A transparent crystal appears to lie motionless in a clear solution and would be what we would describe as a picture of rest. But remember, on the molecular (nanoscopic or itty bitty teeny tiny) level, it is a scene of activity and constant motion. Ions that are stuck in the crystal lattice break away and float off surrounded by water molecules safe in their hydrated cages. Ions which are in solution are caught by the crystal and become stuck back in the solid phase again. Such is the game of dissolving and precipitating a constant state of dynamic equilibrium. This game is responsible for many natural phenomena. For example, the ocean above the Bahama Banks where the water is rich in dissolved calcium (Ca+2 ions) and carbon dioxide (CO2) is often milky white colored due to the formation of CaCO3 crystals in solution. Near Carlsbad, New Mexico, the underground dissolving of calcium carbonate has hollowed out vast caves draped and studded with strange limestone shapes. Less spectacular and slightly more painful are the slow formation of kidney stones from calcium oxalate or the pollution of lakes with heavy metal salts. Ahhhhh, the word salt and the formation of certain compounds certain solid compounds should take us back back to CH 131 and our wonderful solubility rules (

Solubility Rules:

1. All common compounds of Group I and ammonium ions (NH4+1) are soluble.

2. All nitrates (NO3-1), acetates (C2H3O2-1) , and chlorates (ClO3-1) are soluble.

3. All binary compounds of the halogens Group VIIA (other than F) with metals are soluble, except those of Ag, Hg(I), and Pb. Pb halides are soluble in hot water.

4. All sulfates are soluble, except those of barium, strontium, calcium, lead, silver, and mercury (I). The latter three are slightly soluble.

5. Except for rule 1, carbonates (CO3-2), hydroxides (OH-1), oxides (O-2), silicates, and phosphates (PO4-3) are insoluble. Ba(OH)2 is slightly soluble.6. Sulfides (S-2) are insoluble except for calcium, barium, strontium, magnesium, sodium, potassium, and ammonium.

Consider calcium oxalate crystals lying in the bottom of a saturated solution of calcium oxalate. A saturated solution is one in which the dissolved solute is in dynamic equilibrium with undissolved solute in a solution: CaC2O4 (s) Ca+2 (aq) + C2O4-2 (aq)And like any equilibrium system, a saturated solution obeys LeChateliers Principle and is governed by an equilibrium expression and constant.

Concept Test:

CaC2O4 (s) Ca+2 (aq) + C2O4-2 (aq)Which direction will the reaction shift if

[Ca+2] is increased?

[Ca+2] is decreased?

[C2O4-2] is increased?

[CaC2O4] is increased?

There are many ionic compounds that dissolve completely. These would be reactions that go to completion and therefore all the solid would be dissolved. Since we are looking at systems in equilibrium there must be some solid present. Therefore we will be focusing our attention on the aqueous equilibria of slightly soluble ionic compounds. At some point, when you add a salt to a solvent (usually water) you will reach the point where the solvent cannot dissolve any more of that solid at that temperature. At more than this amount (and it is different for each salt), and some of your solute will be left behind undissolved, and will settle at the bottom of the container. In this saturated solution at a particular temperature, equilibrium will exist between the undissolved solid and the dissolved solute. Slightly soluble (which we classified term 1 as insoluble) ionic compounds have very low solubility, so they reach equilibrium with relatively little solute dissolved.

The substances that we will examine are sometimes called sparingly soluble salts. Their concentrations are small, even in a saturated solution. The use of equilibrium expressions is limited to sparingly soluble salts because equilibrium constants are reliable only when the ionic concentrations are low enough such that ion-ion attractions in solution can be ignored (think vant Hoff factor from properties of solutions!!).When a soluble ionic compound dissolves in water, it dissociates completely (100%) into ions.

NaCl (s) + H2O (l) Na+1 (aq) + Cl-1 (aq)For simplicitys sake, we will also assume that a small amount of slightly soluble ionic compound will also dissociate completely into ions. In reality, however, this is not the case. Many slightly soluble salts, particularly those of heavy metals or transition metals, have a significant amount of covalent character in the metal-non-metal bonding (remember, bonding is a continuum not an absolute!! (X (electronegativity values!!) and their solutions often contain other species that are partially dissociated into ions or not dissociated at all. For example, when PbCl2 (lead (II) chloride) is dissolved in water the solution contains Pb+2 and Cl-1 ions but also PbCl2 (aq) dissolved molecules and other species that will not be discussed in this level of chemistry. Just know that the systems are often more complicated than they seem and these calculations are an introduction into reality.

