شیمی تجزیه مبحث: حلالیت رسوب ها دکتر امید رجبی دانشیار گروه شیمی دارویی دانشکده داروسازی مشهد دکتر امید رجبی دانشیار گروه شیمی دارویی دانشکده داروسازی مشهد

The Solubility Product Constant, K sp Many ionic compounds are only slightly soluble in water: ex. Ag salts, sulfides Equations are written to represent the equilibrium between the compound and the ions present in a saturated aqueous solution AgCl(s)Ag + (aq) + Cl – (aq) K sp = [Ag + ][Cl – ]

Ksp’s (25 °C)

Ksp and Molar Solubility The solubility product constant is related to the solubility of an ionic solute K sp = [Ag + ][Cl – ]; solubility given by [Ag + ] From stoichiometry, the ion ratio is 1:1, so [Ag + ] = [Cl – ], both of which are unknown (x) Ag + Cl – Ag + Cl – + K sp = x 2 and [Ag + ] = (K sp ) 1/2

The Common Ion Effect Le Châtelier’s principle is followed for the shift in concentration of products and reactants upon addition of either products or reactants to a solution The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution Ag 2 SO 4 (s)2 Ag + (aq) + SO 4 –2 (aq) Solubility of Ag 2 SO 4  if MgSO 4 is added to solution

Common Ion Effect Illustrated

Does Precipitation Occur? Q ip is the ion product reaction quotient and is based on initial conditions of the reaction Precipitation should occur if Q ip > K sp Precipitation cannot occur if Q ip < K sp A solution is just saturated if Q ip = K sp

Example If 1.00 mg of Na 2 CrO 4 is added to 225 mL of M AgNO 3, will a precipitate form? Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2 – (aq) K sp = 1.1 x 10 – 12

A Conceptual Example Pictured here is the result of adding a few drops of concentrated KI(aq) to a dilute solution of Pb(NO 3 ) 2. What is the solid that first appears? Explain why it then disappears.

Example If L of M MgCl 2 and L of M NaF are mixed, should a precipitate of MgF 2 form? MgF 2 (s) Mg 2+ (aq) + 2 F – (aq) K sp = 3.7 x 10 –8

Example An aqueous solution that is 2.00 M in AgNO 3 is slowly added from a buret to an aqueous solution that is M in Cl – and also M in I –. a. Which ion, Cl – or I –, is the first to precipitate from solution? b. When the second ion begins to precipitate, what is the remaining concentration of the first ion? c. Is separation of the two ions by selective precipitation feasible ? AgCl(s) Ag + (aq) + Cl – (aq) K sp = 1.8 x 10 –10 AgI(s) Ag + (aq) + I – (aq) K sp = 8.5 x 10 –17

Selective Precipitation AgNO 3 added to a mixture containing Cl – and I –

If the anion of a precipitate is that of a weak acid, the precipitate will dissolve somewhat when the pH is lowered: If, however, the anion of the precipitate is that of a strong acid, lowering the pH will have no effect on the precipitate. Added H + reacts with, and removes, F – ; LeChâtelier’s principle says more F – forms. H + does not consume Cl – ; acid does not affect the equilibrium. Effect of pH on Solubility CaF 2 (s) Ca 2+ (aq) + 2 F – (aq) AgCl(s) Ag + (aq) + Cl – (aq)

Example What is the molar solubility of Mg(OH) 2 (s) in a buffer solution having [OH – ] = 1.0 x 10 – 5 M, that is, pH = 9.00? Example What is the molar solubility of Mg(OH) 2 (s) in a buffer solution having [OH – ] = 1.0 x 10 – 5 M, that is, pH = 9.00? Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH – (aq) K sp = 1.8 x 10 –11 A Conceptual Example Without doing detailed calculations, determine in which of the following solutions Mg(OH) 2 (s) is most soluble: (a) 1.00 M NH 3 (b) 1.00 M NH 3 /1.00 M NH 4 + (c) 1.00 M NH 4 Cl.

Equilibria Involving Complex Ions Silver chloride becomes more soluble, not less soluble, in high concentrations of chloride ion.

A complex ion consists of a central metal atom or ion, with other groups called ligands bonded to it. The metal ion acts as a Lewis acid (accepts electron pairs). Ligands act as Lewis bases (donate electron pairs). The equilibrium involving a complex ion, the metal ion, and the ligands may be described through a formation constant, K f : A complex ion consists of a central metal atom or ion, with other groups called ligands bonded to it. The metal ion acts as a Lewis acid (accepts electron pairs). Ligands act as Lewis bases (donate electron pairs). The equilibrium involving a complex ion, the metal ion, and the ligands may be described through a formation constant, K f : Complex Ion Formation Ag + (aq) + 2 Cl – (aq) [AgCl 2 ] – (aq) [AgCl 2 ] – K f = –––––––––– = 1.2 x 10 8 [Ag + ][Cl – ] 2

Complex Ion Formation Concentrated NH 3 added to a solution of pale-blue Cu 2+ … … forms deep-blue Cu(NH 3 ) 4 2+.

