The solubility product is a kind of equilibrium constant and its value depends on temperature. Ksp usually increases with an increase in temperature due to increased solubility. Solubility is defined as a property of a substance called solute

The solubility product is a kind of equilibrium constant and its value depends on temperature. Ksp usually increases with an increase in temperature due to increased solubility. Solubility is defined as a property of a substance called solute to get dissolved in a solvent in order to form a solution

1. Chapter II THE SOLUBILITY EQUILIBRIUM

2. I. solubility and solubility product . • Solubility s ( g L −1 or mol L −1 ) : maximum number of moles of a salt that can be dissolved in one liter of water. We then obtain a saturated solution ⇒ if we add solid: the solubility remains unchanged . Definitions : CH 3 COOAg (s) + water Ag + + CH 3 COO - The solubility product constant : Ksp = [CH 3 COO - ].[Ag + ]

3. The solubility product constant is the equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is denoted by the symbol Ksp . (Van t’ Hoff law ) Ksp ( CH 3 COOAg ) = 2.10 −3 at 25°C Condition for the formation of a precipitate   - - If the solution is obtained by dissolving the salt CH 3 COOAg (s) . Two cases can arise ● [Ag + ] [CH 3 COO - ] < K sp no formation of precipitate equilibrium is not reached Salt in solution water Water and salt

4. ● [Ag + ] [CH 3 COO - ] > K s P : CH 3 COOAg (s) precipitates [ Ag + ] [CH 3 COO - ] = K sp onset of equilibrium onset of precipitation water salt

5. Compound K sp Compound K sp AgOH 1,95.10 -8 Ca(OH) 2 5,5.10 -6 Ag 2 CrO 4 1,12.10 -12 CaSO 4 2,4.10 -5 AgCℓ 1,79.10 -10 Cu(OH) 2 3,16.10 -19 AgBr 4,95.10 -13 Mg(OH) 2 1,82.10 -11 AgI 8,32.10 -17 Pb(OH) 2 8,13.10 -17 BaSO 4 1,07.10 -10 CdS 7,94.10 -27 CaCO 3 3,8.10 -9 FeS 6,31.10 -18

6. The relation between solubility and the solubility product constants is that one can be used to derive the other. In other words, there is a relationship between the solute's molarity and the solubility of the ions because Ksp is the product of the solubility of each ion in moles per liter For example, the lead sulfate dissociates into its ions as given in the equation below: PbSO 4 ⇌ Pb 2+ + SO 4 2- The equilibrium constant for this reaction is given by: Kc = [Pb 2+ ] [SO 4 2- ] / [PbSO 4 ] Relation between Ksp and s

7. Note that , the concentration of lead sulfate almost remains constant. This is because of the low solubility . So, When this solid combines with the K c   it gives a new constant known as the solubility product . K c  [PbSO4] = [Pb 2+ ] [SO 4 2- ] K c  [PbSO4] = K sp K sp  = [Pb 2+ ] [SO 4 2- ] Thus for lead sulfate the solubility and solubility product can be related as follows :

8. Ksp = s 2 K sp  is known as the solubility product of sulfate s is the solubility It is to be noted the solubility of salt is different at different temperatures

9. aqueous solution of

10. Application: Ag 2 CrO 4(s) 2Ag + + Cr O 4 2 - t=0 C o 0 0 t e quilibrium C o - s 2 s s the solubility s Is equal at : We know: K sp = [Ag + ] 2 [CrO 4 2 - ] = 1,12.10 - 12 at 25°C. so : s = 3 1 ) 4 ( s K = 6,54.10 - 5 mol.L - 1

11. II - Effect of the addition of a common ion on the solubility To this solution, Cl − ions are added in the form of solid NaCl or liquid HCl (the volume remains 1 L). t C0 0 c t eq C0 – s’ s’ c+s’ Consider one liter of a saturated AgCl solution : AgCl (s ) Ag (aq) + + Cl (aq) - water 1 - 1

12. AgCl solution NaCl solution

13. According to the law of moderation the reaction ‒1 is favoured , that of the formation of AgCl (s) this reaction is triggered until reaching a new equilibrium state in which the solubility of AgCl (s') is less than (s). S’< s Application : Suppose that one adds to one liter of aqueous solution of Ba 2+ and SO 4 2- ion, in equilibrium with an excess of solid BaSO 4 (s), 0.1 mole of SO 4 2- coming for example fromK 2 SO 4 . It is proposed to determine the solubility "s" before addition and the solubility "s" after addition of SO 4 2- Ksp = 1,07.10 -10 at 25°C

