Ch 17 Solubility Equilibria Brown & LeMay
When a typical ionic solid is dissolved in water that dissolved material can be assumed to be present as separate hydrated anions & cations. For example: CaF 2 (s) dissolving in water: CaF 2 (s) Ca 2+ (aq) + 2F - (aq) The solubility of a substance is the quantity that dissolves to form a saturated solution at a specified temperature. Solubility is often expressed as: grams of solute per liter of solution (g/l). Molar solubility, the number of moles of solute that dissolve in1 L of saturated solution (mol/l).
Solubility is an equilibrium position. As a solid dissolves the concentrations of Ca 2+ and F- increases. As the ions build up in solution it is more likely that these ions will collide and reform the solid phase. These competing processes are occurring: CaF 2 (s) Ca 2+ (aq) + 2F - (aq) Ca 2+ (aq) + 2F - (aq) CaF 2 (s) Eventually dynamic equilibrium is reached, the combined process can be represented as: CaF 2 (s) Ca 2+ (aq) + 2F - (aq) The solubility of a substance and the solubility constant for a substance are not the same thing. Thus, the magnitude of Ksp is a measure of how much of the solid dissolves to form a saturated solution.
* An equilibrium expression can be constructed according to the Law of Mass Action: CaF 2 (s) Ca 2+ (aq) + 2F - (aq) K sp = [Ca 2+ ][F-] 2 CaF 2 (s) is not included, since solids and liquids are never included in the equilibrium expression. This means that the amount of excess solid does not affect the position of the solubility equilibrium nor the solubility of that solid.
Writing Solubility-Product (Ksp) Expressions
Silver Chromate Ag 2 CrO 4 Ag 2 CrO 4 2Ag 1+ + 1CrO 2-4 K sp = [Ag 1+ ] 2 [CrO 2-4 ] 1 Calcium Phosphate Ca 3 (PO 4 ) 2 Ca 3 (PO 4 ) 2 3Ca 2+ + 2PO 3-4 K sp = [Ca 2+ ] 3 [PO 3-4 ] 2
K sp and Precipitate Formation
Ksp values can be used to make predictions as to whether or not a precipitate will form when two solutions are mixed. A quantity, Q, are concentrations at a particular moment. Q has the same mathematical format as Ksp. When comparing the two it will tell us whether a precipitate will form or not.
Equilibrium Curve For Silver Chloride AgCl(s) Ag + +Cl - (aq) When Q=K equilibrium exists (saturated solution) Q>Ksp precipitation exists (shifts to left) Ksp>Q no precipitation exists, solid dissolves (shifts to right) (additional solid dissolves) Ksp = [Ag + ] [Cl - ]
Calculating K sp values from known solubility (1:1 mole ratio)
The solubility of CuBr(s) is 2.0 x 10-4 mol/l (Molarity & not Ksp) at 25ºC. Calculate the value of the K sp for CuBr(s) at 25ºC. R: CuBr(s) Cu 1+ (aq) + Br- (aq) K sp = [Cu 1+ ][Br-] I: 0 0 (concentration before equilibrium) E: + x + x (CuBr dissolves & is at equilibrium) The solubility of CuBr(s) = 2.0 x 10-4 mol/l means that CuBr(s) dissolves to produce 2.0 x 10-4 mol of Cu + and 2.0 x 10-4 mol of Br- per liter of solution: K sp = [Cu + ][Br-] = [2.0 x 10-4 ][2.0 x 10-4 ] Ksp = 4.0 x 10-8 (mol/l) 2
Note the following points: 1. When a K sp value is given, the units are usually deleted. 2. In calculating the K sp value of CuBr(s) it has been assumed that all of the CuBr(s) that dissolves does so to form separate Cu + and Br- ions. 3. If solubility is given in g/100 ml, change it to mol/l before doing calculations.
