Skip to main content
UploadLanguage (EN)Support
BusinessMobileSocial MediaMarketingTechnologyArt & PhotosCareerDesignEducationPresentations & Public SpeakingGovernment & NonprofitHealthcareInternetLawLeadership & ManagementAutomotiveEngineeringSoftwareRecruiting & HRRetailSalesServicesScienceSmall Business & EntrepreneurshipFoodEnvironmentEconomy & FinanceData & AnalyticsInvestor RelationsSportsSpiritualNews & PoliticsTravelSelf ImprovementReal EstateEntertainment & HumorHealth & MedicineDevices & HardwareLifestyle
Uploaded byShaheenPraveen1
PPT, PDF4,491 views

Solubility equilibria

The document discusses precipitation and dissolution processes in chemistry, highlighting their importance in industrial purification and mineral formation. It explains the concepts of saturated and unsaturated solutions, solubility definitions, and solubility rules for ionic materials. Additionally, the document covers solubility product constant (ksp), its calculations, and how factors like common ions and pH affect solubility and precipitation in solutions.

Education◦
Downloaded 49 times
Speaker Shaheen Praveen Department of Soil Science & Agricultural Chemistry COA, IGKV Raipur, C.G.
Precipitation Precipitation - The formation of a solid from solution. The reverse of dissolution. Dissolution - The process by which a substance dissolves. The reverse of precipitation. Importance - (a) Selective precipitation is an important industrial purification process, especially when crystals are formed. (b) Scales that form on boilers and teeth are to be prevented, as are kidney stones. (c) Precipitation forms minerals - dissolution removes them.
Saturation Saturated Solution - One in which a dissolution - precipitation equilibrium exist between a solid and its dissolved form. Here the equilibrium is dynamic and the rate of dissolution is equal to the rate of precipitation. Unsaturated Solution - One in which the concentration of dissolved solid is not sufficient to cause precipitation. Obviously a quantitative description of this type of heterogeneous equilibrium is subject to the law of mass action, and equilibrium expressions can be written and deductions made concerning the concentration of various species at equilibrium.
Solubility Solubility - The greatest amount of a substance that will dissolve in equilibrium in a specified volume of solvent at a particular temperature.  Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution.  Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution. Example - The solubility of silver chloride in water at 25 o C is .0018 g/L or 1.3 x 10-5 M. Most solubility's increase with temperature.
Classification of Ionic Materials by Solubility Soluble Ionic Materials - have solubilities in excess of 10 gL-1 . Insoluble Ionic Materials - have solubilities less than 0.1 gL-1 . Slightly Soluble Materials - have solubilities between 0.1 and 10 gL-1 .
Solubility Rules Salts are generally more soluble in HOT water (Gases are more soluble in COLD water). Alkali Metal salts are very soluble in water. NaCl, KOH, Li3PO4, Na2SO4 etc... Ammonium salts are very soluble in water. NH4Br, (NH4)2CO3 etc… Salts containing the nitrate ion, NO3 - , are very soluble in water. Most salts of Cl- , Br- and I- are very soluble in water - exceptions are salts containing Ag+ and Pb2+ . soluble salts: FeCl2, AlBr3, MgI2 etc... “insoluble” salts: AgCl, PbBr2 etc...
Dissolving a Salt… A salt is an ionic compound - usually a metal cation bonded to a non-metal anion. The dissolving of a salt is an example of equilibrium. The cations and anions are attracted to each other in the salt. They are also attracted to the water molecules. The water molecules will start to pull out some of the ions from the salt crystal.
At first, the only process occurring is the dissolving of the salt - the dissociation of the salt into its ions. However, soon the ions floating in the water begin to collide with the salt crystal and are “pulled back in” to the salt. (precipitation) Eventually the rate of dissociation is equal to the rate of precipitation. The solution is now “saturated”. It has reached equilibrium
Solubility Equilibrium: Dissociation = Precipitation Na+ and Cl - ions surrounded by water molecules NaCl Crystal In a saturated solution, there is no change in amount of solid precipitate at the bottom of the beaker. Concentration of the solution is constant. The rate at which the salt is dissolving into solution equals the rate of precipitation.
Nature of Ionic Equilibria Most salts dissociate into ions when they dissolve. Equilibrium then exists between the solid salt and its aquated ions, and not between the solid salt and dissolved molecules of the salt. For example:  PbSO4(s) = Pb2+ (aq) + SO4 2- (aq) This equilibrium system may be described by the mass-action expression: Ksp = [Pb2+ ][SO4 2- ] Note that the pure solid does not enter into the equilibrium.
Solubility and the Ksp • One may provide solubility information as the solubility, S or as the solubility product, Ksp. • These two quantities are obviously related to each other. • Saturated solutions of salts are another type of chemical equilibrium. • Slightly soluble salts establish a dynamic equilibrium with the hydrated cations and anions in solution.