14. Solution: BaSO 4(s) ⇄ Ba 2+ + SO 4 2- We define the solubility product K s P   : K s = [Ba 2+ ] [SO 4 2- ] BaSO 4(s) ⇄ Ba 2+ + SO 4 2- t=0 C o 0 0 t equilibrium C o - s s s   The solubility s is equal to s = [Ba 2+ ] = [SO 4 2- ] K sp = [Ba 2+ ] [SO 4 2- ] = 1,07.10 -10 at 25°C. And s = 1,03.10 -5 mol.ℓ -1

15. BaSO 4(s) ⇄ Ba 2+ + SO 4 2- Before addition of SO 4 2- s s After addition of SO 4 2- s’ s’ +0,1 K sp = [Ba 2+ ] [SO 4 2- ] = s ’( s ’+ 0,1) = 1,07.10 -10 at 25°C. we know : s > s ’ so s’<< 0,1mol/L K sp = [Ba 2+ ] [SO 4 2- ] =0,1 s ’ = 1,07.10 -10 and  : s’ = 10. K sp = 1,07.10 -9 mol.ℓ -1

16. The Effect of Acid–Base Equilibrium at the Solubility of Salts the effect of pH on the solubility of a representative salt , M + , A − , where A − is the conjugate base of the weak acid HA When the salt dissolves in water, the following reaction occurs: MA (s) ⇌M + ( aq ) +A − ( aq ) with Ksp =[M + ][A − ] The anion can also react with water in a hydrolysis reaction: A − ( aq ) +H 2 O (l) OH - ( aq ) +HA ( aq ) ) III-The effects of pH on solubility The solubility of many compounds depends strongly on the pH of the solution.

17. If instead a strong acid is added to the solution, the added H+ will react essentially completely with A− to form HA. This reaction decreases [A−], which decreases the magnitude of the ion product Q=[M+][A−] A − ( aq ) +H 2 O (l) OH - ( aq ) +HA ( aq ) )

18. According to Le Chatelier’s principle, more MA will dissolve until Q= Ksp . Hence an acidic pH dramatically increases the solubility of virtually all sparingly soluble salts whose anion is the conjugate base of a weak acid. In contrast, pH has little to no effect on the solubility of salts whose anion is the conjugate base of a stronger weak acid or a strong acid, respectively ( chlorides, bromides, iodides, and sulfates). For example, the hydroxide salt Mg(OH)2 is relatively no solubily in water: Mg(OH) 2(s) ⇌Mg 2+ ( aq )+2OH − ( aq ) With Ksp =5.61×10 −12

19. 2- Solubility of hydroxydes Given a solution containing 1 mol.ℓ -1 of magnesium ion Mg 2+ to which a strong base is added, we want to determine the pH at which the precipitation of the hydroxide Mg(OH) 2 begins. Let the balance  : Mg(OH) 2(s) ⇄ Mg 2+ + 2OH - K sp = [Mg 2+ ] [OH - ] 2 = 7.10 - 12 At 25°C. There is a start of precipitation if : K s = [Mg 2+ ] [OH - ] 2 Hence the expression of the pH

20. A solution containing 10 -1 mol l -1 of ferric Fe 3+ ion and 10 -2 mol l -1 of ferrous Fe 2+ ion. Solid sodium hydroxide is gradually added to it. Calculate the precipitation onset pH of each ion. Calculate the concentration of the first ion when the second begins to precipitate. pK sp1(Fe(OH)3(s ) =37,8 et pK sp2(Fe(OH)2(s ) =15,1   ●Solution : Fe(OH) 2(s) ⇄ Fe 2+ + 2OH - ; K s2 = [Fe 2+ ] [OH - ] 2 = 10 -15,1 à 25°C. Fe(OH) 3(s) ⇄ Fe 3+ + 3OH - ; K s1 = [Fe 3+ ] [OH - ] 3 = 10 -37,8 à 25°C. L’hydroxyde Fe(OH) 3(s) ferric hydroxide starts to precipitate when we have : K sp1 = [Fe 3+ ] [OH - ] 3 = 10 -1 [OH - ] 3 = 10 -37,8 and [OH - ] 3 = 10 -36,8 so : pH = 1,73.   L’hydroxyde Fe(OH) 2(s) ferrous hydroxyde starts to precipitate when we have a  : K sp1 = [Fe 2+ ] [OH - ] 2 = 10 -1 [OH - ] 2 = 10 -15,1 et [OH - ] 2 = 10 -14,1 so : pH = 6,95

