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Lecture 35

Lecture 35. Precipitation Titrations Solubility in Acid Solutions. Fractions of Dissociating Species in Polyligand Complexes

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Lecture 35

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  1. Lecture 35 Precipitation Titrations Solubility in Acid Solutions

  2. Fractions of Dissociating Species in Polyligand Complexes When polyligand complexes are dissociated in solution, metal ions, ligand, and intermediates are obtained in equilibrium with the complex. For example, look at the following equilibria Ag+ + NH3D Ag(NH3)+ kf1 = [Ag(NH3)+]/[Ag+][NH3] Ag(NH3)+ + NH3D Ag(NH3)2+ kf2 = [Ag(NH3)2+]/[Ag(NH3)+][NH3] We have Ag+, NH3, Ag(NH3)+, and Ag(NH3)2+ all present in solution at equilibrium where CAg = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+]

  3. The fraction of each Ag+ species can be defined as: b0 = [Ag+]/ CAg b1 = [Ag(NH3)+]/ CAg b2 = [Ag(NH3)2+]/ CAg As seen for fractions of a polyprotic acid dissociating species, one can look at the b values as b0 for the fraction with zero ligand (free metal ion, Ag+), b1 as the fraction of the species having one ligand (Ag(NH3)+) while b2 as the fraction containing two ligands (Ag(NH3)2+). The sum of all fractions will necessarily add up to unity(b0 + b1 + b2 = 1)

  4. For the case of b0, we make all terms as a function of Ag+ since b0 is a function of Ag+. We use the equilibrium constants of each step: CAg = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+] kf1 = [Ag(NH3)+]/[Ag+][NH3] [Ag(NH3)+] = kf1 [Ag+][NH3] Kf1 x kf2 = [Ag(NH3)2+]/[Ag+][NH3]2 [Ag(NH3)2+] = Kf1 x kf2 [Ag+][NH3]2 Substitution in the CAg relation gives: CAg = [Ag+] + kf1 [Ag+][NH3] + Kf1 x kf2 [Ag+][NH3]2 CAg = [Ag+]( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2) [Ag+]/ CAg = 1/( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)

  5. Precipitation Reactions and Titrations

  6. We had previously looked at precipitation equilibria in solution and we are familiar with calculations of solubility of sparingly soluble salts in pure water, in presence of a common ion, and in presence of diverse ions. However, we should remember that a salt is formed from a metal ion (a very weak conjugate Lewis acid) that will not react with water, and a conjugate base that may be weak and will not react with water (like Cl-, Br-, I-) or a strong conjugate base that will react with water (all conjugate bases of weak acids). Calculations presented so far deal with the first situation where the conjugate base is a very weak base. Let us now look at sparingly soluble salts of weak acids:

  7. Effect of Acidity on the Solubility of Precipitates The conjugate base of a weak acid is strong enough to react with water upon dissociation of the precipitate. At low pH values, excessive amounts of the conjugate base will be converted to the weak acid and thus forces the precipitate to further dissociate to produce more anions to satisfy the equilibrium constant (ksp). Look at the general example assuming A-is the conjugate base of a weak acid and in presence of acid solution (low pH):

  8. MA(s)DM++ A-ksp= [M+][A-] A-+ H+DHAka= [A-][H+] /[HA] Since some of the formed A-is converted to HA, [M+] no longer equals [A-]. However, we can write: CT= [A-] + [HA], where CT= [M+] [A-] =a1CT Recall EDTA equilibria in solution when its chelate dissociates in water. Therefore, for the equilibrium reaction: MA(s)DM++ A-

  9. We can write: Ksp= s *a1s S = (ksp/a1)1/2

  10. Example Calculate the solubility of CaC2O4(ka1= 6.5x10-2, ka2= 6.1x10-5, ksp= 2.6x10-9) in a 0.001 M HCl solution. Solution CaC2O4DCa2++ C2O42- The released oxalate will form H2C2O4, HC2O4-, and some will remain as C2O42-where: CC2O42-= s = [Ca2+] = [H2C2O4] + [HC2O4-] + [C2O42-] and: [C2O42-] =a2s

  11. Ksp= s *a2s S = (ksp/a2)1/2 Therefore, we must calculatea2 a2= ka1ka2/ ([H+]2+ ka1[H+] + ka1ka2) a2= (6.5x10-2* 6.1x10-5)/( (0.001)2+ (6.5x10-2* 0.001) + (6.5x10-2* 6.1x10-5) ) a2= 5.7x10-2 s = (2.6x10-9/5.7x10-2)1/2= 2.1x10-4M Compare with solubility in absence of acidity!!