When the solid is first added to water, no ions are initially present. As dissociation continues, the concentration of aqueous ions increases until equilibrium is reached. This process can be represented by the solubility product constant or Ksp expression. Ksp = [M² ][X⁺ ˉ]² (aq)2X(aq)2M(s) 2 MX −++↔
Even “insoluble” salts dissociate a little – their Ksp values range from 10-10 to 10-50 . A Ksp value is unique to a given salt at a given temperature. Why would a change in temperature alter the value of Ksp? Solubility unit- mol/L or, g/L or, mg/L
Solubility indicates the amount of salt that dissociates to form a saturated solution – think solubility curve. In essence, it indicates the equilibrium position for a given set of conditions. We can have different solubilities with the same Ksp.
Ksp Utilization Calculate the effect of a common ion or pH on solubility Determine if a precipitate will form given concentrations of ions and Ksp Determine the order of precipitation in a mixture of ions
How can We Calculate Ksp Write the dissociation equation first. Write the Ksp expression – leaving out the solid (Note: some of these have quite large exponents). For a saturated solution of AgCl, the equation would be: AgCl (s) ↔ Ag+ (aq) + Cl- (aq) The solubility product expression would be: Ksp = [Ag+ ] [Cl- ]
For example (solubility given)- Lead (II) chloride dissolves to a slight extent in water according to the equation: PbCl2 ↔ Pb+2 + 2Cl- Calculate the Ksp if the lead ion concentration has been found to be 1.62 x 10-2 M Solution:- Considering the equation, if lead’s concentration is “x” , then chloride’s concentration is “2x”. So. . . . Ksp = [Pb+2 ] [Cl- ]2 Ksp = (1.62 x 10-2 )(3.24 x 10-2 )2 = 1.70 x 10-5
Calculation of Solubility Write the dissociation equation. Use the equation to consider the amount of ions given that ‘s’ of the solid dissociates. Write the Ksp expression and substitute ‘s’ values and solve (Pay attention to powers and roots). For example:- In a saturated solution of silver carbonate, what is the molar solubility of the salt? Ksp = 8.1 X 10-12
Continued… Ag2CO3(s) ↔ 2 Ag+ + CO3 -2 -s +2s +s Ksp = [Ag+ ]2 [CO3 -2 ] = (2s)2 (s) = 4s3 Ksp = 8.1 X 10-12 = 4s3 s = 1.3 X 10-4 M
Comparing Solubilities The relative solubilities can be deduced by comparing values of Ksp. These comparisons can only be made for salts having the same ION:ION ratio.
Which Salt is More Soluble? Ag2S Ksp = 1.0 X 10-49 Ni(CN)2 Ksp = 3.0 X 10-23 Ag2S ↔ 2 Ag+ + S-2 Ni(CN)2 ↔ Ni+2 + 2 CN- * Since both make 3 ions (4s3 ) – the larger Ksp is the more soluble salt - Ni(CN)2
Common Ion Effect The presence of a common ion in a solution will lower the solubility of a salt. LeChatelier’s Principle: The addition of the common ion will shift the solubility equilibrium backwards. This means that there is more solid salt in the solution and therefore the solubility is lower.
Effect of a Common Ion on pH or Solubility The pH of a solution can also affect solubility if H+ or OH- can interact with the salt’s ions. Q.) How will the solubility of calcium carbonate be affected if it is dissolved in a solution of calcium chloride? presence of Ca2+ ions will shift equilibrium position to left – CaCO3 will be less soluble Q.) How will the solubility of silver phosphate be affected by an decrease in pH? H+ react with PO4 3- thus causing a shift right and increasing solubility
Q.) Would magnesium hydroxide (milk of magnesia) be more soluble in an acid or a base? Why? Mg(OH)2(s) ↔ Mg2+ (aq) + 2 OH- (aq) How will the solubility of magnesium hydroxide be affected by an increase in pH? more OH- will cause shift left – decrease sol. How will the solubility of magnesium hydroxide be affected by an decrease in pH? H+ will react with OH- cause shift right – increase sol.
The Common Ion Effect on Solubility The solubility of MgF2 in pure water is 2.6 x 10-4 mol/L. What happens to the solubility if we dissolve the MgF2 in a solution of NaF, instead of pure water?
Q.) Calculate the solubility of MgF2 in a solution of 0.080 M NaF. MgF2 (s)  Mg2+ (aq) + 2 F- (aq) ---- 0.080 M + s + 2s s 0.080 + 2s Ksp = 7.4 x 10-11 = [Mg2+ ][F- ]2 = (s)(0.080 + 2s)2 Since Ksp is so small…assume that 2s << 0.080 7.4 x 10-11 = (s)(0.080)2 s = 1.2 x 10-8 mol/L
Solubility and pH Q.) Calculate the pH of a saturated solution of silver hydroxide, AgOH. Refer to the table in your booklet for the Ksp of AgOH. AgOH (s)  Ag+ (aq) + OH- (aq) ---- ---- + s + s s s Ksp = 2.0 x 10-8 = [Ag+ ][OH- ] = s2 s = 1.4 x 10-4 M = [OH- ] pOH = - log (1.4 x 10-4 ) = 3.85 pH = 14.00 - pOH = 10.15
Determine if a Precipitate will Form With some knowledge of the reaction quotient (Q), we can decide 1) whether a ppt will form, and 2) what concentrations of ions are required to begin the ppt. of an insoluble salt.
1. Q = Ksp, the system is at equil. (saturated) 2. Q < Ksp, the system is not at equil. (unsaturated – shift right) 3. Q > Ksp, the system is not at equil. (supersaturated – shift left) *Precipitates form when the solution is supersaturated.
Quantitative Analysis of Cations
Solubility equilibria