Chemistry 30

Unit 5:Solutions and

SolubilityTEACHER

Components of a Solution A solution is composed of two or more pure substances mixed together, one of which is a

solvent and the other a solute.

A cup of instant coffee is the solution, the hot water is the solvent, and the instant coffee is the solute.

A saturated solution is a solution that contains the maximum amount of solute dissolved in a solvent. A supersaturated solution is a solution where conditions have been changed to allow more solute to dissolve than would at room temperature (this is done by heating the solvent or solution).

When the solvent and solutes are liquids we can use the terms miscible and immiscible. Miscible means that the two liquids will mix together, and immiscible means that the two liquids will not dissolve in one another.

Remember, the solvent and solute can be solid, liquid OR gas. There are special names for solutions where the solute and solvent are both solids and when the solute is a liquid, and the solvent is a solid.

o Solid-in-solid – alloys (e.g., brass = zinc + copper, sterling silver = silver + copper).

o Liquid-in-solid – amalgams (e.g., older dental fillings = mercury + silver).

Factors that may affect solubility are temperature and pressure:

o Temperature: Gases: Temp , solubility ⬆ ⬇

Solids and liquids: Temp , solubility ⬆ ⬆

o Pressure: Gases: Pressure , solubility ⬆ ⬆

Solids and liquids: unaffected

Like dissolves like. That is polar covalent molecules usually dissolve in other polar covalent molecules and non-polar covalent molecules dissolve in non-polar covalent molecules.

RECALL: Polarity can be determined by subtracting electronegativities. A polar molecule is one where one (or more) atom in the molecule has a stronger pull on the electrons than another atom, so the electrons move closer to the atom with the stronger pull (higher electronegativity). This creates a dipole, where the atom with the stronger pull has a slightly negative charge and the atom with the lower electronegativity (less pull) has a slightly positive charge.

Water as a Solvent: Solutions in which water is the

solvent are called aqueous solutions.

Most covalent liquids are only able to dissolve other covalent compounds; water can dissolve both ionic and covalent compounds.

The attraction of water dipoles for ions pulls ions out of a crystalline lattice and into aqueous solution.

Aqueous Solutions:

Sometimes when we mix two dissolved salts together, we end up with a precipitate. This is because a new ionic compound is formed that is insoluble in water.

We can use solubility charts or solubility tables to determine if a solid will be formed.

Ion Solubility

Exceptions

NO3– soluble none

ClO4– soluble none

Cl– soluble  except Ag+, Hg22+, *Pb2+

I– soluble except Ag+, Hg22+, *Pb2+

Br- soluble except Ag+, Hg22+, *Pb2+

SO42- soluble except Ca2+, Ba2+, Sr2+, Hg2+, Pb2+, Ag+

CO32- insoluble except Group IA and NH4

+

PO43- insoluble except Group IA and NH4

+

OH– insoluble except Group IA, *Ca2+, Ba2+, Sr2+

S2- insoluble except Group IA, IIA and NH4+

Na+ soluble noneK+ soluble noneNH4

+ soluble none

* = slightly soluble

Ex. If we mix sodium ions with hydroxide ions will a precipitate form?

No

Ex. If we dissolved sodium hydroxide in calcium chloride, will a precipitate form?

2NaOH + CaCl2 2NaCl + Ca(OH)2

NaCl is soluble, Ca(OH)2 is not soluble so a precipitate will form.

Ex. Calcium nitrate with sodium carbonate

Ca(NO3)2 + Na2CO3 CaCO3 + 2NaNO3

Calcium Carbonate is insoluble so a precipitate will form, sodium nitrate is soluble

See Components of Solutions Assign

Concentration Concentration refers to the amount of solute dissolved in a specific amount of solvent.

Concentrated and dilute are qualitative terms we use to describe concentration. Dilute means there is not a lot of solute in the solution and concentrated means there is a lot of solute in the solution.

Common concentrations are molarity, molality, ppm, and ppb.

MolarityThis is the number of moles of solute dissolved in one litre of solution.

The formula for molarity is as follows:

c =

n = # moles of solute

V = volume of solution in litres

c = concentration in moles per litre (M)

Ex. If a teaspoon (5.0mL) of a 0.50 M solution of NaCl was evaporated, how many moles of sodium chloride would be left? What mass of NaCl would be left?

c= nVn=cV

n=(0.50M ) (0.0050 L ) = 0.0025 mol

convert ¿mass :0.0025mol x 58.4428g1mol

=0.15g

Ex. Antifreeze is a solution of ethylene glycol, C2H6O2, in water. If 4.50 L of antifreeze contains 2.00 kg of ethylene glycol, what is the concentration of the solution?