  12. Example Find the solubility of Ag3PO4at pH 3.a3at pH 3=3.3x10-14,ksp= 1.3x10-20 Solution Ag3PO4D3 Ag++ PO43- Solid 3s a3s Ksp= (3 s)3*a3s S = (ksp/27a3)1/4 S = (1.3x10-20/27x3.3x10-14)1/4 S = 1.1x10-2M

  13. Effect of Complexation on Solubility We have seen earlier that acidity affects the solubility of precipitates that contains the conjugate base of a weak acid. In the same manner, if a sparingly soluble precipitate was placed in a solution that contains a ligand which can form a complex with the metal ion of the precipitate, solubility of the precipitate will increase due to complex formation. For example, if solid AgCl is placed in an ammonia solution, the following equilibria will take place:

  14. AgCl(s)D Ag+ + Cl- Ag+ + NH3D Ag(NH3)+ Ag(NH3)+ + NH3D Ag(NH3)2+ The dissociated Ag+ ions will form complexes with ammonia, forcing AgCl to further dissolve. Now, the solubility of AgCl will increase and will equal [Cl-]. However, CAg = [Cl-] = s CAg = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+] Therefore, [Ag+] = bo s One can then write: AgCl(s)D Ag+ + Cl- Ksp = [Ag+][Cl-] Ksp = bo s * s S = (ksp/bo)1/2

  15. Example Find the solubility of AgBr (ksp = 4x10-13) in pure water and in 0.10 M NH3 (bo in 0.10 M ammonia = 4.0x10-6). a. In pure water AgBr(s)D Ag+ + Br- Ksp = s * s S = (4.0x10-13)1/2 , S = 6.3x10-7 M

  16. b. In presence of 0.10 M NH3 Ksp = bo s * s S = (ksp/bo)1/2 S = (4.0x10-13/4.0x10-6)1/2 S = 3.2x10-4 M % increase in solubility = {(3.2x10-4 – 6.3x10-7)/6.3x10-7} x 100 = 5.1X104 % Huge increase in solubility is expected as calculation above suggests.

  17. Precipitation Titrations Titrations in which precipitates are formed are called precipitation titrations. The most frequent application of this type of titration uses silver ion to determine chloride. Therefore, these titrations are called argentometric titrations. According to the indicator used, three methods can be described. Chromate is the indicator in Mohr's method while Fajans method makes use of adsorption indicators. Both methods are direct methods. The third method is an indirect method where an excess silver is added to chloride unknown and the remaining silver is back-titrated with a standard thiocyanate solution in presence of Fe(III) as an indicator.

  18. Titration Curves A titration curve for a precipitation titration can be constructed by plotting mL Ag+ against pX (X can be Cl-, Br-, I-, or SCN-) where the sharpness of the end point and the break is directly proportional to: 1. ksp of the silver salt. 2. The concentration of reactants.

  19. Mixtures can also be titrated provided that enough difference in the solubilities of the two silver salts exists (at least 103).

  20. Example Find the pCl in a 20 mL of a 0.10 M Cl- solution after addition of 0, 10, 20, and 30 mL of 0.10 M AgNO3. Ksp = 1.0x10-10. Solution 1. After addition of 0 mL Ag+ [Cl-] = 0.10 M pCl = 1.00 2. After addition of 10 mL Ag+ Initial mmol Cl- = 0.10 x 20 = 2.0 mmol Ag+ added = 0.10 x 10 = 1.0 mmol Cl- excess = 2.0 – 1.0 = 1.0 [Cl-]excess = 1.0/30 = 0.033 M

  21. We should expect that this is the actual concentration present in solution since the solubility of AgCl is very small (as seen from the ksp) and this is especially true since we have a common ion. However, let us try to calculate the [Cl-]dissociation of AgCl. Ksp = s(1/30 + s), Assume 1/30 >> s 1.0x10-10 = 1/30 * s , s = 3x10-9 M Therefore, [Cl-] = 0.033 M, pCl = 1.48

