Solubility Equilibria Will it all dissolve, and if not, how much will?

SOLUBILITY EQUILIBRIA Solubility: Relative term used to describe how much of a particular substance dissolves in a certain amount of solvent. Substances that dissolve very well are said to be soluble Insoluble species don’t dissolve well. All substances are “soluble” to some extent We will look at slightly soluble substances

All dissolving is an equilibrium. If there is not much solid it will all dissolve. As more solid is added the solution will become saturated. Solid ↔ dissolved The solid will precipitate as fast as it dissolves, forming an equilibrium. SOLUBILITY EQUILIBRIA

Watch out Solubility is not the same as solubility product. Solubility product is an equilibrium constant. It doesn’t change except with temperature. Solubility is an equilibrium position for how much can dissolve. A common ion can change this.

K sp Values for Some Salts at 25  C NameFormulaK sp Barium carbonate BaCO x Barium chromate BaCrO x Barium sulfate BaSO x Calcium carbonate CaCO x Calcium oxalate CaC 2 O x Calcium sulfate CaSO x Copper(I) iodide Cu I 1.3 x Copper(II) iodate Cu( I O 3 ) x Copper(II) sulfide CuS 6.0 x Iron(II) hydroxide Fe(OH) x Iron(II) sulfide FeS 6.0 x Iron(III) hydroxide Fe(OH) x Lead(II) bromide PbBr x Lead(II) chloride PbCl x Lead(II) iodate Pb( I O 3 ) x NameFormulaK sp Lead(II) iodide Pb I x Lead(II) sulfate PbSO x Magnesium hydroxide Mg(OH) x Silver bromate AgBrO x Silver bromide AgBr 5.4 x Silver carbonate Ag 2 CO x Silver chloride AgCl 1.8 x Silver chromate Ag 2 CrO x Silver iodate Ag I O x Silver iodide Ag I 8.5 x Strontium carbonate SrCO x Strontium fluoride SrF x Strontium sulfate SrSO x Zinc sulfide ZnS 2.0 x

SOLUBILITY PRODUCT CONSTANTS  Consider the following reaction  The equilibrium constant expression is K sp = [Pb 2+ ][Cl - ] 2  K sp is called the solubility product constant or simply solubility product  For a compound of general formula, M y X z (next page)

K sp = [M z+ ] y [X y- ] z K sp = [Mg 2+ ][NH 4 + ][PO 4 3- ] K sp = [Zn 2+ ][OH - ] 2 K sp = [Ca 2+ ] 3 [PO 4 3- ] 2

 Molar solubility: the number of moles that dissolve to give 1 liter of saturated solution  As with any equilibrium constant the numerical value must be determined from experiment  The K sp expression is useful because it applies to all saturated solutions - the origins of the ions are not relevant  Consider 25  C K sp AgI = 1.5 x

Solving Solubility Problems For the salt AgI at 25  C, K sp = 1.5 x AgI(s)  Ag + (aq) + I - (aq) I C E O O +x x x 1.5 x = x 2 x = solubility of AgI in mol/L = 1.2 x M

Solving Solubility Problems For the salt PbCl 2 at 25  C, K sp = 1.6 x PbCl 2 (s)  Pb 2+ (aq) + 2Cl - (aq) I C E O O +x +2x x 2x 1.6 x = (x)(2x) 2 = 4x 3 x = solubility of PbCl 2 in mol/L = 1.6 x M

Try Problem 81

Relative Solubilities Ksp will only allow us to compare the solubility of solids the that fall apart into the same number of ions. The bigger the Ksp of those the more soluble. If they fall apart into different number of pieces you have to do the math. Which one is most soluble? NameFormulaK sp Iron(II) hydroxide Fe(OH) x Iron(II) sulfide FeS 6.0 x Iron(III) hydroxide Fe(OH) x

Try Problem 86

The Common Ion Effect When the salt with the anion of a weak acid is added to that acid: –it reverses the dissociation of the acid. –lowers the percent dissociation of the acid. The same principle applies to salts with the cation of a weak base.. The calculations are the same as with acid base equilibrium.

Solving Solubility with a Common Ion For the salt AgI at 25  C, K sp = 1.5 x What is its solubility in 0.05 M NaI? AgI(s)  Ag + (aq) + I - (aq) I C E 0.05 O +x x 0.05+x 1.5 x = (x)(0.05+x)  (x)(0.05) x = solubility of AgI in mol/L = 3.0 x M

Try Problem 93

Precipitation The reaction quotient (called ion product) may be applied to solubility equilibria - determines if a substance will precipitate from solution Ion Product, Q =[M + ] a [Nm - ] b If Ksp<Q a precipitate forms, reverse process occurs If Ksp=Q equilibrium solution is just saturated If Ksp>Q No precipitate, forward process occurs

Precipitation Example A solution of 75.0 mL of M BaCl 2 is added to mL of M Na 2 SO 4. Will a precipitate form? (Ksp= 1.5 x M BaSO 4 ) BaSO 4 could form if K sp <Q. For Q you need initial concentrations: [Ba 2+ ] = mmol Ba 2+ / total mL = (0.0750L)(0.020 M)/(0.0750L L) = M [SO 4 2- ] = mmol SO 4 2- / total mL = (0.1250L)(0.040 M)/(0.0750L L) = M Q = [Ba 2+ ] [SO 4 2- ] = ( M)(0.025 M) = 1.9 x K sp <Q so BaSO 4 will form. To figure out concentrations set up an ice table.

Try Problem 98