If we make the assumption that there is complete dissociation of slightly soluble ionic compounds into their component ions, then equilibrium will exist between the solid and the aqueous (dissolved) ions. Thus, for a saturated solution of lead (II) sulfate we would have:PbSO4 (s) Pb+2 (aq) + SO4-2 (aq)

As with all other equilibrium systems, this one can also be expressed using

a reaction quotient (Qc) expression:

Qc =

But solids do not appear (nor do liquids) in an equilibrium expression

And when examining the solubility of a salt we are lookingat the ion-product expression:

Qsp = [Pb+2][SO4-2]

And when the solution is AT equilibrium our Q expression becomes a K

expression, in this case, Qsp would become Ksp (solubility product)at equilibrium: Ksp = [Pb+2][SO4-2]

The Ksp value for lead (II) sulfate at 25oC = 1.6 x 10-8As we have seen with other K values, only temperature will affect the numerical value of K. Adding additional species (increasing concentrations) or removing species (decreasing concentrations) will cause the equilibrium position of the reaction to shift left or shift right but the numerical ratio of products will remain the same.

The form of Qsp and Ksp is the same as all other equilibrium expressions that we have written thus far: each ion concentration will be raised to its exponent which is equal to the coefficient in the balanced dissociation equation. In this case, the exponents will also equal the subscript of each ion given in the formula!

Concept Tests:

Write the dissociation of and the Qsp (ion-product) expression for the following:

Name each salt as well!!

Ca3(PO4)2 Ca+2 (aq) + PO4-3 (aq)Qsp = Cr2(CO3)3 Cr+3 (aq) + CO3-2 (aq)Qsp =

BaSO4 Ba+2 (aq) + SO4-2 (aq)

Qsp =

Of course, sometimes when ions dissociate from their parent ionic compounds we know that they react with water to form acidic or basic solutions. This means, that sometimes ions form and participate in reactions with water that you might not initially expect when examining the Ksp expression. For example, S-2 comes from H2S which is a weak acid. That makes the S-2 ion a strong base. And it will react with water as such:S-2 (aq) + H2O (l) HS-1 (aq) + OH1 (aq)

This creates additional ions in solution that you might not initially predict

Lets examine what happens when a species, such as manganese (II) sulfide is dissolved in water.

MnS (s) Mn+2 (aq) + S-2 (aq)and

S-2 (aq) + H2O (l) HS-1 (aq) + OH-1 (aq)Remember when reactions happen together the true reaction

is the sum of all that is going on:MnS (s) Mn+2 (aq) + S-2 (aq)S-2 (aq) + H2O (l) HS-1 (aq) + OH-1 (aq)MnS(s) + H2O (l) Mn+2 (aq) + HS-1 (aq) + OH-1 (aq)

Which means that the Ksp expression is:

Ksp = [Mn+2][HS-1][OH-1]The magnitude of Ksp is a measure of how far to the right the dissolution proceeds at equilibrium. Larger Ksp values mean that more of the solid dissociated into ions. Smaller Ksp values mean that very little solid dissociated. These values can be used to compare solubilities. (Just like previous Ka values were used to determine the relative [H+1] in solution!) Know that these values will be quite low remember we are talking about species at low concentrations!

Previously we examined two general types of equilibrium problems. For the first type, concentrations were used to find K. For the second type, K was used to find a missing concentration. Since we are still talking about K (!!!) the problems should look very familiar to you.

Determining Ksp from solubility: The solubility of an ionic compound is determined experimentally. Exactly how much solid dissolves is a question that can be investigated and answered in lab. There are many resources or chemical handbooks that contain tables with this information (e.g. the CRC Handbook of Chemistry and Physics). Most solubilities are given in grams of solute dissolved in 100 grams of solvent (H2O). Remember that 1g= 1mL and the error introduced by the mass of the solute will be negligible (we are not able to dissolve much solid into water!!). Thus, knowing grams of solute and mL of solvent we should be able to convert to Molarity (moles solute/L solution). Since we are talking about the amount (moles) of solute that have dissolved in the water, this is also called the molar solubility.

