15.6 Solubility Equilibria Solubility of various compounds plays important roles. –Sugar and table salts ability to dissolve in water help us to flavor food. –Calcium sulfate is less soluble in hot water than cold water, which allows us to coat tubes in boilers. –The low solubility of barium sulfate allows a safe way to X-ray the gastrointestinal tract.
Aqueous Equilibria Assume that when an ionic solid dissolves, that it dissociates completely. CaF 2 (s) Ca 2+ (aq) + 2F - (aq) There is a possibility for the ions to collide and re-form the solid phase. Ca 2+ (aq) + 2F - (aq) CaF 2 (s) Finally, equilibrium is reached, and no more solid will dissolve (saturated). CaF 2 (s) Ca 2+ (aq) + 2F - (aq)
Solubility Product Constant Remember that solids are left out of equilibrium expressions K sp =[Ca 2+ ][F - ] 2 K sp is the solubility product constant (sometimes referred to as just the solubility product). *Notice that the amount of solid present nor the size of particles effect the equilibrium.
Solubility vs. Solubility Product Solubility is an equilibrium position. Solubility product is an equilibrium constant and has ONE value for a particular solid at a particular temperature.
Calculating K sp from Solubility Calculate K sp for bismuth sulfide (Bi 2 S 3 ), which has a solubility of 1.0 x mol/L at 25°C. K sp = [Bi 3+ ] 2 [S 2- ] 3 =(2.0 x ) 2 (3.0 x ) 3 = 1.1 x HW odd on page 743 Bi 2 S 3 (s) 2Bi 3+ (aq) +3S 2- (aq) INot needed00 C-1.0 x = -x+2x+3x ENot needed2.0 x x
Calculating Solubility from K sp The K sp value from Cu(IO 3 ) 2 is 1.4 x at 25°C. Calculate its solubility. K sp =[Cu 2+ ][IO 3 - ] 2 =[x][2x] 2 =(x)(4x 2 )=4x x =4x 3 X=3.3 x mol/L= Cu(IO 3 ) 2 solubility HW on page 743 Cu(IO 3 ) 2 (s) Cu 2+ (aq) +2 IO 3 - (aq) INot needed00 C-x+x+2x ENot needed+x+2x
Relative Solubilities Use K sp values to predict relative solubilities of a group of salts. –Salts producing the same number of ions Ex. AgI (s), CuI (s), CaSO 4 (s) Solubility is directly proportional to K sp because each compound produces 2 ions. »Largest K sp = most soluble »Smallest K sp = least soluble –Salts producing differing numbers of ions. Ex. CuS (s), Ag 2 S (s), Bi 2 S 3 (s) Solubility NOT directly proportional to K sp and you have to calculate the solubilities.
Practice Problems HW 15.83,15.87 on page 743
Common Ion Effect Consider, the solubility of solid Ag 2 CrO 4 (K sp =9.0x ) in a M aqueous solution of AgNO 3. –What is the common ion? –What is the [Ag + ] 0 ? –What is the [CrO 4 -2 ] 0 ? –What is the K sp expression?
K sp = [Ag + ] 2 [CrO 4 ] x = ( x) 2 (x) * Since K sp is small, x is considered negligible compared to M; therefore x x = (0.100) 2 (x) X=9.0 x M= solubility of Ag 2 CrO 4 (s) [Ag + ]=0.100 M and [CrO 4 - ]=9.0 x M Ag 2 CrO 4 (s) 2 Ag + (aq) +CrO 4 2- (aq) INot needed0.100 M0 C-x+2x+x ENot needed x+x
Solubility of Ag 2 CrO 4 in pure water = 1.3 x M Solubility of Ag 2 CrO 4 in M AgNO 3 =9.0x M * Note that solubility of a solid is lowered due to the common ion effect. Example on page 743 HW and on page 743.
pH and Solubility If the anion (X - ) is an effective base (or HX is a weak acid), the salt MX will increase solubility in an acidic solution. –Effective bases include OH -, S 2-, CO 3-, C 2 O 4 2-, and CrO –Salts with these anions are more soluble in acidic solutions that pure water. HW (in reference to 15.81) on page 743
Precipitation Precipitation: the formation of a solid from solution Ion Product (Q)=defined the same as K sp except initial concentrations are used Ex. Ca(NO 3 ) 2 (aq) is mixed with NaF (aq). What is the ion product for CaF 2 ? Q=[Ca 2+ ] 0 [F - ] 0 2
To predict if precipitation will occur: If Q>K sp, precipitation occurs until concentrations are reduced to satisfy K sp. If Q
Determining Precipitation Conditions A solution is prepared by adding mL of 4.00 x M Ce(NO 3 ) 3 to mL of 2.00 x M KIO 3. Will Ce(IO 3 ) 3 (K sp =1.9 x ) precipitate from this solution? 1.Calculate initial conditions. 2.Solve for Q. Q=[Ce 3+ ] 0 [IO 3 - ] 0 3 =(2.86x10 -3 )(5.71x10 -3 ) 3 = 5.32 x Compare Q to K sp. Q>K sp, so Ce(IO 3 ) 3 will precipitate. Example on page 743 (Ksp values on page 718)
Precipitation and Equilibrium Concentrations A solution is prepared by mixing mL of 1.00x10 -2 M Mg(NO 3 ) 2 and mL of 1.00x10 -1 M NaF. Calculate the concentrations of Mg 2+ and F - at equilibrium with solid MgF 2 (K sp =6.4x10 -9 ).