  22. 3. After addition of 20 mL Ag+ Initial mmol Cl- = 0.10 x 20 = 2.0 mmol Ag+ added = 0.10 x 20 = 2.0 mmol Cl- excess = 2.0 – 2.0 = 0 This is the equivalence point Ksp = s * s S = (1.0x10-10)1/2 S = 1.0x10-5 M

  23. 4. After addition of 30 mL Ag+ Initial mmol Cl- = 0.10 x 20 = 2.0 mmol Ag+ added = 0.10 x 30 = 3.0 mmol Ag+ excess =3.0 – 2.0 = 1.0 [Ag+]excess = 1.0/50 = 0.02 M Therefore, once again we have a common ion situation. We should expect that this is the actual Ag+ concentration present in solution since the solubility of AgCl is very small (as seen from the ksp) and this is especially true since we have a common ion. However, we can calculate the Cl- concentration from dissociation of AgCl in presence of excess Ag+ since this is the only source of Cl-.

  24. Ksp = (0.02 + s) * s Assume that 0.02 >> s 1.0x10-10 = 0.02 S S = 5.0X10-9 M = [Cl-] pCl = 8.30

  25. Methods for Chloride Determination a. Mohr Method This method utilizes chromate as an indicator. Chromate forms a precipitate with Ag+ but this precipitate has a greater solubility than that of AgCl, for example. Therefore, AgCl is formed first and after all Cl- is consumed, the first drop of Ag+ in excess will react with the chromate indicator giving a reddish precipitate. 2Ag+ + CrO42- (Yellow)= Ag2CrO4 (Red)

  26. In this method, neutral medium should be used since, in alkaline solutions, silver will react with the hydroxide ions forming AgOH. In acidic solutions, chromate will be converted to dichromate. Therefore, the pH of solution should be kept at about 7. There is always some error in this method because a dilute chromate solution is used due to the intense color of the indicator. This will require additional amount of Ag+ for the Ag2CrO4 to form.

  27. The amount of chromate added to the solution is critical for the success of the titration. As a first estimate, consider the calculation of the initial concentration of CrO42– that must be present so that Ag2CrO4 just begins to form at the equivalence point in the titration of 0.1 M Cl– with 0.1 M Ag+. The value of the solubility product for AgCl is 1.82 × 10–10, so that at the equivalence point: [Ag+] = [Cl–] = (1.82 × 10–10)1/2 = 1.35 × 10–5 M The value of [CrO42–] at the equivalence point required for Ag2CrO4 to just begin forming is given by the following: Ksp = [Ag+]2 [CrO42–] = 1.29 × 10–12

  28. [CrO42–] = Ksp/[Ag+]2 = 1.29 × 10–12/(1.35 × 10–5)2 [CrO42–] = 7.08 × 10–3 M However, since a volume of titrant equal to the volume of analyte solution was added, the original concentration of CrO42– was halved, so that its concentration of CrO42– was: Original [CrO42–] = 2 × 7.08 × 10–3 M = 0.0142 M In practice, a concentration of the order of 7 × 10–3 M CrO42– at the equivalence point is not practical because the yellow color of the chromate ions masks the color change at the end point. A concentration of CrO42– at the end point of about 2.5 × 10–3 M is generally used for the Mohr titration.

  29. This introduces a positive titration error because a concentration of Ag+ higher than 1.35 × 10-5 M must be reached, and a finite amount of Ag2CrO4 must be precipitated before the color of the silver chromate can be observed. Compensation may be made for the Mohr chloride titration error by running an indicator blank.

  30. Example Find the concentration of chloride in a 25 mL solution to which few drops of K2CrO4 were added, if the end point required 20 mL of 0.10 M AgNO3. Solution Ag+ + Cl-D AgCl(s) The reaction between silver ions and chloride is 1:1 mmol Ag+ = mmol Cl- 0.10 x 20 = MCl- * 25 MCl- = 0.08 M

  31. b. Volhard Method This is an indirect method for chloride determination where an excess amount of standard Ag+ is added to the chloride solution containing Fe3+ as an indicator. The excess Ag+ is then titrated with standard SCN- solution untill a red color is obtained which results from the reaction: Fe3+ (Yellow) + SCN- = Fe(SCN)2+ (Red) The indicator system is very sensitive and usually good results are obtained. The medium should be acidic to avoid the formation of Fe(OH)3 .