Moles of C2H6O2 = 2.00 × 103 g C2H6O2 × = 32.2 mol C2H6O2

c= = = 7.16 mol/L

Standard solution: A standard solution is a solution of a known concentration. This means it has a precise mass of

solute in a specific volume of solution.

Standard solutions are used in experiments where the concentration of a solution must be known.

We use volumetric flasks to prepare standard solutions as they have a small margin of error when compared to other pieces of lab equipment. Volumetric flasks come in a variety of sizes (volumes).

To prepare a standard solution we follow the following steps:

1. Calculate the required mass of solute needed using the volume and concentration you want to end up with.

2. Weigh out the mass of the solute needed and add it to a volumetric flask of the appropriate size.

3. Dissolve the solid in pure water using less than half of the final solution volume.

4. Once the solute is dissolved, add the rest of the water. Be sure to use a medicine dropper for the final few milliliters of water. Use the calibration line to set the meniscus in the appropriate spot.

Ex. Describe the preparation of 2.000 L of a standard aqueous solution containing 0.1000 mol/L potassium nitrate.

n = Vc

= 2.000 L × 0.1000 mol/L = 0.2000 mol

mass of KNO3 = 0.2000 mol KNO3 × = 20.22 g KNO3

To prepare the solution we place 20.22 g of KNO3 in a 2.000 L volumetric flask. About half the required amount of water is added. When all the potassium nitrate is dissolved, the solution is diluted with the remaining water.

Dilution Calculations: When making a solution in chemistry laboratories you usually only have access to substances

with high concentration solutions (called stock solutions) and are then required to dilute the stock solutions. A calculation needs to be completed in order to determine the amount of distilled water that needs to be added to a certain volume of stock solution in order to create the desired concentration.

Since the number of moles of solute in a solution does not change when you dilute it, the equation for dilution is as follows:

V1 C1 = V2 C2

before after

Ex. Water is added to 200.mL of 2.40M ammonia cleaning solution (NH3), until the final volume is 1.00L. Find the molar concentration of the final diluted solution.

Vi Ci = Vf Cf

(0.200L)(2.40M) = (1.00L)(Cf)

Cf = 0.480M

Ex . What volume of concentrated sulphuric acid (containing 18.0 M H2SO4) is required to prepare 5.00 L of 0.150 M aqueous sulphuric acid solution by dilution with water?

V1 = ? V2 = 5.00 L

c1 = 18.0 mol/L c2 = 0.150 mol/L

V1 =

= = 0.0417 L = 41.7 mL

Ion Concentration:Consider a 0.20M aqueous solution of sodium carbonate. The sodium carbonate will be completely dissociated into ions:

Na2CO3(s) 2 Na+(aq) + CO3

2-(aq)

The concentration of the sodium ions can be calculated using the conversion factors from the balanced equation (similar to mol- mol stoichiometry). Remember: concentration is represented by using square brackets.

[Na+] = 0.20 mol/L Na2CO3 × = 0.40 mol/L [Na+]

[CO32-] = 0.20 mol/L Na2CO3 × = 0.20 mol/L [CO3

2-]

Ex. What are the concentrations of the ions in an aqueous solution containing 0.15 mol/L iron(III) nitrate?

Fe(NO3)3(s) Fe3+(aq) + 3 NO3

−(aq)

[Fe3+] = 0.15 mol/L Fe(NO3)3 × = 0.15 mol/L Fe3+

[NO3−] = 0.15 mol/L Fe(NO3)3 × = 0.45 mol/L NO3

−

Ex. 250mL of 0.30M K2SO4 and 250mL of 0.80M MgCl2 are mixed and no reaction results. What is the concentration of each substance in the final solution, and the concentration of each individual ion?

[K2SO4]: C f=C iV iV f

C f=(0.30M )(250mL)

(500mL)

Cf = 0.15M

[MgCl2]: C f=C iV iV f

C f=(0.80M)(250mL)

(500mL)

Cf = 0.40M

[K+]=0.30M [SO42-]=0.15M [Mg2+]=0.40M [Cl-]= 0.80M

Ppm & ppb: Usually in chemistry we work with molarity when discussing concentration. When working with

very dilute solutions (solutions with low concentrations) it is more convenient to work with ppm or ppb.