Complex Ion Formation and Solubilities AgCl is insoluble in water. But if the concentration of NH 3 is made high enough … … the AgCl forms the soluble [Ag(NH 3 ) 2 ] + ion.

Example Calculate the concentration of free silver ion, [Ag + ], in an aqueous solution prepared as 0.10 M AgNO 3 and 3.0 M NH 3. Example Calculate the concentration of free silver ion, [Ag + ], in an aqueous solution prepared as 0.10 M AgNO 3 and 3.0 M NH 3. Ag + (aq) + 2 NH 3 (aq) [Ag(NH 3 ) 2 ] + (aq) K f = 1.6 x 10 7 Example If 1.00 g KBr is added to 1.00 L of the solution described in Example 16.13, should any AgBr(s) precipitate from the solution? AgBr(s) Ag + (aq) + Br – (aq) K sp = 5.0 x 10 –13

حلالیت اگسالات کلسیم را در محلولی با غلظت یون هیدرونیوم 1.00x10 -4 محاسبه کنید CaC 2 O 4 (S) Ca 2+ + C 2 O 4 2- C 2 O H 3 O + HC 2 O H 2 O HC 2 O H 3 O + H 2 C 2 O 4 + H 2 O CaC 2 O 4 (S) Ca 2+ + C 2 O 4 2- C 2 O H 3 O + HC 2 O H 2 O HC 2 O H 3 O + H 2 C 2 O 4 + H 2 O

Solubility = [Ca 2+ ] = [C 2 O 4 2- ]+[HC 2 O 4 - ]+[H 2 C 2 O 4 ] Solubility = [Ca 2+ ] = [C 2 O 4 2- ]+[HC 2 O 4 - ]+[H 2 C 2 O 4 ]

K sp =[Ca 2+ ][C 2 O 4 2- ]=2.3x10 -9 [Ca 2+ ] = [C 2 O 4 2- ]+[HC 2 O 4 - ]+[H 2 C 2 O 4 ] K b1 =[C 2 O 4 2- ][H 3 O + ]/[HC 2 O 4 - ]=5.42x10 -5 K b2 =[HC 2 O 4 - ][H 3 O + ]/[H 2 C 2 O 4 ]=5.36x10 -2 [H 3 O + ]=1.0 x10 -4 [Ca 2+ ] = [C 2 O 4 2- ]+[HC 2 O 4 - ]+[H 2 C 2 O 4 ] K b1 =[C 2 O 4 2- ][H 3 O + ]/[HC 2 O 4 - ]=5.42x10 -5 K b2 =[HC 2 O 4 - ][H 3 O + ]/[H 2 C 2 O 4 ]=5.36x10 -2 [H 3 O + ]=1.0 x10 -4

(1.0x10 -4 )[C 2 O 4 2- ]/ [HC 2 O 4 - ]=5.42x10 -5 [HC 2 O 4 - ]=1.84[C 2 O 4 2- ] (1.0x10 -4 ) x1.84[C 2 O 4 2- ]/ [H 2 C 2 O 4 - ]=5.36x10 -2 [H 2 C 2 O 4 - ]= [C 2 O 4 2- ] (1.0x10 -4 )[C 2 O 4 2- ]/ [HC 2 O 4 - ]=5.42x10 -5 [HC 2 O 4 - ]=1.84[C 2 O 4 2- ] (1.0x10 -4 ) x1.84[C 2 O 4 2- ]/ [H 2 C 2 O 4 - ]=5.36x10 -2 [H 2 C 2 O 4 - ]= [C 2 O 4 2- ]

[Ca 2+ ] = [C 2 O 4 2- ]+[HC 2 O 4 - ]+[H 2 C 2 O 4 ] [Ca 2+ ]=[C 2 O 4 2- ]+ 1.84[C 2 O 4 2- ] [C 2 O 4 2- ] [C 2 O 4 2- ]= [Ca 2+ ]/2.84 [Ca 2+ ]=[C 2 O 4 2- ]+ 1.84[C 2 O 4 2- ] [C 2 O 4 2- ] [C 2 O 4 2- ]= [Ca 2+ ]/2.84 K sp =[Ca 2+ ][C 2 O 4 2- ]=2.3x10 -9 [Ca2+] [Ca2+]/2.84= 2.3x10 -9 [Ca2+]=8.1x10 -5 Solubility of CaC 2 O 4 =8.1x10-5

محاسبات حلالیت در حالتی که غلظت یون هیدرونیوم متغیر است حلالیت PbCO 3 را در آب محاسبه کنید:

PbCO 3 Pb 2+ + CO 3 2- CO H 2 O HCO OH - HCO H 2 O H 2 CO 3 + OH - 2H 2 O H 3 O + + OH - CO H 2 O HCO OH - HCO H 2 O H 2 CO 3 + OH - 2H 2 O H 3 O + + OH - Solubility= [Pb2+] =[CO 3 2- ]+[HCO 3 - ]+[H 2 CO 3 ]

