – 3 mol dm – 3 MgCl2 [Ksp MgF2 = 6.4 x 10 - 9] 12. Will any solid Ba(IO3)2 form when 6.5 mg of BaCl2 is dissolved in 0.500 dm3 of 0.033 mol dm – 3 NaIO3?[Ksp Ba(IO3)2 = 1.5 x 10 – 9, BaCl2 = 208.23] (Ans = 6.2 x 10 - 5) 13. (a) what concentration of Cu2+ is necessary to initiate precipitation from a solution containing 0.0015 mol dm – 3 KOH? (7.1 x 10 – 14) (b) Suppose more solid Cu(NO3)2 was added until [Cu2+] = 0.0015 mol dm – 3 , what would be the concentration of OH – in solution?(1.0 x 10 – 8) (c) Determine the pH of the solution. (6.01) [Ksp = 1.6 x 10 – 19] (Ignore the effect of Cu2+, a weak acid) 14. If silver ions are to be removed from solution by precipitation of Ag2S, what final concentration of sulphide ions are required to reduce the Ag+ concentration to 1.0 x 10 – 12 mol dm – 3[Ksp = 1.0 x 10 – 49] (1.0 x 10 – 25) As more and more S 2 – (from Na2S) added, more and more Ag2S precipitates out, but there is a tiny amount of S 2 – and Ag+ ions that are in solution at all times.  First, determine the overall and the net-ionic equations for the reaction that occurs when the two solutions are mixed. K2CrO4(aq) + Pb(NO3)2(aq) ⟶ 2 KNO3(aq) + PbCrO4(s) Pb2+(aq) + CrO42-(aq) ⟶ PbCrO4(s) The latter reaction can be written in terms of Ksp as: PbCrO4(s) ⟶ Pb2+(aq) + CrO42-(aq) Ksp = [Pb2+][CrO42-] Using the dilution equation, C1V1 = C2V2, determine the initial concentration of each species once mixed (before any reaction takes place). (0.0020)(25.0) = (C2)(100.0) C2 for K2CrO4 = 0.00050 mol dm- 3 Similar calculation for the lead(II) nitrate yields: C2 for Pb(NO3)2 = 0.0000938 M Using the initial concentrations, calculate the reaction quotient Q, and compare to the value of the equilibrium constant, Ksp. QC = (0.0000938)(0.00050) = 4.69 x 10-8 QC > Ksp so a precipitate of lead(II) chromate will form. 9. The molar solubility of BaSO4 is 1.05 x 10 – 5 mol dm – 3. What is the Ksp of BaSO4? (Ans: 1.10 x 10 – 10 ) 10. A 1.00 dm3 of a saturated solution of Ba3(PO4)2 contains 3.96 x 10 –4 g of the salt. Calculate the Ksp of the salt at 25 0 C [Ba3(PO4)2 = 601.8 g mol – 1] (Ans: 1.30 x 10 - 29) 11. Given a saturated solution of Mg(OH)2 (a) What is the molar solubility of the salt? (1.55 x 10 - 4) (b) What is the concentration of Mg2+ and OH – and the pH of the saturated solution? ([Mg2+ = 1.55 x 10 - 4], [OH – = 3.11 x 10 – 4], pH = 10.49) (c) What is the solubility of Mg(OH)2 in g dm – 3?[8.99 x 10 – 3] [9 marks] 8. X g of BaF2 are dissolved in 75.0 cm3 of a saturated solution. Determine the value of X [Ba = 137; F = 19, Ksp of BaF2 = 10 – 6] End of solubility For your meditation: Pv 2:6 8. X g of BaF2 are dissolved in 75.0 cm3 of a saturated solution. Determine the value of X [Ba = 137; F = 19, Ksp of BaF2 = 10 – 6] End of solubility For your meditation: Pv 2:6