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Solid phase – aqueous phase equilibrium

Solid phase – aqueous phase equilibrium When the multiplication of ion activities belonging to insoluble compounds surpasses a certain value, the compound will tend to precipitate as a solid. For example: CaCO 3(s) ⇌ Ca 2+ ( aq ) + CO 3 2- ( aq ). K sp = (Ca 2+ ) (CO 3 2- ).

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Solid phase – aqueous phase equilibrium

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  1. Solid phase – aqueous phase equilibrium When the multiplication of ion activities belonging to insoluble compounds surpasses a certain value, the compound will tend to precipitate as a solid. For example: CaCO3(s)⇌ Ca2+(aq)+CO32-(aq) Ksp = (Ca2+) (CO32-) Precipitation in clear solutions is not immediate but is comprised of two stages: Nucleation (creation of small particles) Growth of particles as a result of a kinetic preference for further precipitation on the solid that had already precipitated As stated, In order for precipitation to occur, the multiplication of concentrations of ions have to be higher than the solubility product (supersaturation). In solutions that are in low supersaturation, the rate of solid creation is slow. The precipitation rate depends on the deviation from equilibrium and on physical and chemical conditions such as stirring properties and the nucleation rate.

  2. The solubility product constant Ksp CaAb(s)⇌aC(aq) + bA(aq) In Equilibrium: Ksp = (C)a(A)b = IAP (ion activity product) K’sp = [C]a[A]b = Q The addition of CaAb(s) to saturated water will not change the concentration of dissolved ions SI - the Saturation Index = (supersaturation) Q > K’sp SI > 0 (undersaturation) Q >K’sp SI < 0

  3. Examples What is the dissolution reaction and IAP of: CaCO3? CaSO4 Ca5(PO4)3OH ? Ca3(PO4)2 Fe(OH)2 Fe(OH)3

  4. The effect of ionic strength on the solubility constant Non-ideal activity coefficients (lower than unity) increase the “solubility” of solids which dissolve to ions. CA(s) C(aq) + A(aq) What are the consequences for water with high ionic strength (such as seawater)?

  5. The effect of temperature on solubility constant CA(s) C(aq) + A(aq) If the dissolution reaction is endothermic ΔH > 0 Increasing T will increase solubility (e.g. NaCl) IfΔH < 0 solubility will decrease upon heating

  6. Example 1 Calcium sulfate was added to distilled water. The solubility constant of CaSO4(s)at 250is 1.9610-4. Calculate the equilibrium concentration of the calcium ion (in mg/L) at saturation, assuming negligible ionic strength and 250. CaSO4(s) Ca2+(aq) + SO42-(aq) Ksp = 1.96 10-4 = )Ca2+()SO42-( Solution 1.96 •10-4 = X2 X = [Ca2+] = 1.4 •10-2 M conversion to mg/L: In mg Ca2+/L = 1.4 •10-2 M (1000)(40) = 561 mg/l In mg/L as CaSO4 = 561 * 136/40 = 1907 mg/l

  7. Example 2 Calculate the solubility (in mg/L) of cadmium hydroxide in distilled water at 250 . Ksp = 5.9 • 10-15 Solution: Cd(OH)2 Cd2+(aq) + 2OH- K’sp = [Cd2+] [OH-] 2 = 5.9 • 10-15 Charge balance equation: 2[Cd2+] = 1[OH-] Therefore: [Cd2+] [OH-]2 = 5.9 • 10-15 = [Cd2+] (2[Cd2+])2 [Cd2+] = 1.14 • 10-5 M (Mw = 112.41 g/mol)

  8. Example 3 (taken from an exam) • You are required to add 24.3 mg/l of Mg to distilled water. You have two chemicals at your disposal: • Mg(OH)2(s) (as salt)andCO2(g) (in a gas cylinder) • What is the maximum Mg concentration using only Mg(OH)2(s)? And what will be the pH in this case? • Explain how CO2(g) addition can increase Mg(OH)2(s) dissolution • Calculate the limiting pH needed for sustaining 24.3 mg/l Mg2+ in solution • Calculate the amount of CO2(g) required • Mg(OH)2(s)⇌ Mg2+ + 2OH⁻ • Ksp = (Mg(OH)2(s)) = 8.9  10-12

  9. Example 3 solution • a. Mg(OH)2(s)⇌ Mg2+ + 2OH⁻ • b. CO2 dose allows lower pH since it is an acid. • c.

  10. Example 3 solution • d. What is the alkalinity? • 2mM OH⁻ were added for the 1mM of Mg2+ • CO2 addition did not change alkalinity • What is the CT? • What is the actual CO2 concentration in the water after adding the chemicals? • [H2CO3*] = M • Which one is the CO2 addition?

  11. The influence of additional reactions on the solubility of insoluble solids Any reaction that involves the products of the solid precipitation equation decreases the free product concentration and therefore increases the solubility of the solid. CaAb(s) aC(aq) + bA(aq) 1. Reactions that involve weak acid species are influenced by the pH Reactions with carbonates(CO32-) Phosphates (PO43-) and sulfides (S2-).