Calculating K sp values from known solubility (1:2 mole ratio)
The solubility of CaF 2 (s) is 2.15 x 10-4 mol/l at 25ºC. Calculate the K sp for CaF 2 (s) at 25ºC. R: CaF 2 (s) Ca 2+ (aq) + 2 F- (aq) K sp = [Ca 2+ ][F-] 2 I: 0 0 E: + x + 2x NOTE: xcaf 2 xca 2+ + 2xF- [Ca 2+ ] = x = 2.15 x 10-4 mol/l [F-] 2 = (2x) 2 = 2(2.15 x 10-4 mol/l) = (4.30 x 10-4 mol/l) 2 = 1.85 x 10-7 K sp = [Ca 2+ ][F-] 2 Place calculations into the expression = (2.15 x 10-4 )(1.85 x 10-7 ) K sp = 3.98 x 10-11
Calculating Solubility from the K sp
The K sp for Cu(IO 3 ) 2 (s) is 1.4 x 10 7 at 25ºC. Calculate the solubility of Cu(IO 3 ) 2 (s) at 25ºC. R: Cu(IO 3 ) 2 (s) Cu 2+ (aq) + 2 IO 3 -(aq) K sp = [Cu 2+ ][IO 3 -] 2 I: 0 0 E: x 2x K sp = [Cu 2+ ][IO 3 -] 2 = (x)(2x) 2 = 4x 3 1.4 x 10-7 = 4x 3 1.4 x 10-7 = 3.5 x 10-8 = x 3 4 3.5 x 10-8 = x 3 x = 3 3.5 x 10-8 = 3.3 x 10 3 mol/l x = mol/l of Cu(IO 3 ) 2 that dissolves = solubility
Calculating K sp values from known solubility of one ion
Solid silver chromate (Ag 2 CrO 4 ) is added to pure water and analysis of the equilibrated solution shows that its silver ion (Ag+) concentration is 1.3x10-4 M. Calculate Ksp for this compound. R: Ag 2 CrO 4 (s) 2Ag + (aq)+cro 4 2- (aq) K sp = [Ag + ] 2 [Br-] 1 Noting the equation there are twice as many Ag+ ions or half the CrO 4 2- ions. So this becomes a stoichiometry problem: 1.3x10-4 mol Ag + x 1 mol CrO 4 2- = 6.5x10-5 mol/l CrO 4 2-1 L 2 mol Ag + K sp = [Ag + ] 2 [Br-] = [1.3x10-4 ] 2 [6.5x10-5 ] 1 = 1.1x10-12
Common Ion Effect Now we will consider solutions which contain an ion in common with one in the salt.
Calculate the solubility Ag 2 CrO 4 (s) in 0.100 M solution of AgNO 3. K sp for Ag 2 CrO 4 = 9.0 x 10 12. Before any Ag 2 CrO 4 dissolves, the solution contains Ag+, NO 3 -, and H 2 O. The Ag 2 CrO 4 (s) dissolves as: K sp = [Ag+] 2 [CrO 4 2- ] = 9.0 x 10 12 R: Ag 2 CrO 4 (s) 2Ag + (aq) + CrO 4 2- (aq) I: 0.100 0 (AgNO 3 solution) C: 2x x (Ag 2 CrO 4 solution) E: 0.100+2x x (at equilibrium) [Ag+] 2 = 0.100 + 2x K sp = [Ag+] 2 [CrO 4 2- ] [CrO 4 2- ] = x 9.0 x 10 12 = (0.100 + 2x) 2 (x) (+2x is negligible since K sp = 9.0 x 10 12 is much smaller than 0.100 M)
9.0 x 10 12 = (0.100) 2 (x) x 9.0 x 10-12 = 9.0 x 10-10 mol/l = solubility (0.100) 2 Solubility of Ag 2 CrO 4 (s) In Pure Water: 1.3 x 10-4 M Solubility of Ag 2 CrO 4 (s) In 0.100 M AgNO 3 : 9.0 x 10-10 M The common ion effect lowers the solubility of the Solid & the equilibrium shifts to left: Ag 2 CrO 4 (s) 2Ag+ (aq) + CrO 4 2- (aq)
Precipitation Conditions (When Solutions Are Mixed)