1.Find Initial Concentrations 2. Find Q 3. Compare Q vs. K sp Q>K sp, so solid MgF 2 will form. * The next steps are used to determine equilibrium concentrations.
4.Run the reaction to completion. (BRA) 5.Calculate the concentration of excess reactant. [F - ] excess =22.0 mmol / mL = 5.50 x M Mg F - MgF 2 (s) B efore rxn (150.0 mL) (1.00x10 -2 M) =1.50 mmol (250.0 mL) (1.00x10 -1 M) =25.0 mmol 0 R eaction -x-2 x+x A fter rxn =0 *limiting reactant (1.50)=22.0 mmol *excess 1.50 mmol
6.Determine concentrations at equilibrium (ICE). K sp = 6.4x10 -9 =[Mg 2+ ][F - ] 2 =(x)(5.50x x) 2 (x)(5.50x10 -2 ) 2 X=2.1 x M = [Mg 2+ ] [F - ]=5.50x10 -2 M HW15.99 on page 743 MgF 2 (s) Mg 2+ (aq) +2F - (aq) INot needed0 M5.50 x M C-x+x+2x ENot needed+ x5.50 x x
Selective Precipitation Using anions that form precipitates with only one or a few metal ions in a mixture in order to separate the metal ions. Most insoluble sulfide salts can be precipitated in an acidic solutions. Soluble sulfide salts can be precipitated by making the solution slightly basic.
A solution contains 1.0x10 -4 M Cu + and 2.0x10 -3 M Pb 2+. If a source of I - is added gradually to this solution, will PbI 2 (K sp =1.4x10 -8 ) or CuI (K sp =5.3x ) precipitate first? Specify the [I - ] necessary to begin precipitation of each salt. -For PbI 2 : K sp =1.4x10 -8 =[Pb 2+ ][I - ] 2 1.4x10 -8 =(2.0x10 -3 ) [I - ] 2 [I - ]=2.6 x M is necessary to begin precipitation -For CuI: K sp =5.3x =[Cu + ][I - ] 5.3x =(1.0x10 -4 )[I - ] [I - ]=5.3x10 -8 M is necessary to begin precipitation *CuI will precipitate first since [I - ] required is less. HW on page 744
Groups: 1. Insoluble Chlorides (Ag +, Pb 2+, Hg 2 2+ ) 2. Sulfides insoluble in acid solution (Hg 2+, Cd 2+, Bi 3+, Cu 2+, Sn 4+ ) 3. Sulfides insoluble in basic solution (Co 2+, Zn 2+, Mn 2+, Ni 2+, Fe 2+, Cr 3+, Al 3+ ) 4. Insoluble Carbonates (Group 2A) 5. Alkali Metal and ammonium ions (flame test)
Complex Ion Equilibria Complex Ion: a charged species with a metal ion surrounded by ligands (Ag(NH 3 ) + ) –The ligand donates its lone electron pair to an empty orbital of the metal ion to form a covalent bond Ligand: a Lewis base (H 2 O, NH 3, Cl -, CN - ) Coordination number: the number of ligands attached to the metal ion
Formation constants or stability constants: –Metal ions add ligands one at a time in steps characterized by their own equilibrium constants Ag + + NH 3 Ag(NH 3 ) + K 1 =2.1x10 3 Ag(NH 3 ) + + NH 3 Ag(NH 3 ) 2 + K 2 =8.2x10 3 –All species (Ag +, NH 3, Ag(NH 3 ) +, Ag(NH 3 ) 2 + ) exist at equilibrium. HW on page 744
Usually, [ligand] is much larger than [metal ion] and approximations are used to simplify problems. Assume both reactions go to completion Ag + + NH 3 Ag(NH 3 ) + K 1 =2.1x10 3 Ag(NH 3 ) + + NH 3 Ag(NH 3 ) 2 + K 2 =8.2x10 3 Ag NH 3 Ag(NH 3 ) 2 + K=K 1 x K 2 From here, use BRA to find concentrations and the equilibrium constant. –Assume the ligand amount consumed is neglible Examples a, 105, 109 on page 744 HW b, 107on page 744
Strategies for Dissolving Water- Insoluble Ionic Solids 1.If the anion of the solid is a good base, solubility is increased with the addition of an acid 2.If the anion is NOT a good base, solids can be dissolved in a solution containing a ligand to form stable complex ions with its cation.
Aqueous Ammonia is Added to Silver Chloride (white). Silver Chloride, Insoluble in Water, Dissolves to Form Ag(NH 3 ) 2 + (aq) and Cl - (aq)