A “part” refers to a gram.

A concentration of 1ppb is equivalent to 1g of solute dissolved in 1,000,000,000g of solvent.

The allowable concentration of toxic chemicals is usually measured in ppm or ppb.

For example the allowable concentration of mercury in air is 82ppm, this means there are 82 g Hg

1billion gramsof air . The allowable concentration of copper in drinking water is 1ppm, this

means 1gCu

1million gramsof water .

To do ppm or ppb calculations we assume that 1g of water has a volume of 1mL.

Ex. Change 2.00M KNO3 to ppm.

2.00M=2.0mol KNO31Lsolution

change to grams KNO3

1million gramsof solution

Convert 2.00mol KNO3 to mass: 2.0mol KNO3×101.10332g KNO31mol KNO3

= 202g KNO3

1L solution = 1000mL = 1000g

202 gKNO3

1000g solution× 10001000

= 20200 g1000000g

=20200 ppm

Ex. The concentration of gold in the ocean is 0.005ppb. What would be the molarity of gold in a sample of ocean water.

0.005 ppb= 0.005 g Au÷10000001000000000g solution÷1000000

=0.000000005g Au1000 gsolution

1000g solution = 1000mL solution = 1 L solution

Determine moles Au: 0.000000005 g Au×1mol Au

196.9665694g Au=3×10−11mol Au

[Au]=3×10−11mol Au

1 Lsolution= 3 x 10-11M

See Concentration and Dilution Assignment

The Solubility Product Constant: All salts have different solubilities. Even ‘insoluble’ salts will dissolve in water to some extent.

When a salt is dissolved in water, some of the salt dissolves and some may not. The amount of salt that dissociates into ions is different for every salt.

Ksp is a number that represents how much of the salt will dissociate into ions.

Ksp is equal to the product of the concentration terms each raised to the power of the coefficients of the substance in the dissociation equation.

Ksp values @ 25oC can be found on the table on the next page.

The equilibrium constant Ksp called the solubility constant indicates how soluble a substance is.

o If the Ksp is low, the solid is not very soluble.

o If the Ksp is high, the solid is soluble.

An expression for the solubility product constant, Ksp, of calcium phosphate would be as follows:

Balanced Equation: Ca3(PO4)2(s) ⇌ 3 Ca2+(aq) + 2 PO4

3-(aq)

Ksp is equal to the product of the concentrations of the products, each raised to the power of the coefficient in the balanced equation, over the concentration of the reactant to the power of the coefficient.

When using Ksp we do not include the concentrations of solids, liquids or gases, as they are not dissolved. Therefore, the concentration of the reactant should not be included, as it is a solid.

Ksp = [Ca2+]3[PO43-]2

Example: Calculate the solubility product constant for calcium fluoride at 20oC given that it`s solubility is 1.6 × 10−2 g/L.

First write the chemical equation:

CaF2 Ca2+ + 2F-

Then convert to molarity (the number of moles of CaF2 that dissolved in 1L):

mm CaF2 = 78.08g/mol

1.6×10−2g1 L

× 1mol78.1g

=2.0×10−4M = x

Find Ksp using solubility constant expression:

Ksp = [Ca2+][F-]2 = (x) (2x)2 = (2.0 x 10-4 )(4.0 x 10-4 )2 = 3.2 x 10-11

Solubility Product Constants near 25 oC

Formula Ksp

Al(OH)3 1.3×10–33

AlPO4 6.3×10–19

BaCO3 5.1×10–9

BaCrO4 1.2×10–10

BaF2 1.0×10–6

Ba(OH)2 5×10–3

BaSO4 1.1×10–10

BaSO3 8×10–7

BaS2O3 1.6×10–6

BiOCl 1.8×10–31

BiOOH 4×10–10

CdCO3 5.2×10–12

Cd(OH)2 2.5×10–14

CdC2O4 1.5×10–8

CdS 8×10–28

CaCO3 2.8×10–9

CaCrO4 7.1×10–4

CaF2 5.3×10–9

CaHPO4 1×10–7

Ca(OH)2 5.5×10–6

CaC2O4 2.7×10–9

Ca3(PO4)2 2.0×10–29

CaSO4 9.1×10–6

CaSO3 6.8×10–8

Cr(OH)2 2×10–16

Cr(OH)3 6.3×10–31

CoCO3 1.4×10–13

Co(OH)2 1.6×10–15

Co(OH)3 1.6×10–44

CoS 4×10–21

CuCl 1.2×10–6

CuCN 3.2×10–20

CuI 1.1×10–12

Cu3(AsO4)2 7.6×10–36

CuCO3 1.4×10–10

CuCrO4 3.6×10–6

Cu[Fe(CN)6] 1.3×10–16

Formula Ksp

Cu(OH)2 2.2×10–20

CuS 6×10–37

FeCO3 3.2×10–11

Fe(OH)2 8.0×10–16

FeS 6×10–19

FeAsO4 5.7×10–21

Fe4[Fe(CN)6]3 3.3×10–41

Fe(OH)3 4×10–38

FePO4 1.3×10–22

Pb3(AsO4)2 4×10–36

Pb(N3)2 2.5×10–9

PbBr2 4.0×10–5

PbCO3 7.4×10–14

PbCl2 1.6×10–5

PbCrO4 2.8×10–13

PbF2 2.7×10–8

Pb(OH)2 1.2×10–15

PbI2 7.1×10–9

PbSO4 1.6×10–8

PbS 3×10–28

Li2CO3 2.5×10–2

LiF 3.8×10–3

Li3PO4 3.2×10–9

MgNH4PO4 2.5×10–13

Mg3(AsO4)2 2×10–20

MgCO3 3.5×10–8

MgF2 3.7×10–8

Mg(OH)2 1.8×10–11

MgC2O4 8.5×10–5

Mg3(PO4)2 1×10–25

MnCO3 1.8×10–11

Mn(OH)2 1.9×10–13

MnS 3×10–14

Hg2Br2 5.6×10–23

Hg2Cl2 1.3×10–18

Hg2I2 4.5×10–29

HgS 2×10–53

Formula Ksp

NiCO3 6.6×10–9

Ni(OH)2 2.0×10–15

NiS 3×10–19

ScF3 4.2×10–18

Sc(OH)3 8.0×10–31

AgC2H3O2 2.0×10–3

Ag3AsO4 1.0×10–22

AgN3 2.8×10–9

AgBr 5.0×10–13

AgCl 1.8×10–10

Ag2CrO4 1.1×10–12

AgCN 1.2×10–16

AgIO3 3.0×10–8

AgI 8.5×10–17

AgNO2 6.0×10–4

Ag2SO4 1.4×10–5

Ag2S 6×10–51

Ag2SO3 1.5×10–14

AgSCN 1.0×10–12

SrCO3 1.1×10–10

SrCrO4 2.2×10–5

SrF2 2.5×10–9

SrSO4 3.2×10–7

TlBr 3.4×10–6

TlCl 1.7×10–4

TlI 6.5×10–8

Tl(OH)3 6.3×10–46

Sn(OH)2 1.4×10–28

SnS 1×10–26

ZnCO3 1.4×10–11

Zn(OH)2 1.2×10–17

ZnC2O4 2.7×10–8

Zn3(PO4)2 9.0×10–33

ZnS 2×10–25

* Sulfide equilibrium of the type:MS(s)  +  H2O(l)  ⇌  M2+(aq)  +  HS–(aq)  +  OH–(aq)

Determining Individual Ion Concentration from Ksp:

Ex. Calculate the concentrations of barium ions and sulphate ions in a saturated aqueous solution of barium sulphate at 25oC.

BaSO4(s) ⇌ Ba2+(aq) + SO4

2-(aq)

Ksp = [Ba2+][ SO42-] = 1.1 × 10-10 from table

Since [Ba2+] = [SO42-],

then Ksp = [Ba2+][ SO42-] can be written as Ksp = x2

where x = [Ba2+] = [SO42-]

Ksp = 1.1 × 10-10 = x2

x = 1.0 × 10-5 mol/L

[Ba2+] = [SO42-] = 1.0 × 10-5 mol/L

Ex. When 10.0 L of a saturated solution of magnesium carbonate, at 22°C is evaporated to dryness, 0.12g of solid magnesium carbonate is obtained. Calculate the Ksp of magnesium carbonate.