[Pb 2+ ][CO 3 2- ]=K sp =3.3x10-14 [HCO 3 - ][OH - ]/[CO 3 2- ]=K 1b =1.0x /4.7x =2.13x10 -4 [H 2 CO 3 ][OH - ]/[HCO 3 - ]=K 2b =1.0x /4.45x10 -7 =2.25x10 -8 [H 3 O + ][OH - ]=1.00x معادله موازنه بار یا Charge balance : 2[Pb 2+ ]+[H 3 O + ]=2[CO 3 2- ]+[HCO 3 - ]+[OH - ] [Pb 2+ ]=[CO 3 2- ]+[HCO 3 - ]+[H 2 CO 3 ] 6 معادله و 6 مجهول داریم!

[Pb 2+ ]=[CO 3 2- ]+[HCO 3 - ]+[H 2 CO 3 ] 2[Pb 2+ ]+[H 3 O + ]=2[CO 3 2- ]+[HCO 3 - ]+[OH - ] 0=[OH - ]+[HCO 3 - ] [OH - ] = [HCO 3 - ] [HCO 3 - ][OH - ]/[CO 3 2- ] = K 1b = 1.0x /4.7x = 2.13x10 -4 =[HCO 3 - ] 2 /[CO 3 2- ] [HCO3-]=√K 1b [CO 3 2- ] [Pb 2+ ]=[CO 3 2- ]+[HCO 3 - ] [Pb 2+ ]=[CO 3 2- ]+√K 1b [CO 3 2- ]

[Pb 2+ ][CO 3 2- ]=K sp =3.3x [ CO 3 2- ]=K sp / [Pb 2+ ] [Pb 2+ ]=[CO 3 2- ]+√K 1b [CO 3 2- ] [Pb 2+ ]= Ksp/ [Pb 2+ ] +√K 1b Ksp/ [Pb 2+ ] [Pb 2+ ] = 1.9x10 -6

حلالیت Fe(OH) 3 را در آب محاسبه کنید: Ksp=[Fe 3+ ][OH - ] 3 =4x Charge Balance: 3[Fe 3+ ]+[H 3 O + ]=[OH - ] [Fe 3+ ](3[Fe 3+ ]) 3 =4x Ksp=[Fe 3+ ][OH - ] 3 =4x Charge Balance: 3[Fe 3+ ]+[H 3 O + ]=[OH - ] [Fe 3+ ](3[Fe 3+ ]) 3 =4x [Fe3+]=2x [OH-]=6x [H3O+] = 1.7x10 -5 فرض ما در مورد حذف H3O+ در مقابل Fe3+ اشتباه بوده است!

3[Fe3+] « [H3O+] 3[Fe 3+ ]+[H 3 O + ]=[OH - ] [Fe 3+ ]=4x /(1.00x10 -7 ) 3 = 4x10 -17

اثر غلظت الکترولیت روی حلالیت Effects of Ionic strength on Solubility Effects of Ionic strength on Solubility

Activity dilute solution

Activity Ions in concentrated solution interact

Activity In concentrated solution, ions interaction makes it appear that there are fewer ions than there really are a A =  A [A] where: a A is activity of A  A is the activity coefficient [A] is molarity (as usual) In concentrated solution, ions interaction makes it appear that there are fewer ions than there really are a A =  A [A] where: a A is activity of A  A is the activity coefficient [A] is molarity (as usual)

Significance K sp for BaSO 4 in water: in 0.01 M KNO 3 : 2.9 x Solubility has NOT really increased but K + and NO 3 - act to shield Ba 2+ and SO 4 2+ from each other in solution K sp for BaSO 4 in water: in 0.01 M KNO 3 : 2.9 x Solubility has NOT really increased but K + and NO 3 - act to shield Ba 2+ and SO 4 2+ from each other in solution

Do All Ions Work Equally Well? 2 factors important: –concentration of ions –charge on ions 2 factors important: –concentration of ions –charge on ions

Ionic Strength  = 0.5  c i Z i 2 where: c i is molarity of ith ion Z i is charge on ith ion Note: effect of square is to remove sign of charge so + and - don’t cancel out!  = 0.5  c i Z i 2 where: c i is molarity of ith ion Z i is charge on ith ion Note: effect of square is to remove sign of charge so + and - don’t cancel out!

Examples Calculate and compare the ionic strength of 0.1 M NaCl, 0.1 M Na 2 SO 4, and 0.1 M MgSO M NaCl = M Na 2 SO 4 = M MgSO 4 = 0.4 Calculate and compare the ionic strength of 0.1 M NaCl, 0.1 M Na 2 SO 4, and 0.1 M MgSO M NaCl = M Na 2 SO 4 = M MgSO 4 = 0.4

K sp =a A m.a B n =[A] m [B] n.γ A m.γ B n [A] m [B] n =K sp / γ A m.γ B n