Lecture 35 Precipitation Titrations Solubility in Acid Solutions

Lecture 35 Precipitation Titrations Solubility in Acid Solutions 1

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+ + NH 3 D Ag(NH 3)+ + NH 3 D Ag(NH 3)2+ kf 1 = [Ag(NH 3)+]/[Ag+][NH 3] kf 2 = [Ag(NH 3)2+]/[Ag(NH 3)+][NH 3] We have Ag+, NH 3, Ag(NH 3)+, and Ag(NH 3)2+ all present in solution at equilibrium where CAg = [Ag+] + [Ag(NH 3)2+] 2

The fraction of each Ag+ species can be defined as: b 0 = [Ag+]/ CAg b 1 = [Ag(NH 3)+]/ CAg b 2 = [Ag(NH 3)2+]/ CAg As seen for fractions of a polyprotic acid dissociating species, one can look at the b values as b 0 for the fraction with zero ligand (free metal ion, Ag+), b 1 as the fraction of the species having one ligand (Ag(NH 3)+) while b 2 as the fraction containing two ligands (Ag(NH 3)2+). The sum of all fractions will necessarily add up to unity (b 0 + b 1 + b 2 = 1) 3

For the case of b 0, we make all terms as a function of Ag+ since b 0 is a function of Ag+. We use the equilibrium constants of each step: CAg = [Ag+] + [Ag(NH 3)2+] kf 1 = [Ag(NH 3)+]/[Ag+][NH 3] [Ag(NH 3)+] = kf 1 [Ag+][NH 3] Kf 1 x kf 2 = [Ag(NH 3)2+]/[Ag+][NH 3]2 [Ag(NH 3)2+] = Kf 1 x kf 2 [Ag+][NH 3]2 Substitution in the CAg relation gives: CAg = [Ag+] + kf 1 [Ag+][NH 3] + Kf 1 x kf 2 [Ag+][NH 3]2 CAg = [Ag+]( 1 + kf 1 [NH 3] + Kf 1 kf 2 [NH 3]2) [Ag+]/ CAg = 1/( 1 + kf 1 [NH 3] + Kf 1 kf 2 [NH 3]2) 4

Precipitation Reactions and Titrations 5

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: 6

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 p. H 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 p. H): 7

MA(s) D M+ + A- ksp = [M+][A-] A- + H+ D HA ka = [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-] = a 1 CT Recall EDTA equilibria in solution when its chelate dissociates in water. Therefore, for the equilibrium reaction: MA(s) D M+ + A 8

Before equil Solid 0 0 Equation MA(s) M+ A- After Equil Solid s a 1 s We can write: Ksp = s * a 1 s S = (ksp/a 1)1/2 9

Example Calculate the solubility of Ca. C 2 O 4 (ka 1 = 6. 5 x 10 -2, ka 2 = 6. 1 x 10 -5, ksp = 2. 6 x 10 -9) in a 0. 001 M HCl solution. Solution Ca. C 2 O 4 D Ca 2+ + C 2 O 42 The released oxalate will form H 2 C 2 O 4, HC 2 O 4 -, and some will remain as C 2 O 42 - where: CC 2 O 42 - = s = [Ca 2+] = [H 2 C 2 O 4] + [HC 2 O 4 -] + [C 2 O 42 -] and: [C 2 O 42 -] = a 2 s 10

Before equil Equation After Equil Solid Ca. C 2 O 4 (s) Solid Ksp = s * a 2 s S = (ksp/a 2)1/2 Therefore, we must calculate a 2 = ka 1 ka 2 / ([H+]2 + ka 1[H+] + ka 1 ka 2) 0 Ca 2+ s 0 C 2 O 42 a 2 s a 2 = (6. 5 x 10 -2 * 6. 1 x 10 -5)/( (0. 001)2 + (6. 5 x 10 -2 * 0. 001) + (6. 5 x 10 -2 * 6. 1 x 10 -5) ) a 2 = 5. 7 x 10 -2 s = (2. 6 x 10 -9/5. 7 x 10 -2)1/2 = 2. 1 x 10 -4 M Compare with solubility in absence of acidity!! 11