Concept Test:

When lead (II) fluoride is shaken with water at 25oC the solubility is found to be

0.64 grams/L. Calculate the Ksp of lead (II) fluoride.Step 1: Write the dissociation equation and solubility product equation:PbF2 (s) Ksp =

Step 2: Convert the solubility into molar solubility:

Step 3: Determine the concentrations of each ion

in order to plug them into the Ksp expression

Remember to include the stoichiometric ratios!!

Step 4: Calculate the numerical value of Ksp for lead (II) fluoride

Ksp =

Ksp =

Ksp =

Remember that K values are unitless!!

The low Ksp values is consistent with the low solubility of seemingly insoluble salts. Remember that we are making assumptions that species are behaving ideally in solution. If you look up the actual value in a table, the Ksp for lead (II) fluoride is 3.6 x 10-8. Notice that our value is greater than the actual (experimentally determined value). This is an example of the complexity of these types of systems. The reason that the experimentally determined value is lower than our calculated value is because some of the PbF2 dissolves in water but remains as a molecule. Remember that vant Hoff factors also assume complete dissociation and we used exact whole numbers, but in reality, the true vant Hoff factors are always less than (smaller) than what we assume as species do not behave ideally! Because other species are present in solution (which we did not include in the calculation), our calculated Ksp value (which was a pretty simple calculation) is used as an approximation.

Determining Solubility from KspThe reverse of the problem type above is that the Ksp value is given and instead, the solubility of the compound based on its formula is found. These problems are similar to our ICE tables that we used for studying other systems at equilibrium.

Concept Test:

Calcium hydroxide is a major component in mortar, plaster, cement, and solutions of calcium hydroxide are used in industry as a cheap, strong base. Calculate the solubility of calcium hydroxide in water if Ksp = 6.5 x 10-6. Also, what are the equilibrium concentrations of Ca+2 and OH-1?Step 1: Write the dissociation equation and set up the ICE table

I

C

E

Step 2: Set up Ksp expression and solve for xKsp =

x = x = [Ca(OH)2] = [Ca+2]eq =

[OH-1]eq = Using Ksp values to compare solubility: Ksp values are a guide to relative solubility as long as you compare species that contain the same number of TOTAL ions!! In such cases, the greater the Ksp value, the more soluble the species. It does not matter if the number of cations and anions are the same, it is the total number of ions in solution that matter. Thus, one can compare the solubilities of Ca(OH)2 with Ag2CrO4 because the total number of ions for each is 3. For calcium hydroxide, there is one cation and two anions while for silver chromate there are two cations and 1 anion, but the total number of ions is still 3 (Lets examine this in more detail: Comparing silver chloride with silver carbonate:

AgCl (s) Ag+1 (aq) + Cl-1 (aq) Ksp = 1.8 x 10-10Ag2CO3 (s) 2Ag+1 (aq) + CO3-2 (aq) Ksp = 8.5 x 10-12It would appear that AgCl is more soluble than Ag2CO3. But remember, when we are talking about solubilities, what we are really concerned about are the concentration of species in solution. The chemical species that has the largest concentration (M) of ions in solution will be the salt that is more soluble. That should make sense!!

So, lets solve the concentration of Ag+1, Cl-1, CO-3, AgCl, and Ag2CO3 in the samples above:

AgCl (s) Ag+1 (aq) + Cl-1 (aq)I-00

C -+x+x

E-xX

Set up Ksp expression and solve for xKsp = [Ag+1][Cl-1]

1.8 x 10-10 = [x][ x]

1.8 x 10-10 = [x2]

x = [Ag+1] = [Cl-1] = 1.3 x 10-5 M = [AgCl] Ag2CO3 (s) 2Ag+1 (aq) + CO3-2 (aq)I-00

C -+2x+x

E-2xX

Set up Ksp expression and solve for xKsp = [Ag+1]2[CO3-2]

8.5 x 10-12 = [x][2x]28.5 x 10-12 = x[4x2]

8.5 x 10-12 = 4x3x = [CO3-2] = 1.3 x 10-4 M = [Ag2CO3][Ag+1] = 2(1.3 x 10-4 M) = 2.6 x 10-4 MNotice that [AgCl] is less than the [Ag2CO3] that is in solution, as are the individual concentrations of ions!! So again, only comparisons between species that contain the same number of total ions can be made!