When solutions are mixed, reactions occur. In this case a precipitation occurs. Find TWO common ions. The ion product, another name for a Reaction Quotient (Q), will be used to solve these problems. They are identical expressions except the initial concentrations are used. This differs from the K sp expression which involves equilibrium concentrations only. For example, for the compound CaF 2 (s), the ion product expression is Ion Product = Q = [Ca 2+ ] 0 [F-] 0 2
When a solution of Ca 2+ is added to a solution of F-, precipitation may or may not occur. Precipitation will occur if Q > K sp (Shifts to left) To predict whether precipitation will occur calculate the ion product for the solid using the ions in solution. Then apply the following rules: If Q > K sp ppt occurs (remember shifts left) If K sp Q no ppt occurs (shifts right)
Example Problem: Precipitation Conditions
A solution is prepared by mixing 100.0 ml of 1.0 x 10-3 M Ca(NO 3 ) 2 & 100.0 ml of 1.0 x 10-3 M NaF. Does CaF 2 (s) precipitate from this solution? K sp of CaF 2 (s) is 4.0 x 10-11 To see if CaF 2 forms compute the concentrations of Ca 2+ and F- (Which has a volume of 200.0 ml, Additive Volumes) (Find the TWO common ions!) [Ca 2+ ] = mmol Ca 2+ = (100.0mL)(1.0 x 10-3 M) = ml of soln [Ca 2+ ] = 0.1 mmol 200.0 ml 5.0 x 10-4 M [F - ] = mmol F - = (100.0mL)(1.0 x 10-3 M) = 0.1 mmol ml of soln 200.0 ml [F - ] = 5.0 x 10-4 M
CaF 2 (s) Ca 2+ (aq) + 2 F-(aq) Ion product is: Q = [Ca 2+ ] [F - ] 2 = (5.0 x 10-4 M)( 5.0 x 10-4 M) 2 = 1.25 x 10-10 Given K sp = 4.0 x 10-11 (Compare Q & Ksp) Q>K sp and CaF 2 (s) will form (Shifts to the left)
Stop Here!!
17.5 Factors That Affect Solubility
Three factors that have a significant impact on solubility are: The presence of a common ion The ph of the solution The presence or absence of complexing agents.
Common-Ion Effect
The solubility of a slightly soluble salt is decreased when a common ion is added. This is an application of Le Chatelier s principle. Remember how the common ion places a stress on the product side? The equilibrium shifts to the left forming CaF 2 (solid) and precipitation occurs. The solubility of the CaF 2 (s) decreases. CaF 2 (s) Ca 2+ (aq) + 2 F-(aq) 1.00 M CaF 2 and 0.50 M NaF
Solubility & ph
Again apply Le Chatelier s principle: Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH-(aq) 1. If OH- is removed, then equilibrium shifts to the right & Mg(OH) 2 (s) dissolves. 2. The solubility of slightly soluble salts containing basic ions increases as ph decreases. a) In this case, OH- is removed due to the addition of the acid, H+. b) The basic the anion, the greater the effect.
Red Book Ch 4 Solving Typical Solubility Equilibrium Problems
Common solubility problems: 1. Calculation of the value of K sp from the solubility of a solid. 2. Calculation of the solubility of a solid, given the K sp value.