21. L’hydroxyde Fe(OH) 2(s) begins to precipitate when the concentration of Fe 3+ ions is equal to:

22. How much solid Pb (NO 3)2 must be added to 1.0 L of 0.0010 M Na 2 SO 4 solution for a precipitate of PbSO 4 , [ Ksp = 1.6 × 10 —8 ] to form ? Assume no change in volume when the solid is added. We need to mixt Application: Solution : To precipate PbSO 4 We need to mixt SO 4 2- + Pb 2+ PbSO 4 So the precipitation equilibrium PbSO 4 SO 4 2- + Pb 2+ And [ Ksp = 1.6 × 10 —8 ] Ksp = [ SO 4 2- ] x [ Pb 2+ ] When the Na 2 SO 4 is added means that [ SO 4 2- ] is 0.001 M In moment when precipate forms the Q= Ksp

23. 0.001 x [ Pb 2+ ] = 1.6 × 10 —8 [ Pb 2+ ]= 1.6 × 10 —5 mol/L In 1.0 L of water the mol Pb 2+ = mol Pb (NO 3)2 =1.6 × 10 —5 n Pb (NO 3)2 = m / 331.22 g/mol m = 0.0053 g

24. – Solubility Equilibrium - Exercice -1 The solubility product of lead iodate (II) Pb(IO 3 ) 2 is such that pKsp = 12.58. Calculate the solubility of this salt : -In pure water

25. In sodium nitrate solution C = 0.1 mol.L -1 . The NaNO 3 nitrate solution does not contain common ions, Pb (IO3)2 is not influenced by the presence of NaNO3. In lead nitrate (II) solution at C=3.5 . 10 -2 mol.L -1 . In a lead nitrate solution, there is a common ion which is Pb 2+ with C= 3,5.10 -2 mol.L -1 . And Ksp = (c+s’)(2s’) 2 and s’<<< C et s’<s

26. We neglect s' before C Ksp = C(4s’ 2 ) s’= ( ksp /4C) 1/2 s’ = 1,3.10 -6 mol.L -1 - In sodium iodate solution at C = 0.1 mol.L -1 . Ksp = [Pb 2+ ][IO 3 - ] 2 = s’’(C+2 s’’) 2 K sp = s’’(C) 2 s’’= Ksp /C 2 = 10 -10,58 mol.L -1

27. Exercice -2 A solution containing the Cl - and I - ions is available at the same concentration C = 10 -3 mol.L -1 to which a solution of silver nitrate is gradually added . Two precipitates may appear , AgCl and AgI . Precipitation is said to be competitive . They are successive if, at the time the second precipitation begins , the remaining concentration of the first anion is less than 1% of its initial value, and if not simultaneous . Data: Ksp ( AgCl ) = 1.6 .10 -10 and Ksp ( AgI ) = 7.9. 10 -17 . 1 Determine the minimum Ag+ concentration from which each precipitate is formed . Deduce the one that appears first in the beaker .

28. 1- to calculate the concentration in [Ag+], at the beginning of precipitation , that is when the equilibrium has just been reached but the concentration in I - and Cl - is still almost equal to C . At the beginning of AgCl precipitation . Ksp ( AgCl )= [Ag+]. C where [Ag+] AgCl = Ksp ( AgCl )/C [Ag+] AgCl = 1.6.10 -7 mol.L -1 Similarly , at the beginning of the precipitation of AgI Ksp ( AgI )= [Ag+]. C where [Ag+] AgI = Ksp ( AgI )/C [Ag+] AgI = 7.9.10 -14 mol.L-1 So it is the silver iodide precipitate that appears first.

29. 2 - Is the precipitation successive or simultaneous ? 2- Precipitation is successive if when the precipitation of AgCl begins [Ag+] AgCl (max), the concentration in I - is less than 1% of its initial value, that is less than 0.01C As the solid AgI is of course present then at this moment Ksp ( AgCl )= [Ag+] AgCl .[I-] so [I-] Remaining = Ksp ( AgI )/ [Ag+] AgCl = 4.9.10 -10 mol.L -1 .   As [I-] remaining <C /100 then both precipitation is succesous

30. Exercice -3 Nickel solution 0.01 M at pH=1, saturated with hydrosulphic acid by bubbling of this gas (solution A). Knowing that the solubility of H 2 S is equal to 0.1 M a – Does nickel sulphide precipitate in solution A?