Chapter 19 Solubility and Complex Ion Equilibria "if you are not part of the solution, then you are part of the precipitate" - all solutions of salts exist as a balance between the dissolved cations and anions and the solid material that precipitates from the salt - an equilibrium between the "solution" and the "precipitate" - for example, precipitation of mineral salts is fundamental to the development of geological land forms
- the book gives calcium carbonate as an example CaCO 3 (s) is in equilbrium with Ca 2+ (aq), H + (aq), CO 3 2- (aq), and both CO 2 (g) and CO 2 (aq) - calcium carbonate makes up limestone (which can undergo metamorphosis to give marble) - limestone is the backbone of the Rocky Mountains - in essence, the Rockies are the dead bodies of animals that lived millions of years ago which precipitated out of the ocean
19.1 The Solubility Product, K sp - in order to quantify solubility, we need to be able to write the equilibrium - and we need an equilbrium constant Example: CaCO 3 (s) W Ca 2+ (aq) + CO 3 2- (aq) since calcium carbonate is a solid, we can express this equilibrium as: K sp = [Ca 2+ (aq)][co 3 2- (aq) ] = 2.8 x10-9 K sp is called the "solubility product"
-K sp is the quantification of the relationship between a solid salt and its constituent ions - like any other equilibrium, the exponents of the concentration terms depend upon the stoichiometric factors Example: PbCl 2 (s) W Pb 2+ (aq) + 2Cl - (aq) K sp = [Pb 2+ (aq)][cl - (aq)] 2 = 1.6 x10-5 Ag 2 CrO 4 (s) W 2Ag + (aq) + CrO 4 2- (aq) K sp = [Ag + (aq)] 2 [CrO 4 2- (aq) ] = 1.1 x10-12 Note: always written as a dissociation of the solid
19.2 The Relationship between Solubility and K sp - there is a relationship between the "saturation" concentration and a compound's K sp - that is, there is a relationship between the maximum concentration that can be obtained in solution before precipitation occurs and K sp Example: What is saturation concentration for Ag 2 CrO 4?
solubility equilibrium: Ag 2 CrO 4 (s) W 2Ag + (aq) + CrO 2-4 (aq) K sp = [Ag + (aq)] 2 [CrO 2-4 (aq) ] = 1.1 x10-12 to find the maximum solubility, let's assume that the concentration of chromate is given by "x" (that is, x = [CrO 4 2- (aq) ]) then the stoichiometric factors tell us that the concentration of silver ions is "2x" ([Ag + (aq)] = 2x). hence, K sp = [Ag + (aq)] 2 [CrO 2-4 (aq) ] = 1.1 x10-12 = (2x) 2 (x) = 4x 3 = 1.1 x10-12 and x 3 = (1.1 x10-12 )/4 = 2.75 x10-13 x = 6.5 x10-5 M
That is, in an aqueous solution of silver ions and chromate ions, the maximum soluble concentration is [CrO 4 2- (aq) ] = 6.5 x10-5 M and [Ag + (aq)] = 1.3 x10-4 M or, if solid Ag 2 CrO 4 is added to water, then it will produce a solution that has ionic concentrations of [CrO 4 2- (aq) ] = 6.5 x10-5 M and [Ag + (aq)] = 1.3 x10-4 M For any salt, there is a simple algebraic expression for its maximum solubility: MX ö (x)(x) = x 2 M 2 X ö (2x) 2 (x) = 4x 3 MX 2 ö (x)(2x) 2 = 4x 3 M 3 X ö (3x) 3 (x) = 27x 4 M 2 X 3 ö (2x) 2 (3x) 3 = 102x 5 etcetera
19.3 The Common Ion Effect in Solubility Equilibria - again, single sources of cations and anions behave ideally (i.e. dissolving AgCl (s) ) but a more complex solution must take into account all of the ions present - adding one of the components of a solubility equilibrium will shift the equilibrium (Le Chatelier's Principle) - that is, adding Ag + (aq) to a solution of Ag + and CrO 4 2- shifts the equilibrium to a new position and precipitation
- this means that the overall solubility of a solid compound is lower in the presence of a common ion Example: What would be the consequence of adding 0.10M AgNO 3 to our saturated solution of Ag 2 CrO 4? Ag 2 CrO 4 (s) W 2Ag + (aq) + CrO 2-4 (aq) initially: 2x x from AgNO 3 : 0.10 equilibrium: (2x + 0.10) x K sp = (0.10 + 2x) 2 (x) = 1.1 x10-12