Next Example
The K sp value for AgBr(s) is 7.7 x 10-13 at 25 C. Calculate the solubility of AgBr(s) in water at 25 C. AgBr(s) Ag+(aq) + Br-(aq) K sp = [Ag+][Br-] = 7.7x10-13 I: 0 0 C: x x E: x x K sp = [Ag+][Br-] = 7.7x10-13 = (x)(x) X 2 = 7.7x10-13 X = 8.8 x 10-7 mol/l = solubility
Next Example
Rank the following solids from most to least soluble: AgI(s) (K sp = 1.5 x 10-16 ), AgBr(s) (K sp = 7.7 x 10-13 ), AgCl(s) (K sp = 1.6 x 10-10 ), SrSO 4 (s) (K sp = 2.9 x 10-7 ), CuI(s) (K sp = 5.0 x 10-12 ), CaSO 4 (s) (K sp = 6.1 x 10-5 )
Each solid produces two ions: MA(s) M n+ (aq) + A n- (aq) K sp = [M n+ ][A n- ] AgI(s) (K sp = 1.5 x 10-16 ), AgBr(s) (K sp = 7.7 x 10-13 ), AgCl(s) (K sp = 1.6 x 10-10 ), SrSO 4 (s) (K sp = 2.9 x 10-7 ), CuI(s) (K sp = 5.0 x 10-12 ), CaSO 4 (s) (K sp = 6.1 x 10-5 ) CaSO 4 (s)> SrSO 4 (s)> AgCl(s)>CuI(s)> AgBr(s)>AgI
Rank the following solids from most to least soluble: CuS(s) (K sp = 8.5 x 10-45 ), Ag 2 S(s) (K sp = 1.6 x 10-49 ), Bi 2 S 3 (s) (K sp = 1.1 x 10-73 ),
Each solid has a different number of ions so the K sp values cannot be compared directly to determine the relative solubilities. CuS(s) (K sp = 8.5 x 10-45 ), Ag 2 S(s) (K sp = 1.6 x 10-49 ), Bi 2 S 3 (s) (K sp = 1.1 x 10-73 ),
Selective Precipitation
Separation of ions by precipitation is called selective precipitation (note solubility rules handout). For example, if you wanted to separate out Ag+ from a mixture of Ba 2+, Ag+, NO 3 - Adding Na 2 SO 4 to the mixture forms a ppt of BaSO 4 (s) and leaves Ag+ in solution (Ag 2 SO 4 is very soluble & all NO 3 - are soluble). Sulfide ions are often used to separate metal ions
A solution contains 1.0 x 10-3 M Fe(NO 3 ) 2 and 1.0 x 10-3 M MnSO 4. Both Fe 2+ and Mn 2+ form sulfide salts: FeS (K sp = 3.7 x 10-19 ) and MnS (K sp = 1.4 x 10-15 ). If sulfide ion is added to the solution containing 10-3 M Fe2+ and 10-3 M Mn 2+, which sulfide salt will precipitate first and at what [S 2- ] will precipitation occur?
Precipitation will occur when the ion product (Q) is greater than the K sp for that salt: For FeS: FeS(s) Fe 2+ (aq) + S 2- (aq) Q = [Fe 2+ ] 0 [S 2- ] 0 = [1.0 x 10-3 ] 0 [S 2- ] 0 Let Q = K sp = 3.7 x 10-19 = [1.0 x 10-3 ] 0 [S 2- ] 0 [S 2- ] 0 = 3.7 x 10-16 M When [S 2- ] = 3.7 x 10-16 M no ppt of FeS(s) will occur, but if [S 2- ] > 3.7 x 10-16 M, FeS(s) will form, (shifts to the left).