MgCO3(s) ⇌ Mg2+(aq) + CO3

2-(aq)

Ksp = [Mg2+][ CO32-]

Calculate the concentration of MgCO3:

Moles of MgCO3 = 0.12 g MgCO3 ×

= 1.4 × 10-3 mol MgCO3

c = [MgCO3] = = 1.4 × 10-4 mol/L

if [MgCO3] = 1.4 × 10-4 mol/L, then [Mg2+] and [ CO32-] each = 1.4 × 10-4 mol/L

Ksp = (1.4 × 10-4 )(1.4 × 10-4 )

Ksp = (1.4 × 10-4)2 = 2.0 × 10-8

Ex. A saturated solution of ionic calcium hydroxide has a hydroxide ion concentration of 3.0×10 -3mol/L. Calculate the Ksp of calcium hydroxide at the same temperature.

Ca(OH)2(s) ⇌Ca2+(aq) + 2 OH−

(aq)

Calculate [Ca2+] : 3.0 × 10-3 mol/L OH- × = 1.5 × 10-3 mol/L Ca2+

Ksp = [Ca2+][ OH-]2 = (1.5 × 10-3)( 3.0 × 10-3)2 = 1.4 × 10-8

See Solubility Product Constant Assignment

The Common Ion Effect: Remember that when a salt dissolves in water, some of it dissolves and some may stay as a solid.

There is a point at which the salt dissolving reaches equilibrium. This means that for each formula unit of salt that dissolves, a pair of ions come together to create a formula unit.

When we dissolve ions in water we know that the product of the concentrations of the ions, to the power of their coefficients in a balanced equation, is equal to the Ksp.

As long as the product of the concentration of the ions does not exceed the Ksp value, no precipitate forms.

Let’s look at the following example:

Copper (I) iodide is not very soluble in water (Ksp = 4 × 10−19): CuI(s) ⇌Cu+(aq) + I−

(aq)

Since Ksp = [Cu+][I-], if the product of the concentrations of Cu+ and I- stays below 4 × 10−19, no precipitate will form, however, if we add extra iodine or copper ions (the common ion) things will change. Let`s say we added some iodide ions; this will increase the concentration of iodide and the product of the copper ions and iodide ions would be higher than the Ksp value. This is not possible so the concentrations would need to be lowered. In order to lower the concentration of iodide ions, some copper (I) iodide would need to precipitate out. This follows Le Chatelier`s Principle because we increased the concentration of the products (iodide ions) so the reaction shifts to the reactants.

After adding iodide ions to the solution, copper (I) iodide will precipitate out, but once the reaction reaches equilibrium again, the concentration of copper ions and iodide will no longer be equal (there will be more iodide ions in solution than copper ions). This also means the concentration of Cu+ will be lower than at the original equilibrium and the concentration of I-

will be higher than at the original equilibrium.

Overview: If a slightly soluble ionic compound is dissolved in water, you can force precipitation of that salt by adding a readily soluble ionic compound that has an ion in common with the slightly soluble salt. This shift is known as the common ion effect.

Ex. The Ksp of silver iodide is . What is the iodide ion concentration of a 1.00 L saturated solution of AgI to which 0.020 mol of AgNO3 is added?

AgI(aq) Ag+(aq) + I-(aq)⇌ Ksp =

If the equilibrium concentration of iodide ion from the dissociation is x, then the equilibrium concentration of silver ion is x + 0.020. Because of the small value of Ksp, x will be small compared to 0.020. Therefore, the [Ag+] at equilibrium equals x + 0.020 ~ 0.020.

Ksp =

(0.020)(x) =

x = M = [I−]

Selective Precipitation Selective precipitation is when you separate ions from two solutes to create a precipitate.

To calculate selective precipitation we use initial concentrations (Q), instead of the equilibrium concentrations.

In selective precipitation there are restraints in forming a precipitate; a precipitate doesn't always form in a solution. It depends on the relationship between K and Q:

o If Q>K  a precipitate forms and the reaction proceeds to the left

o If Q<K  no precipitate forms and the reaction proceeds to the right

o If Q=K  the reaction is at equilibrium

Example: If 40L of 0.0050M KCl is mixed with 60L of 0.0030M Pb(NO3)2. Will the precipitate PbCl2 (Ksp= 1.6 X 10-5) form?

First we calculate the initial concentrations of Pb2+ and Cl-:

[Pb2+]0= (40L)(.005M)/(40L+ 60L)= .002M

[Cl-]0= (60L)(.003M)/(40L+60L) = .0018M

Calculate value of Q:

Q=[Pb2+][Cl-]2

Q=[.002M][.0018M]2 

Q= 6.48 X 10-9 M

Q<K no ppt

Example: If 100L of 0.03M Pb(NO3)2 is mixed with 200L of 0.09M KCl. Does a precipitate (PbCl2) form? (Ksp = 1.6 X 10-5). What are the equilibrium concentrations of Pb2+  and Cl-?