Previously we examined the common ion effect with regard to LeChateliers Principle (adding a chemical species that was the same in the dissociation reaction would cause the equilibrium to shift away from the added species). We also examined the common ion effect with regard to buffer system. In fact, that was how a buffer system was made. We used a weak acid and the salt of anion. For example, we discussed acetic acid/sodium acetate buffer systems (CH3COOH/CH3COONa) , as well as sodium dihydrogen phosphate/ sodium monohydrogenphosphate buffer systems (NaH2PO4/Na2HPO4).

Remember that the presence of a common ion would suppress the acid from dissociating, again, driving he equilibrium away from the species that is added:

Fe(SCN)+2 Fe+3 + SCN-1Adding Fe+3 would cause the equilibrium to shift: LEFT

CH3COOH CH3COO-1 + H+

Adding acetate (in the form of sodium acetate) would cause the equilibrium

to shift: LEFT

The same result is seen when dealing with ionic species! The presence/addition of a common ion will suppress (decrease) the solubility of a slightly soluble/insoluble ionic compound.

PbCrO4 (s) Pb+2 (aq) + CrO4-2 (aq) Ksp = 2.3 x 10-13Remember that at a given temperature, the product of the ions must always equal this Ksp value. Thus, if the concentration of one of the ions is increased, the other must go down to maintain the same Ksp value. Suppose Na2CrO4 (sodium chromate), a very soluble salt is added to the solution: predict what will happen?

PbCrO4 (s) Pb+2 (aq) + CrO4-2 (aq) Ksp = 2.3 x 10-13

(increases)

Since the reaction is at equilibrium, the reaction must counteract this stress of the added ion so Pb+2 will combine with the excess CrO4-2 to form more PbCrO4 (s). This decreases the concentration of Pb+2 ions in solution. The overall effect is the same that we have seen with other equilibrium systems: the reaction will shift left. After the addition of the sodium chromate, the [CrO4-2] is higher but the [Pb+2] is lower, thus we have essentially decreased the amount of PbCrO4 that has dissolved so we have suppressed its ionization.

Of course, we can calculate the effect of a common ion on solubility and it will be very similar to previous calculations that we have seen! Again ICE tables will be used, but instead of the starting concentrations of ions being zero, we now have a concentration for one of the species initially!

Concept Test:

We previously calculated the solubility of Ca(OH)2 in water. What will its solubility in 0.10 M Ca(NO3)2 be? Ksp = 6.5 x 10-6Step 1: Write the dissociation reaction for Ca(OH)2Step 2: Identify the common ion: Ca(NO3)2 added:

Step 3: Calculate the concentration (M) of the common ion added

Step 4: Set up an ICE table with Ca(OH)2 and calculate [Ca(OH)2]I

C

E

Step 5: Solve for missing variables and [Ca(OH)2]Ksp =

assume x is small!

x = 4.0 x 10-3 M

assumption check!

[Ca(OH)2] = In the initial problem, [Ca(OH)2] = 1.2 x 10-2 M but after Ca(NO3)2 was added the[Ca(OH)2] is 4.0 x 10-3 M.

The hydronium ion concentration [H3O+1] can have a profound effect on the solubility of ionic compounds. If the ionic compound contains the anion of a weak acid, then adding H3O+1 will increase the solubility of the ionic solid. Again, it is LeChateliers Principle that comes into play:CaCO3 (s) Ca+2 (aq) + CO3-2 (aq)Calcium carbonate is a sparingly soluble salt. The carbonate ion is the anion of a weak acid carbonic acid (H2CO3). If you remember, we have talked about carbonic acid previously concerning what happens to carbonic acid in water:

H2CO3 (aq) H2O (l) + CO2 (g)Thus, if acid is added to a solution containing calcium carbonate, the equilibrium shifts:

CaCO3 (aq) Ca+2 (aq) + CO3-2 (aq)Add H3O+1 which will react with CO3-2H3O+1 + CO3-2 H2CO3 H2O (l) + CO2 (g)Thus, the added acid will react with the carbonate ion forming carbonic acid which decomposes to form carbon dioxide and water. As the carbon dioxide leaves the system, this decreases the amount of CO3-2 in the system. Back to the original equation. Which way will the equilibrium shift if the [CO3-2] is decreased?