Example Find the solubility of Ag 3 PO 4 at p. H 3. a 3 at p. H 3=3. 3 x 10 -14, ksp= 1. 3 x 10 -20 Solution Ag 3 PO 4 D 3 Ag+ + PO 43 Solid 3 s a 3 s Ksp = (3 s)3 * a 3 s S = (ksp/27 a 3)1/4 S = (1. 3 x 10 -20/27 x 3. 3 x 10 -14)1/4 S = 1. 1 x 10 -2 M 12

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 Ag. Cl is placed in an ammonia solution, the following equilibria will take place: 13

Ag. Cl(s) D Ag+ + Cl. Ag+ + NH 3 D Ag(NH 3)+ + NH 3 D Ag(NH 3)2+ The dissociated Ag+ ions will form complexes with ammonia, forcing Ag. Cl to further dissolve. Now, the solubility of Ag. Cl will increase and will equal [Cl-]. However, CAg = [Cl-] = s CAg = [Ag+] + [Ag(NH 3)2+] Therefore, [Ag+] = bo s One can then write: Ag. Cl(s) D Ag+ + Cl. Ksp = [Ag+][Cl-] Ksp = bo s * s S = (ksp/bo)1/2 14

Example Find the solubility of Ag. Br (ksp = 4 x 10 -13) in pure water and in 0. 10 M NH 3 (bo in 0. 10 M ammonia = 4. 0 x 10 -6). a. In pure water Ag. Br(s) D Ag+ + Br- Ksp = s * s S = (4. 0 x 10 -13)1/2 , S = 6. 3 x 10 -7 M 15

b. In presence of 0. 10 M NH 3 Ksp = bo s * s S = (ksp/bo)1/2 S = (4. 0 x 10 -13/4. 0 x 10 -6)1/2 S = 3. 2 x 10 -4 M % increase in solubility = {(3. 2 x 10 -4 – 6. 3 x 10 -7)/6. 3 x 10 -7} x 100 = 5. 1 X 104 % Huge increase in solubility is expected as calculation above suggests. 16

Precipitation 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. 17

Titration Curves A titration curve for a precipitation titration can be constructed by plotting m. L Ag+ against p. X (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. 18

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Mixtures can also be titrated provided that enough difference in the solubilities of the two silver salts exists (at least 103). 21

Example Find the p. Cl in a 20 m. L of a 0. 10 M Cl- solution after addition of 0, 10, 20, and 30 m. L of 0. 10 M Ag. NO 3. Ksp = 1. 0 x 10 -10. Solution 1. After addition of 0 m. L Ag+ [Cl-] = 0. 10 M p. Cl = 1. 00 2. After addition of 10 m. L 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 22

We should expect that this is the actual concentration present in solution since the solubility of Ag. Cl 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 Ag. Cl. Ksp = s(1/30 + s), Assume 1/30 >> s 1. 0 x 10 -10 = 1/30 * s , s = 3 x 10 -9 M Therefore, [Cl-] = 0. 033 M, p. Cl = 1. 48 23

3. After addition of 20 m. L 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. 0 x 10 -10)1/2 S = 1. 0 x 10 -5 M 24

4. After addition of 30 m. L 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 Ag. Cl 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 Ag. Cl in presence of excess Ag+ since this is the only source of Cl-. 25

Ksp = (0. 02 + s) * s Assume that 0. 02 >> s 1. 0 x 10 -10 = 0. 02 S S = 5. 0 X 10 -9 M = [Cl-] p. Cl = 8. 30 26

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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 Ag. Cl, for example. Therefore, Ag. Cl 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. 2 Ag+ + Cr. O 42 - (Yellow) = Ag 2 Cr. O 4 (Red) 28

In this method, neutral medium should be used since, in alkaline solutions, silver will react with the hydroxide ions forming Ag. OH. In acidic solutions, chromate will be converted to dichromate. Therefore, the p. H 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 Ag 2 Cr. O 4 to form. 29

