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Solubility Equilibria

Solubility Equilibria. Calculating Equilibrium Concentrations. CO (g) + 3H 2 (g) = CH 4 (g) + H 2 O(g) 0.850M 1.333M ?M 0.286M Keq = 3.933 Keq = [ CH 4 ] [H 2 O] [ CO ] [ H 2 ] 3 Rearrange and solve for [CH 4 ] 27.7 mol /L or 27.7 M. Solubility Rules.

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Solubility Equilibria

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  1. Solubility Equilibria

  2. Calculating Equilibrium Concentrations • CO (g) + 3H2(g) = CH4(g) + H2O(g) 0.850M 1.333M ?M 0.286M Keq = 3.933 Keq= [CH4] [H2O] [CO ] [H2 ]3 Rearrange and solve for [CH4] 27.7 mol/L or 27.7 M

  3. 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...

  4. 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.

  5. 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.

  6. Na+ and Cl -ions surrounded by water molecules NaCl Crystal Solubility Equilibrium: Dissociation = Precipitation 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. Dissolving NaCl in water

  7. Dissolving silver sulfate, Ag2SO4, in water • When silver sulfate dissolves it dissociates into ions. When the solution is saturated, the following equilibrium exists: Ag2SO4(s) 2 Ag+(aq) + SO42-(aq) • Since this is an equilibrium, we can write an equilibrium expression for the reaction: Ksp = [Ag+]2[SO42-] Notice that the Ag2SO4 is left out of the expression! Why? Since K is always calculated by just multiplying concentrations, it is called a “solubility product” constant - Ksp.

  8. Writing solubility product expressions... • For each salt below, write a balanced equation showing its dissociation in water. • Then write the Ksp expression for the salt. • BaSO4(s)  Ba 2+ (aq) + SO42- (aq) • Mg(OH)2 (s)  Mg 2+ (aq) + 2OH- (aq) Ksp = [Ba2+] [ SO4 2-] Ksp = [Mg2+] [ OH-]2

  9. Some KspValues(Table 18-3 in your text) Note: These are experimentallydetermined, and may be slightly different on a different Ksp table.

  10. Using solubility product constants to determine the solubility of a sparingly soluble compound • Use the Ksp value from Table 18-3 to calculate the solubility in mol/L of Copper(II) carbonate at 298K. • Known: Ksp CuCO3 = 2.5 x 10-10 • Unknown: solubility of CuCO3 = mol/L • 1. CuCO3 = Cu2+ (aq) + CO32- (aq) = 2.5 x 10-10 • 2. plug in “s” for single ions • 3. s x s = s2 = 2.5 x 10-10. Solve for S! • 4. S= 1.6 x 10-5 mol/L

  11. Predicting Precipitates • Solubility products are useful for finding the solubility of an ionic compound in a SATURATED SOLTUION. • You can also use Ksp to predict whether a precipitate will form or not. • Precipitate- a solid that is sometimes formed when 2 aqueous solutions are added together. • Remember, these precipitates are a result of a double replacement reaction.

  12. Mixing Solutions - Will a Precipitate Form? If 15 mL of 0.024-M lead nitrate is mixed with 30 mL of 0.030-M potassium chromate - will a precipitate form? Pb(NO3)2(aq) + K2CrO4(aq) PbCrO4(s) + 2 KNO3(aq)

  13. Rules • If Qsp < Ksp. The solution is UNSATURATED. NO precipitate. • If Qsp= Ksp. The solution is SATURATED and no precipitate is formed. • If Qsp> Ksp. A precipitate WILL form. The reaction will SHIFT left and the solid will be formed as opposed to ions.

  14. Example of Predicting Precipitates • 4FeCl3 + 3K4Fe(CN)6 12KCl + Fe4(Fe(CN)6)3 • KCl is soluble, and so you will be looking at the OTHER product. • Fe4(Fe(CN)6)3 = 4Fe3+(aq) + 3Fe(CN)64-(aq) • 1. Take the original M and reduce them by ½ as the two solutions combined have twice the volume.

  15. Fe4(Fe(CN)6)3 = 4Fe3+(aq) + 3Fe(CN)64-(aq) • Ksp = [Fe3+]4 [Fe(CN)64- ] 3 • Qsp = [Fe3+] 4 [Fe(CN)64- ] 3= (0.050) 4 (0.050)3 = 7.8 x 10-10 • Ksp for Fe4(Fe(CN) 6)3 is 3.3 x 10-41 • Remember the rules! Qsp > Ksp then a precipitate will form. The reaction will shift left and Fe4(Fe(CN) 6)3 will form a precipitate.

  16. 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?

  17. 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 10-8mol/L vs 2.6 x 10 -4

  18. Explaining the 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!

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