CaCO3 (aq) Ca+2 (aq) + CO3-2 (aq)

The reaction will shift to the right thus causing MORE CaCO3 to dissolve,

Thus it increases the solubility of CaCO3This ONLY works for the anions of WEAK ACIDS. Remember, the anions of strong acids are weak bases and have no propensity to be attracted to H+1 ions in solution. Therefore the concentration of the base will not be affected so no more solid will be dissolved!

Q and K values are related to one another. In fact, we talked about how K is just a special type of Q value when the system is at equilibrium. For other situations Q values are determined which show us the reaction as is proceeds and as the concentrations of reactants and products change in a chemical reaction. At any particular time during a reaction at some temperature we could monitor the concentrations of species perhaps by doing a titration, or by spectrophotometry. And given the value of Q we could determine if the ratio of concentrations of products is greater than the concentration of reactants, or if the ratio of concentrations of reactants is greater than the concentration of products, or if the reaction has reached a state of equilibrium.

We know that Q can be less than K, Q can be greater than K, or Q can be equal to K. But what does that mean?

The same ideas that we have discussed previously concerning Q and K apply here!

Qsp < Ksp: Remember that the formula to determine Qsp is the same as the formula to determine Ksp. Therefore, if Qsp is smaller than Ksp, then the concentrations of the products, the ions, must not be large enough. If we are comparing this situation to equilibrium, then we have the ability to dissolve more of our solid (reactant) and not a high enough [product]. The reaction would have to shift towards the formation of more product (ions) in order to attain equilibrium. This means that our solution is unsaturated and we will not have a ppt (or solid) present.Qsp > Ksp: If Qsp is larger than Ksp that means that the concentration of products must be greater and the solution is temporarily supersaturated. There is too much [ion]. In order to reach equilibrium, the concentration of the products must decrease and the concentration of the reactants must increases. This indicates a shift towards the left (or towards the reactants) in order to attain equilibrium. This means that a precipitate will form until the concentration of the ions is lower and the solution is saturated (instead of supersaturated).Qsp = Ksp: The special situation which the system is at equilibrium. This situation exists when the solution is saturated and no change occurs. Thus, despite the system being dynamic with changes occurring on the nanoscopic scale, no net change is occurring on the macroscopic level.

Previously, we have examined simple ions in solution, such as Na+1 or even SO4-2 ions that are dissolved in water. These simple ions consist of one or a few (sulfur bonded to four oxygens) atoms that either have an excess number of electrons (they are anions) or they are electron deficient (they are cations). A complex ion is exactly what it sounds like. It is an ion that it a little bit more complex in nature/structure than anything we have examined before. Typically they are ions that consist of a central metal ion that is covalently bonded to two or more anions or molecules which are called ligands. Chloride, hydroxide, and cyanide ions are some of the more complex ion ligands while water, carbon monoxide, and ammonia are some of the more common molecular ligands. There is a more general, slightly more recent theory of acids and bases, the Lewis acid base theory.

Gilbert Newton Lewis is one of the great chemists in history. His greatest discovery may well be the theory of the covalent bond in 1916, but he made many other contributions. One was his theory of acids and bases. What Lewis wanted was a general definition of an acid and a base, one that was universal no matter what the chemical environment. He knew that the current ideas (the Arrhenius theory - dating from the late 1800's and the Brnsted-Lowry theory - announced in January 1923) were not sufficient. Lewis wrote:

Lewis then gives his definition of an acid and a base: The basic substance furnishes a pair of electrons for a chemical bond, the acid substance accepts such a pair."

It is important to make two points here:

NO hydrogen ions need to be involved.