if we assume that x << 0.10, then this becomes: K sp = (0.10) 2 (x) = 1.1 x10-12 x = 1.1 x10-12 = 1.1 x10-10 0.01 and our assumption is valid (1.1 x10-10 << 0.01)! - our new concentration of CrO 4 2- (aq) is 1.1 x10-10 M compared to 6.5 x10-5 M for our saturated solution - this is about 500,000 times less soluble
19.4 Limitations of the K sp Constant - every salt has a K sp but for some, it is very large and the salt is essentially 100% soluble. (i.e. K sp for NaCl is about 40 - the solubility limit for NaCl is about 369 grams per litre) - at these levels "concentrations" and "activity" can differ significantly and solubility is really based on activities - hence, while recognizing solubility limits, the K sp values are slightly erroneous and we simply treat the salts as if they will completely dissociate
The Diverse ("Uncommon") Ion Effect: The Salt Effect The presence of a non-interfering ion in solution can still alter the solubility of a salt: i.e. adding NaNO 3 to a solution of Ag 2 CrO 4 will increase the solubility of the precipitate. The effect is not large and is a consequence of the organization of the water molecules in solution - sometimes referred to as "the salt effect" K sp depends upon the ionic atmosphere of the solution
Incomplete dissociation of the Solute into Ions Normally, we speak of ionic salts as separate ions that each go their own way and do not interact in solution. This is what we imply with the "net ionic equation" that we write for the overall balanced equation. However, ions in solution sometimes form "ion pairs". These result from the formation of a solvent cage that can trap the ions together. Formation of ion pairs alters the true solubility making it appear greater than K sp would suggest.
Simultaneous Equilibria For a slightly soluble salt of a weak acid or base, we need to take into account the formation of the respective conjugates and possibility of the formation of a complex ion - both of which can change the solubility and make the value of K sp "wrong". All of these (Diverse Ion, Ion Pairing, Simultaneous Equilbria) represent limitations on the practical use of the K sp values obtained in reference books. That said, the numbers are still useful and provide valuable insight into what might happen in solution. Industrially, this is very important!
19.5 Criteria for Precipitation and Its Completeness Working out the "equilibrium quotient" is actually useful! Q sp can tell us whether or not a solution is saturated or even supersaturated. -Q sp is calculated based on the concentrations given -K sp is the maximum possible values if K sp > Q sp then precipitation should not occur K sp < Q sp precipitation should occur K sp = Q sp solution is saturated
Note: it is important to remember that real solutions have real volumes and dilution occurs during mixing Example: 1.0 ml of a 0.50M NaCl solution is added to 100 ml of a 0.01M AgNO 3 solution. Will precipitation occur? need to find the concentration of the chloride ions in the combined solution: (0.50 M)(0.001 L) = (x)(0.101 L) x = (0.50)(0.001) = 0.00495M (0.101)
need to find the concentration of silver ions in the combined solution: (0.010M)(0.100 L) = (x)(0.101 L) x = (0.010)(0.100) = 0.0099 M (0.101) and Q sp = [Ag + (aq)][cl - (aq)] = (0.0099)(0.00495) = 4.9 x10-5 which is well beyond the K sp value of 1.8 x10-10 which means that a precipitate will form.
- another question that we might ask is: Is precipitation complete? - the answer depends upon what is meant by complete as precipitation is an equilibrium and there will always be some of both sides of the equation. - usually, define complete as 99.9% precipitated - in the above example, [Ag + (aq)] = [Cl - (aq)] = 1.34 x10-5 M - so, we started with 0.0099 M Ag + (aq) and finished with 0.0000134M - which is 0.135% remaining. This precipitation has almost gone to completion.