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 Cr. O 42– that must be present so that Ag 2 Cr. O 4 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 Ag. Cl 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 [Cr. O 42–] at the equivalence point required for Ag 2 Cr. O 4 to just begin forming is given by the following: Ksp = [Ag+]2 [Cr. O 42–] = 1. 29 × 10– 12 30

[Cr. O 42–] = Ksp/[Ag+]2 = 1. 29 × 10– 12/(1. 35 × 10– 5)2 [Cr. O 42–] = 7. 08 × 10– 3 M However, since a volume of titrant equal to the volume of analyte solution was added, the original concentration of Cr. O 42– was halved, so that its concentration of Cr. O 42– was: Original [Cr. O 42–] = 2 × 7. 08 × 10– 3 M = 0. 0142 M In practice, a concentration of the order of 7 × 10– 3 M Cr. O 42– 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 Cr. O 42– at the end point of about 2. 5 × 10– 3 M is generally used for the Mohr titration. 31

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 Ag 2 Cr. O 4 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. 32

Example Find the concentration of chloride in a 25 m. L solution to which few drops of K 2 Cr. O 4 were added, if the end point required 20 m. L of 0. 10 M Ag. NO 3. Solution Ag+ + Cl- D Ag. Cl(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 33

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 Fe 3+ as an indicator. The excess Ag+ is then titrated with standard SCN- solution untill a red color is obtained which results from the reaction: Fe 3+ (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. 34

The color of the Fe 3+ indicator is intense yellow, which prevents observation of the Fe(SCN)2+ red color. Therefore, a dilute Fe 3+ indicator solution should be used. It is important to be able to calculate theoretical concentration of Fe 3+ indicator to appreciate the existence of an error. This can be done using the following arguments: At the equivalence point, we have Ag. SCN only, therefore [Ag+] = CSCN- : 35

[Ag+] = [SCN-] + [HSCN] + [Fe(SCN)2+] [Ag+] = ksp/[SCN-] ka = [H+][SCN-]/[HSCN] Ka = 0. 1, at p. H = 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 36

Fe 3+ + SCN- = Fe(SCN)2+ Kf = [Fe(SCN)2+]/[Fe 3+][SCN-] 1. 4*102 = (6. 4*10 -6)/[Fe 3+](1. 77*10 -7) [Fe 3+] = 0. 26 M However, the [Fe 3+] 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 theoretical [Fe 3+] is used. 37

Another problem arises at the end of the titration, where we have both Ag. SCN(s) and Ag. Cl(s). The solubility of Ag. Cl (Ksp = 1*10 -10) is greater than that of Ag. SCN (Ksp = 1*10 -12) , therefore the following reaction takes place when all Cl- is over: Ag. Cl(s) + SCN- D Ag. SCN + Cl. This is an important reason for error, if not taken care of. 38

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 39

Example A 10 m. L of a chloride sample was treated with 15 m. L of 0. 1182 M Ag. NO 3. The excess silver was titrated with 0. 101 M SCN- requiring 2. 38 m. L to reach the red Fe(SCN)2+ end point. Find the concentration of chloride (At. Wt = 35. 5) in g/L. Solution mmol Ag+ reacted = mmol Ag+ taken - mmol Ag+ back-titrated mmol Ag+ reacted = mmol Clmmol. Ag+ back-titrated = mmol SCNmmol Cl- = 0. 1182 x 15 - 0. 101 x 2. 38 = 1. 53 MCl- = 1. 53/10 = 0. 153 M g/L Cl- = 0. 153 x 35. 5 = 5. 44 g 40

C. Fajans Method Fluorescein and its derivatives are adsorbed to the surface of colloidal Ag. Cl. After all chloride is used, the first drop of Ag+ will react with fluorescein (Fl-) forming a reddish color. Ag+ + Fl- (Yellowish green) = Ag. Fl (Red) 41

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 42

Beyond the equivalence point (end point as well), Ag+ is in excess and 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. 43

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 p. H 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 Ag. OH. Fluorescein derivatives that are stronger acids than fluorescien (like eosin) can be used at acidic p. H without problems. 44