Pb2+     +    2Cl-      ----->     PbCl2

[Pb2+]=(100L)(.03M)/(100L+200L) = .01M

[Cl-] = (200L)(.09M)/(100+200L) = .06M

Q=[.01M][.06M]2 ------->    Q=3.6 X 10-5

Because Q>K a precipitate forms.

By using selective precipitation, we can calculate the amount of a reagent necessary to react with the salt.

Example: If a solution contains  1.0 x 10-5 M  of  Pb2+  and 2.0 x 10-5 M  Ag+ and Cl-  is slowly added to the  solution, will AgCl (Ksp=1.6 x 10-10)  or PbCl2 (Ksp=1.6 x 10-5) precipitate first? Which will be in excess?

First we plug the given values into the Ksp expression for PbCl2 and solve for the concentration of Cl-.1.6 x 10-5 = Ksp =[Pb2+][Cl-]2 = [1.0X10-5 M]  [Cl-]2

[Cl-]2= 1.6 M   

[Cl-] = 1.3 M

Next we plug in the given values to the Ksp expression for AgCl.

Ksp=1.6 X 10-10 = [Ag+][Cl-]

Ksp = 1.6 X 10-10  = [2.0 X 10-5 ] [Cl-]

[Cl-]= 8 X 10-6  M

AgCl will precipitate first because Cl- concentration @ equilibrium is smaller than PbCl2.

See Common Ion and Selective Precipitation Assignment

Lab: Analyzing for Anions

PURPOSE

In this experiment you will observe and record some characteristic chemical reactions of several negatively charged ions: SO4

2-, CO32-, and I- using selected reagents. From a study of these observations

you will be able to develop a method for identifying each of these anions in the presence of the other ions.

Your work will be typical of the work done by chemists who have developed elaborate schemes which can be used to identify dozens of cations and anions in a given sample. Organize a data table in which you can record your observations for the results obtained by adding each of the reagents to each of the solutions of anions to be studied.

MATERIALS

Anion obtained

Na2SO4, sodium sulfate, 0.1 M SO42-

Na2CO3, sodium carbonate, 0.1 M CO32-

NaCI, sodium chloride, 0.1 M CI-

Nal, sodium iodide, 0.1 M I-

Cation obtained

Ba(NO3)2 , barium nitrate, 0.1 M Ba2+

AgNO3, silver nitrate, 0.1 M Ag+

HNO3, nitric acid, 1.0 M H+

PROCEDURE

a. The tests can easily be made by putting a few drops of a solution on a plastic film and adding to this a few drops of one of the reagents. DO NOT use the tip of a dropper from a stock bottle for stirring. Doing so could contaminate the solution contained in the bottle. Use a thin glass rod.

b. Test 2-3 drops of each anion solution separately with 1-2 drops of the 0.1 M Ba(NO3)2 reagent. Record the results observed. To each mixture that contains a precipitate add 2-3 drops of 1.0 M HNO3 and record any changes you observe.

c. Test 2-3 drops of each anion solution separately with 1-2 drops of the 0.1 M AgNO3 reagent. Record the results observed. To each mixture that contains a precipitate add 2—3 drops of 1.0 M HNO3 and record any changes you observe.

d. Prepare new samples of the silver precipitates which formed in step c. Add 2—3 drops of 6 M aqueous ammonia, NH3(aq), to the precipitates and record any changes you observe.

e. Test 2—3 drops of each anion solution separately with 1—2 drops of 1.0 M HNO3 and record the results.

f. Examine the pattern of reactions shown by your table of observations. Develop a scheme by which you could identify either one or a combination of two of the anions in a single solution. Organize your scheme in such a way that another student could follow it. You may wish to try it out by making up your own "trial unknowns" containing various combinations of the anions.

g. Obtain an unknown from your teacher and analyze it for the four anions. Report your results. Include supporting evidence for your conclusions.

ANALYSIS

1) Write net ionic equations for the precipitation reactions which occurred when solutions of the anions were mixed with the reagent containing Ba2+ ions.

2) Do the same for the precipitation reactions which occurred in step c with the Ag+ ions.

3) Write equations for the reactions which occurred when the precipitates were acidified with 1.0 M HNO3 .

4) Write equations for the reactions which occurred when the silver precipitate reacted with aqueous ammonia to form the complex ion, Ag(NH3)2+.