Are you wondering. What conditions favour equilibrium? a) a small value of K sp b) high initial concentration of the target ion c) the presence of a large concentration of a common ion What conditions favour equilibrium, period? a) low temperature b) pressure c) concentration
19.6 Fractional Precipitation - also called selective precipitation - the idea is to separate out one component of a mixture based on the relative solubilities of the species involved (the relative magnitudes of K sp ) - important for industry and for chemical analysis - fractional precipitation requires a significant difference in K sp values for the compounds involved - it is accomplished by the slow and careful addition of one of the components
Example: How do we selectively precipitate out phosphate from a mixture containing 0.10M CO 3 2- (aq) and 0.10M PO 4 3- (aq)? - need to use Table 19-1 - K sp values MgCO 3 W Mg 2+ (aq) + CO 3 2- (aq) K sp = 3.5 x10-8 -MgCO 3 precipitates when: [Mg 2+ (aq)] = 3.5 x10-8 0.10 = 3.5 x10-7 M
whereas Mg 3 (PO 4 ) 2 precipitates when: [Mg 2+ (aq)] = (1 x10-25 / 0.1 2 ) ⅓ = 2.1 x10-8 M so, the magnesium phosphate will precipitate first and will keep on precipitating exclusively, as long as the concentration of magnesium ions does not exceed 3.5 x10-7 M. That is, as long as the concentration of Mg 2+ (aq) exceeds 2.1 x10-8 M but not 3.5 x10-7 M, then Mg 3 (PO 4 ) 2 will precipitate out exclusively. Of course, as Mg 3 (PO 4 ) 2 precipitates, this will remove phosphate from solution.
We need to know at what point will we stop exclusively precipitating Mg 3 (PO 4 ) 2 : for [Mg 2+ (aq)] = 3.5 x10-7 M (when MgCO 3 starts to appear), the concentration of PO 4 3- (aq) will be given by: so, K sp = [Mg 2+ (aq)] 3 [PO 3-4 (aq) ] 2 = 1 x10-25 [PO 3-4 (aq) ] = (1 x10-25 /[Mg 2+ (aq)] 3 ) ½ = (1 x10-25 /[3.5 x10-7 ] 3 ) ½ = 0.00153 M This means that 99.85% of the phosphate will have precipitated before MgCO 3 starts to form. This is good selectivity but not perfect.
19.7 Solubility and ph -H + (aq) and OH - (aq) are ions in solution just like any other ions in solution and they can act as common ions thereby shifting equilibria. - particularly true if the anion is the conjugate base of a weak acid or OH - itself Example: iron hydroxide Fe(OH) 3 (s) W Fe 3+ (aq) + 3OH - (aq) K sp = [Fe 3+ (aq)][oh - (aq)] 3 = 4 x10-38
this compound is a major contributing factor to the formation of rust but it can be washed away or dissolved with acids: Fe(OH) 3 (s) W Fe 3+ (aq) + 3OH - (aq) K sp = 4 x10-38 3OH - (aq) + 3H + (aq) W 3H 2 O K w -3 = 1 x10 42 Fe(OH) 3 (s) + 3H + (aq) W Fe 3+ (aq) + 3H 2 O K eq = K sp x K w -3 = (4 x10-38 )(1 x10 42 ) = 4 x10 4 the value of K eq tells us that the iron hydroxide is reasonably soluble at low ph
- other slightly soluble compounds are also more soluble in acidic media i.e. limestone or marble - CaCO 3 - dissolves in acid which is why acid rain is so damaging to buildings
19.8 Equilibria involving Complex Ions The concept of Lewis acids and bases is critical to understanding complexation of ions in solution. Any cation can act as a Lewis acid by accepting an electron pair from a Lewis base. Any anion can act as a Lewis base by donating an electron pair. But even neutral molecules can act as Lewis acids and Lewis bases. In particular, water is a Lewis base capable of supporting the dissolution of cations and anions in aqueous solution. This is why salts are soluble.