  32. The color of the Fe3+ indicator is intense yellow, which prevents observation of the Fe(SCN)2+ red color. Therefore, a dilute Fe3+ indicator solution should be used. It is important to be able to calculate the theoretical concentration of Fe3+ indicator to appreciate the existence of an error. This can be done using the following arguments: At the equivalence point, we have AgSCN only, therefore [Ag+] = CSCN- :

  33. [Ag+] = [SCN-] + [HSCN] + [Fe(SCN)2+] [Ag+] = ksp/[SCN-] ka = [H+][SCN-]/[HSCN] Ka = 0.1, at pH = 2 ksp/[SCN-] = [SCN-] + [H+][SCN-]/ka + [Fe(SCN)2+] An average observer can distinguish the red color of Fe(SCN)2+ when the concentration of the complex is about 6.4*10-4 M. Therefore, one can find the concentration of SCN-. 1.1*10-12 /[SCN-] = [SCN-] + 10-2 [SCN-]/0.1 + 6.4*10-6 1.1[SCN-]2 +6.4*10-6[SCN-] – 1.1*10-12 = 0 [SCN-] = 1.77*10-7 M

  34. Fe3+ + SCN- = Fe(SCN)2+ Kf = [Fe(SCN)2+]/[Fe3+][SCN-] 1.4*102 = (6.4*10-6)/[Fe3+](1.77*10-7) [Fe3+] = 0.26 M However, the [Fe3+] used in Volhard method should not exceed 0.002 M, which results in an error. Such an error is usually much less than that obtained when the theoretical [Fe3+] is used.

  35. Another problem arises at the end of the titration, where we have both AgSCN(s) and AgCl(s). The solubility of AgCl (Ksp = 1*10-10) is greater than that of AgSCN (Ksp = 1*10-12) , therefore the following reaction takes place when all Cl- is over: AgCl(s) + SCN-D AgSCN + Cl- This is an important reason for error, if not taken care of.

  36. This problem had been overcome by two main procedures: The first method includes addition of some nitrobenzene, which surrounds the precipitate and shields it from the aqueous medium. The second procedure involves filtration of the precipitate directly after precipitation, which protects the precipitate from coming in contact with the added SCN- solution

  37. Example A 10 mL of a chloride sample was treated with 15 mL of 0.1182 M AgNO3. The excess silver was titrated with 0.101 M SCN- requiring 2.38 mL to reach the red Fe(SCN)2+ end point. Find the concentration of chloride (AtWt = 35.5) in g/L. Solution mmol Ag+ reacted = mmol Ag+ taken - mmol Ag+ back-titrated mmol Ag+ reacted = mmol Cl- mmolAg+ back-titrated = mmol SCN- mmol Cl- = 0.1182x15 - 0.101x2.38 = 1.53 MCl- = 1.53/10 = 0.153 M g/L Cl- = 0.153 x 35.5 = 5.44g

  38. C. Fajans Method Fluorescein and its derivatives are adsorbed to the surface of colloidal AgCl. After all chloride is used, the first drop of Ag+ will react with fluorescein (Fl-) forming a reddish color. Ag+ + Fl- (Yellowish green) = AgFl (Red)

  39. Principle of adsorption: Consider the titration of Cl- with Ag+. 1. Before the equivalence point, Cl- is in excess and the primary layer is Cl-. This repulses the indicator anions; and the more loosely held secondary (counter) layer of adsorbed ions

  40. Beyond the equivalence point (end point as well), Ag+ is in excessand the surface of the precipitate becomes positively charged, with the 1 layer being Ag+. This will now attract the indicator anion and adsorb it in the 2 (counter) layer.

  41. The color of the adsorbed indicator is different from that of the un-adsorbed indicator, signaling the completion of the titration. The degree of adsorption of the indicator can be decreased by increasing the acidity. Since fluorescein and its derivatives are weak acids, the pH of the solution should be slightly alkaline to keep the indicator in the anion form but, at the same time, is not alkaline enough to convert Ag+ into AgOH. Fluorescein derivatives that are stronger acids than fluorescien (like eosin) can be used at acidic pH without problems.

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