  12. The LSI Will a given solid dissolve or precipitate in a given solution? Firstly we’ll study the question with regard to CaCO3(s). Langelier index method (Langelier, 1936) An historical method that is still in use in order to answer whether natural water tends to precipitate or dissolve CaCO3. Langelier Saturation Index (LSI) LSI = pH – pHL If LSI < 0 water is undersaturated If LSI = 0 water is saturated If LSI > 0 water is supersaturated pHL - defined as the pH at which the water is exactly at saturation.

  13. pHL was defined by Langelier for “natural” waters, i.e. at the range 5.0 < pH < 9.5, at which it can be assumed that Alkalinity (H2CO3*) [HCO3-] Assuming that natural water is controlled by the carbonate system: At saturation: [Ca2+] [CO32-] = K’sp  

  14. A logarithm is applied at both sides: log KC2 = -pHL – pKsp – log [Alk] – log [Ca2+] pHL = pKC2 – pKsp – log [Alk] – log [Ca2+] Therefore, in this method there is a need for knowing only the alkalinity and calcium concentrations in order to know whether CaCO3 will dissolve or precipitate. • Inherent problems with the Langelier method • Gives no information on the deviation from saturation, but only on the direction (there is no quantitative information regarding the amount of material which would dissolve or precipitate) • The assumption [HCO3-] = Alkalinity is not true in every case. At relatively high pH values above 9 it is inaccurate. • Ionic strength is not taken account of in many cases, a situation which could lead to mistakes in solutions with high ionic strengths

  15. Quantitative determination of Precipitation Potential from equilibrium considerations When water is at equilibrium with CaCO3, four equilibrium equations can be defined: 1. 2. 3. 4. Four equations and six variables. Meaning, if there is equilibrium there is a need to measure 2 parameters for full water characterization (alkalinity and pH, or pH and Ca2+, or alkalinity and Ca2+)

  16. When a solution is not at equilibrium with the solid and there is interest in knowing the solution’s condition in relation to equilibrium, three parameters are to be measured. Ca2+ must be known, and two additional parameters are needed for characterizing the carbonate system. By knowing the three parameters it is possible to obtain a clear situation in relation to saturation. Meaning, it is possible to obtain information that would have been similar to what had been deduced from the Langelier index, without the simplifying assumptions. However, with the current method it is still not possible to calculate the precipitation/dissolution potential which is necessary for planning the chemical dosage for softening or stabilizing water. Why is it impossible? – because during the precipitation process the pH and alkalinity necessarily change as a result of CO32- removal from the aqueous phase.

  17. Precipitation potential determination (numeric method) • Data: [Ca2+], alkalinity, pH, EC, and temperature • Procedure • Calculate the product: [Ca2+] · [CO32-]and compare to Ksp’ • Determine saturation, oversaturation or undersaturation • In cases where there is “oversaturation” the computer “assumes” infinitesimal CaCO3 precipitation and recalculates ionic strength (activity coefficients), alkalinity, dissolved calcium andCT. • New pH and CO32- concentration are calculated from the new data, and the section (1) is repeated. • (2) – (4) are also repeated until the product of both ions equals Ksp’. The CaCO3 concentration which had precipitated till this point constitutes the precipitation potential (PP) or CCPP. This is a general method and it can be used for calculating the PP for any solid

  18. Example: The following results were obtained from surface water analysis H2CO3* alkalinity = 140 mg/L as CaCO3 Ca2+ = 125 mg/L as CaCO3 pH = 8.0 Temperature = 200C, TDS = 200 mg/L Ksp’ (CaCO3) = 10-8.05 Determine saturation state using LSI pHL= pKc2 – pKsp – log Alk – log [Ca] pHL= 10.35 – 8.05 – log 0.0028– log 0.00125 = 7.76 LSI = pH – pHL = 8.0 – 7.76 = 0.24 > 0L Supersaturation

  19. b. Perform one iteration with 1 mg/l CaCO3 to find the CCPP. Assume constant Ionic strength • 1. • 2. • 3. Now calculate new pH  new [CO32-] and go back to 1. • 4. Finish when Q=Ksp

  20. Relation between SI and CCPP CCPP is not always correlated to SI SI can be small while CCPP is large and the opposite SI is a driving force for dissolution / precipitation CCPP is the capacity of the reaction to occur Both are important for evaluating equilibrium state and predict the reaction In real life, kinetics is very important

  21. 2. Reactions that involve complexation (ion pairing) Complexes are dissolvable species comprised of a metal with a positive valence and ligands with a negative valence. Metals appear in the aqueous phase as hydrated ions (an ion complex + water molecules only) or as a complex with a different ligand. In order to predict the metal cation concentration which participates in the precipitation reaction, there is a need to conduct speciation for the metal and to determine the concentration of the free ion and its various complexes.

  22. An example of species distribution according to pH as a result of hydroxide complexation of divalent iron ions Complexation reactions are significant for understanding solids solubility, process kinetics and toxicity of metals in aqueous solutions. This material is studied partially in advanced undergraduate courses and mainly in graduate courses.

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