from the book: - addition of a moderately concentrated solution of ammonia to AgCl (s) results in the silver compound dissolving - the key is the formation of the complex ion, [Ag(NH 3 ) 2 ] + AgCl (s) + 2NH 3 (aq) W [Ag(NH 3 ) 2 ] + (aq) + Cl - (aq) - a complex ion is a cation bound by ligands - coordination compounds are substances composed of complex ions
All salts are actually complex ions when dissolved in aqueous media. This is what the (aq) actually means - that the species is complexed by water molecules. i.e. Na + (aq) means that there are six water molecules surrounding the sodium ion: Na +
- however, we generally ignore this complexation by water and do not write it out in the formation constant expression. (And many problems arise because of this!) - we do consider complexation for other species - examples of the formation constants are provided in Table 19-2. - this is one of the most important concepts in chemistry - complexation affects many different areas - from biochemistry (bone loss) to mining (cyanide extraction of metals) to medicine (chelating agents to deal with mercury poisoning) to dishwashing liquids (surfactants)
Examples: If a 0.20 mol sample of AgNO 3 is dissolved in 1.0 L of a 0.5M NaCN solution and 0.10 mol of NaCl is added, will AgCl (s) precipitate? To start, assume that the initial reaction between cyanide and silver ions goes to completion. That is, Ag + (aq) + 2CN - (aq) W Ag(CN) 2 - (aq) initial: 0.20 0.50 0 change: - 0.20-0.40 0.20 equilibrium:. 0 0.10 0.20
but, of course, this is really an equilibrium, which we can now work backwards to find the free silver ion concentration: Ag + (aq) + 2CN - (aq) W Ag(CN) 2 - (aq) initial: 0.10 0.20 change: x x equilibrium: x 0.10 + x 0.20 - x - x substituting in to the equilibrium expression gives: [Ag(CN) 2- ] = (0.20 - x) = 5.6 x10 18 [Ag + ][CN - ] 2 (x)(0.10 + x) 2
assuming that x is small compared to 0.10 gives, (0.20) / (x)(0.10) 2 = 5.6 x10 18 and x = (0.20) / (0.10) 2 (5.6 x10 18 ) = 3.57 x10-18 the molarity of the silver ions that are not complexed by the cyanide solution is 3.57 x10-18 And working out the Q sp for the AgCl (s) gives: Q sp = [Ag + (aq)][cl - (aq)] = (3.57 x10-18 )(0.10) = 3.57 x10-19 and this is considerably less than K sp = 1.8 x10-10
hence, complexation results in there being not nearly enough free silver ions in solution to cause precipitation more importantly, any silver salts in contact with the solution will dissolve - this is why this is used in mining and purification of silver ores
What is the minimum concentration of cyanide ions required to prevent AgCl (s) from forming in a 1.00 L solution containing 0.01 mol AgNO 3 and 1.00 mol NaCl? - since the solution has [Cl - (aq)] = 1.0 M, the maximum solubility of the silver ions is: [Ag + (aq)] = K sp /[Cl - (aq)] = (1.8 x10-10 )/(1.0) = 1.8 x10-10 - hence, virtually all of the silver ions must be complexed by cyanide - except 1.8 x10-10 M - need 0.020 mol of cyanide and a little more to ensure that the equilibrium will remain in balance
- excess cyanide is given by: K f = [Ag(CN) 2 - (aq) ] = 5.6 x10 18 [Ag + (aq)][cn - (aq)] 2 = (0.010) = 5.6 x10 18 (1.8 x10-10 )[CN - (aq)] 2 [CN - (aq)] 2 = 0.010 = 9.92 x10-12 (1.8 x10-10 )(5.6 x10 18 ) [CN - (aq)] = 3.417 x10-6 M - the total concentration is therefore: 0.020 M + 3.417 x10-6 M. 0.020 M (stoichiometric)
What is the solubility of AgCl in 0.01 M NaCN? combine equilibria: AgCl (s) W Ag + (aq) + Cl - (aq) K sp = 1.8 x10-10 Ag + (aq) + 2CN - (aq) W Ag(CN) 2 - (aq) K f = 5.6 x10 18 AgCl (s) + 2CN - (aq) W Ag(CN) 2 - (aq) + Cl - (aq) K eq = K sp x K f = 1.01 x10 9 accordingly, AgCl (s) + 2CN - (aq) W Ag(CN) - 2 (aq) + Cl - (aq) initial: 0.01 0 0 change: - 2x x x equilibrium: 0.01-2x x x
K eq = [Ag(CN) 2 - (aq) ][Cl - (aq)] = (x)(x) = (x/0.01-2x) 2 [CN - aq] 2 (0.01-2x) 2 (0.01-2x) x = (1.01 x10 9 ) 1/2 = 3.176 x10 4 x = 3.176 x10 4 (0.01-2x) = 3.176 x10 2-6.3529 x10 4 x 6.343 x10 4 x = 317.6 x = 5.0 x10-3 Hence, the molar solubility of AgCl (s) in 0.010M NaCN is 0.005M (stoichiometric amount). It s not really necessary to do all of the math if we know that all of the cyanide will complex silver ions! (But good practice anyway!)
19.9 Qualitative Cation Analysis - this is an old method of determining the actual contents of an unknown solution - it shows up in experiments on semi-quantitative analysis -pre-dates modern instrumentation such as ICP and MS - still has some use if faced with a complete unknown - takes advantage of precipitation to separate out the cations of individual groups - further reactions can eventually tells us exactly what is present.