NO solvent needs to be involved.The Lewis theory of acids and bases is more general than the "one sided" nature of the Brnsted-Lowry theory. Keep in mind that Brnsted-Lowry, which defines an acid as a proton donor and a base as a proton acceptor, REQUIRES the presence of a solvent, specifically a protic solvent, of which water is the usual example. Since almost all chemistry is done in water, the fact that this limits the Brnsted-Lowry definition is of little practical consequence.

The Lewis definitions of acid and base do not have the constraints that the Brnsted-Lowry theory does and, as we shall see, many more reactions were seen to be acid base in nature using the Lewis definition than when using the Brnsted-Lowry definitions.

To recap, Lewis acids and bases are defined as:Acid: an electron acceptor.

Base: an electron donor.

A "Lewis acid" is any atom, ion, or molecule that can accept electrons and a Lewis base is any atom, ion, or molecule capable of donating electrons. You may read donating or accepting an electron pair. This is not always the case; it could be a single electron.

It turns out that it may be more accurate to say that Lewis acids are substances which are electron deficient or low electron density and Lewis bases are substances which are electron rich or high electron density.

Examples of Lewis acids:

positive ions

having less than a full octet in the valence shell

polar double bonds (one end)

expandable valence shells

Examples of Lewis bases:

negative ions

one of more unshared pairs in the valence shell

polar double bonds (the other end)

the presence of a double bond

All complex ions are Lewis adducts. That is, the metal ion acts like the Lewis acid (it accepts the electron pair) while the ligands behave as the Lewis bases (donate the electron pair to the metal).

Formation of Complex Ions:Complex ions are prevalent in chemistry and are the basis of such biologically important substance such as hemoglobin and vitamin B12. They are important also because a water soluble complex ion can often be formed in preference to an insoluble salt.

Whenever a metal ion enters water, a complex ion forms, with water acting as the ligand. The hydrated cage that we have previously spoken of. When the hydrated cation is treated with another solution which contains a different ligand, the bound water molecules will exchange places and the new ligand will move in and become bound to the metal cation. Any negative ion or polar molecule can be attracted to a metal ion:M(H2O)4+2 (aq) + 4NH3 (aq) M(NH3)4+2 (aq) + 4H2O (l)

At equilibrium, the K expression looks like:

Kc =

But remember, water does not appear in the overall K expression, and since we

are talking about the formation of a complex ion, these equilibrium expressions

are known as formation constants.

Kf =

At the nanoscopic level, the actual process is stepwise, with each new ligand replacing the water molecules one at a time which gives a series of intermediate species, each with its own formation constant. M(H2O)4+2 (aq) + NH3 (aq) M(H2O)3(NH3)+2 (aq) + H2O (l)Kf1 =

M(H2O)3(NH3)+2 (aq) + NH3 (aq) M(H2O)2(NH3)2+2 (aq) + H2O (l)Kf2 =

M(H2O)2(NH3)2+2 (aq) + NH3 (aq) M(H2O)(NH3)3+2 (aq) + H2O (l)Kf3=

M(H2O)(NH3)3+2 (aq) + NH3 (aq) M(NH3)4+2 (aq) + H2O (l)Kf4 =

The sum of each of the above equations give us the target equation:

M(H2O)4+2 (aq) + 4NH3 (aq) M(NH3)4+2 (aq) + 4H2O (l)This means that the overall Kf expression is the PRODUCT of each individual step along the way:

Kf = Kf1 x Kf2 x Kf3 x Kf4

Complex ions can also be formed from the dissociation of sparingly soluble salts in solution. Calcium carbonate is not readily soluble in water, but we have seen how acidic solutions make calcium carbonate more soluble. Silver chloride is not really soluble in water nor is it readily soluble in acidic solutions. It is, however, soluble in ammonia. AgCl (s) + 2NH3 (aq) Ag(NH3)2+1 (aq) + Cl-1 (aq)Again, dissolving AgCl (s) can be viewed as a multi-step process. Again, LeChateliers Principle comes into play. First, some AgCl dissolves creating some Ag+1 and Cl-1 ions in solution. However, the Ag+1 ions are attracted to the ammonia molecules forming the complex, which removes some of the Ag+1 from solution. The lowering of the [Ag+1] means that the equilibrium will shift to the right thus allowing more of the AgCl (s) to dissolve, increasing its solubility. AgCl (s) Ag+1 (aq) + Cl-1 (aq) Ksp = 1.8 x 10-10 Ag+1 (aq) + 2NH3 (aq) Ag(NH3)2+1 (aq) Kf = 1.6 x 107Again, the overall equilibrium expression will be the sum of the above:

AgCl (s) + Ag+1 (aq) + 2NH3 (aq) Ag+1 (aq) + Cl-1 (aq) + Ag(NH3)+1 (aq)AgCl (s) + 2NH3 (aq) Ag(NH3)+1 (aq) + Cl-1 (aq)

Knet = Ksp x Kf = = 2.9 x 10-3

The table above shows the formation constants of some complex ions. Notice that the formation constants are all large, this means that the formation of the complex ion is highly favored. Because of this behavior, complex ion formation is used to retrieve a metal from its ore, eliminate toxic or unwanted metal from solution, or convert the metal ion into a different form.

Many of the ideas discussed so far are used to analyze the ions in a mixture. This is a brief introduction to controlling and selectively precipitating ions in the presence of other ions and how complex mixtures of ions are separated into smaller groups and each ion separated from one another. You can often select one ion in a solution from another by exploiting differences in the solubility of their compounds with a given precipitating ion. In the process of selective precipitation, a precipitating ion is added until all of the less soluble compound precipitates while the soluble compounds remain behind in solution. Thus, an ion has been removed and identified from solution and then you are left with soluble species still in solution. You can decant off the solution and theoretically, all of an ion has been removed from solution. Typically, these sorts of reactions are carried out on solutions that contain species that are unknown to the chemist. Designing an experiment to selectively remove ions from solution is limited by the solubility rules and also by the chemists ingenuity. Think carefully when separating species, you cannot choose a precipitating species that would precipitate all of your ions at once. Then you would not be able to identify the species that were in solution!The general procedure is to begin separating the unknown solution into ion-groups based on their reactivity. Two test tubes, one which contains the unknown ions, and another which contains water are treated in the same manner. This helps the chemist to identify positive or negative results. Ion Group 1: Insoluble chlorides: The entire mixture is treated with 6.0 M HCl. Most metal chlorides are soluble so only a few ions will selectively precipitate. If there is a precipitate, it is either Ag+1, Hg2+2, or Pb+2. If the solution contains all three of these ions, performing this step first would make analysis difficult. If no precipitate occurs, then none of these ions are present. If the solution contains one of these ions, then the solution that remains is decanted.

Ion Group 2: Insoluble sulfides: The remaining mixture is acidic, so we are specifically looking for acid insoluble sulfides. This will ppt out several ions if one is present:

Cu+2, Cd+2, Hg+2, As+3, Sb+3, Bi+3, Sn+2, Sn+4, Pb+2Again, if no ppt forms, then none of these ions are present. If a ppt forms then the remaining solution is decanted.

Ion Group 3: Base insoluble sulfides and hydroxides: The remaining solution is made basic using ammonia and now basic insoluble sulfides would ppt out: Zn+2, Mn+2, Ni+2, Fe+2, Co+2. Then additional hydroxide is added which would ppt out Al+3 and Cr+3. If no ppt occurs then none of these ions are present. If a ppt is formed, then the solution is decanted.Ion group 4: Insoluble phosphates: To the basic solution phosphate ions are added which would ppt out Mg+2, Ba+2, or Ca+2 if present. If no ppt forms, then these ions are not present. If a ppt forms, then the solution is decanted.

What remains now, at the very end, is a solution that contains the following possible ions: Na+1, K+1, NH4+1. These species are all always soluble. Since these ions may have been added as slat throughout the qual scheme, you will want to test your ORIGINAL solution for these species. Identifying them might appear difficult but it is actually quite simple: flame test. Sodium will test orange, potassium will test purple and ammonia will not test a color. The ammonium ion can be tested for however using additional base, which will converts NH4+1 to NH3 which can be tested for using litmus paper.

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