Now do the same for MnS: MnS(s) Mn 2+ (aq) + S 2- (aq) Q = [Mn 2+ ] [S 2- ] = [1.0 x 10-3 ] [S 2- ] Let Q = K sp = 1.4 x 10-15 = [1.0 x 10-3 ] [S 2- ] [S 2- ] 0 = 1.4 x 10-12 M If [S 2- ] >1.4 x 10-12 M then MnS(s) will ppt out (shifts to the left)
Ppt of each sulfide will occur when the ion product (Q) > Ksp for that salt. If [S 2- ] >3.7 x 10-16 M then FeS(s) will ppt out Ksp = 3.7 x 10-19 (Q>Ksp) If [S 2- ] > 1.4 x 10-12 M then MnS(s) will ppt out Ksp = 1.4 x 10-15 (Q>Ksp) So which will ppt out first? Answer: FeS(s) will ppt out first (least soluble of the two) and then MnS(s) at the above concentrations
Old Stuff
Selective Precipitation of Sulfides
The key to the selective precipitation of sulfide salts is the basicity of S 2-, which allows its concentration to be controlled by controlling the ph of the solution. Sulfide forms the acid H 2 S, for which the following equilibrium expression can be written: [H+] 2 [S 2- ] = K = 1.32 x 10-20 [H 2 S]
To form sulfide precipitates, the solution is saturated with H 2 S, which produces a 0.10 M concentration of H 2 S. Thus, for a saturated solution, the equilibrium expression can be written: [H+] 2 [S 2- ] = [H+][S 2- ] = 1.32 x 10-20 [H 2 S] 0.10 [H+][S 2- ] = (0.10)(1.32 x 10-20 ) = 1.3 x 10-21 Note from this expression that the concentration of S 2- can be regulated by controlling the [H+]. A large [H+] means a small [S 2- ] and vice-versa.
Calculate the [S 2- ] in a solution saturated with H 2 S where the ph is 1.00 The ph = 1.00. To find the [H+], take the antilog of ph: [H+] = 10 -ph = 1.0 x 10-1 M The equilibrium expression for H 2 S in a saturated solution is: [H+] 2 [S 2- ] = 1.3 x 10-21 [S 2- ] = 1.3 x 10-21 = 1.3 x 10-21 = 1.3 x 10-19 M [H+] 2 1.0 x 10-1 M
Each solid produces two ions: MA(s) M n+ (aq) + A n- (aq) K sp = [M n+ ][A n- ] If x = solubility then each case at equilibrium [M n+ ] = x [A n- ] = x K sp = [M n+ ][A n- ] = (x)(x) x 2 = K sp x = K sp = solubility The solid with the largest K sp has the highest solubility CaSO 4 (s) >SrSO 4 (s) >AgCl(s) >CuI(s) >AgBr(s) >AgI(s)
Le Chatelier s Principle
A system will remain at equilibrium unless disturbed in some way. This prediction may be made in terms of Le Chatelier s Principle: If a stress is applied to a system at equilibrium, the position of the equilibrium will shift in the direction which reduces the stress.
Several types of stresses will be considered: *Addition or removal of reactants or products *Change in pressure by; -Change in volume -Addition of an inert gas *Change in temperature
A 1.000 L vessel contains 0.921 M N 2 (g), 0.763 M H 2 (g), and 0.158 M NH 3 (g) at equilibrium. The reaction is exothermic (heat is produced and acts as a product) 3 H 2 (g) + N 2 (g) <--> 2 NH 3 (g) + heat Predict effect by adding 1.000 mol H 2 Answer: Stress is added to H 2 and cause a shift to the right, consuming H 2 and relieving stress Predict effect by adding 1.000 mol NH 3 Answer: The shift will be to the left, since stress is applied to the right
3 H 2 (g) + N 2 (g) <--> 2 NH 3 (g) + heat Predict the effect by halving the volume (increasing the pressure) Answer: The system will shift in a direction that reduces its own volume. This can be done by a shift to the right, since the right side of the equation involves fewer molecules of gas and thus lower volume Predict the effect by adding He(g). Assume ideal behavior for all gases. Answer: There will be no effect. Ideal gases (in this case an inert gas) have negligible volumes and thus He is not involved in the reaction
3 H 2 (g) + N 2 (g) <--> 2 NH 3 (g) + heat Predict the effect of increasing the temperature. The equilibrium will shift to the left. The reaction is exothermic, which can be shown by treating heat as a product: 3 H 2 (g) + N 2 (g) <--> 2 NH 3 (g) + heat The temperature is increased by adding heat and will thus shift to the left, consuming some of the added heat.
Catalysts have no effect on the total final pressures at equilibrium, though it would cause the system to reach the same equilibrium state more quickly.