Designer

This molar product of Ag+ and Cl - at equilibrium is called solubility product . In expressing the solubility product of the ions involved , the stoichiometric coefficients are raised .

Ag2S 2Ag+ + S2- , Ksp = [Ag+]2[S2-]

Table 2.2 , shows the solubility product (Ksp) of certain slightly soluble salts at 25oC .

It must be understood that the smaller the value , the less soluble salt in water . We can determine the Ksp from the molar concentration at equilibrium and vice versa . Ksp is comparatively easy to calculate with respect to other equilibrium calculations .

Example 1.

An aqueous saturated solution of silver carbonate was analysed and found to contain 8 mg of the salt dissolved in 250 ml. Calculate the Ksp of the Ag2CO3 .

Solution:

Ag2CO3 2 Ag+ + CO32- ,

it shows that in every one mole of silver chloride dissolved two moles of silver ions and one mole of carbonate ion will be produced.

Ksp = [2Ag+]2[CO32-] , the molecular weight of Ag2CO3 = 275 . 8

Amount of mg of Ag2CO3 = = 32 mg /l

Molar solution of Ag2CO3 = = 1.16 × 10 -4 M

Ksp = [2Ag+]2[CO32-] = [2 X 1.16 x 10 – 4]2 [1.16 X 10 – 4] = 6.4X 10 – 12

We said that the reverse is true , that is we can calculate the molar concentration from the solubility product , Ksp .

Example 2. Calculate the molar solubility of silver sulphate solution whose solubility product at 25 0C is 1.4 X 10 -5 .

Ag2SO4 2 Ag + + SO42-

Ksp = [2Ag +] 2[SO42-] = 1.4 X 10 -5

Let us assume that x moles per litter dissolved in the solution , so [Ag +] = 2 x, [SO42] = x So substituting these values into the equation , Ksp = [2Ag +] 2[SO42-] = 1.4 x 10 -5

(2x)2(x) = 1.4 X 10 -5 and 4 X3 = 1.4 X 10 -5 , X = = 1.5 X 10-2

[SO4-2] = [X] = 1.5 X 10-2 , [Ag+] = [2X] = 3.0 X 10-2

Determination of end-point in precipitation titration

There are different methods to determine the end-points of the precipitation titrations. One of these methods is the application of chemical indicator which can be presented in the following ways:

a. Appearance of colored precipitate; good example of this end – point detection is the application of Mohr's method of titration . In this method , the indicator used is dilute solution of potassium chromate (K2CrO4) . Silver chloride is less soluble than potassium chromate and for the silver chloride to precipitate first in the process , its initial concentration must be high .

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

[2Ag+]2[CrO42-] = 1.7 X 10-12

Based on their solubility product , more dilute solution of K2CrO4 must be used to give red sharp end-point . It must be noted that concentrations of chromate ion as large as 0.006 M masks the red color of the Ag2CrO4 due to persistence of the color of K2CrO4 .

The pH of the solution in which this reaction takes place must be considered. In more acidic solution , dichromate is formed which is more soluble than chromate and may need large silver concentration .

2CrO42- + 2H+ Cr2O72-+ H2O

In strong alkaline solution , the silver may precipitate as silver hydroxide which changes into silver oxide .

2Ag+ + 2OH - 2AgOH(s) Ag2O(s) + H2O

b. Formation of soluble colored complex .

In Volhard method, standard solution of thiocyanate may be used for silver ion titration .

Ag+ + SCN - AgSCN

The indicator for this titration is Fe3+ solution , so that the last excess drop turns the solution red due to complex formation ,

Fe 3+ + SCN - Fe(SCN)2+

To avoid the precipitation of Fe3+ hydroxide or hydrated oxide , the titration must be carried out in acidic solution .

c. Adsorption indicators.

Fajan's method of chloride titration utilizes specific indicators called adsorption indicators . An adsorption indicator is an organic substance that has the property of being adsorbed on or desorbed from the surface of the precipitate . The adsorption and desorption takes places theoretically near the equivalence point accompanied with color change. Titrations that employ adsorption indicators are generally accurate , fast and reliable

Fluorescein is typical organic dye used for the titration of chloride ions with silver nitrate .

yellow red Na-form

derivatives of fluorescein like 2,7- Dichlorofluorescein is preferably used in Fajan's method of chloride determination .

2.7- Dichlorofluorescein

Experiment (2-1) Chloride determination by Fajans method.

Reagents:

A. 0.1N AgNO3 (standardized) 17.0 g dissolved in H2O , dilute to liter .

B. Dichlorofluorescein solution ( 0.1% ethanol solution ) .

C. Dextrin (D-glucose polymer) .

Procedure: Prepare a suitable chloride solution from an analytical reagent salt in one liter volumetric flask . Take 50 ml aliquot of the solution into 250 ml conical flask ; add five drops of the indicator and 0.1 g of dextrin .

Titrate against the standardized AgNO3 solution from the burette until permanent pink color is obtained . Repeat the titration three times and report the chloride content in mg/l .

Experiment (2-2): Determination of chloride by Mohr method

The most significant idea in this method is the application of K2CrO4 solution as an indicator to detect the end point . Even though both AgCl and Ag2CrO4 are insoluble salts , the solubility product of Ag2CrO4 is far more greater than that of AgCl ; thus AgCl precipitates first during the titration process and first silver ions after the total precipitation of of chloride ion react with the chromate ions forming brick-red color as an end point .

Reagents :

A. Standard AgNO3 solution (0.1 N)

B. 0.25 M K2CrO4 solution

Procedure: Take 50 ml aliquot of the prepared solution in the preceding experiment (unknown) into 250 ml conical flask ; add 2 ml of the K2CrO4 solution. Start the titration by adding AgNO3 solution from the burette to the chloride solution until the permanent brick-red color appears . repeat the process three times and report as before .

Experiment (2-3): Determination of Ag in a solution (Volhard’s method)

In this method , the amount of silver in a sample is determined by titrating against standardized KSCN solution . Ferric alum (ammonium iron(III) sulfate ) is used to indicate the end point . In this titration :

Ag+ + SCN - AgSCN

And Fe3+ + SCN - Fe(SCN)2+ (reddish-brown)

Reagents:

a. 0.1 N KSCN solution

b. Ferric alum , saturation solution containing 6M HNO3 free of nitrogen oxides.

c. Solution containing Ag+ (unknown) .

Transfer 50 ml aliquot of the sample solution into 250 ml conical flask . Add 2 ml of the ferric alum solution . Start the titration by adding KSCN solution from the burette with constant vigorous agitation until reddish brown color is formed . Report the result as before and calculate the of Ag .

3. Complexometric titrations – EDTA titrations

The most important complexing agent in terms of stoichiometric titrations is the EDTA (Ethylenediaminetetraacetic acid ) . This organic complexing agent has more than two chelating sites . Titrations involving these chelating agents are called Complexometric titrations. EDTA forms stable complexing compound with many metal cations at different pH values . Calcium and magnesium ions are some of the metal cations that are relatively stable EDTA complex ; their formation constants are too close together so that they can be titrated together using Eriochrome black T (EBT) as an end point indicator , so this titration can be used to determine the hardness of water .

EBT forms too weak complex with calcium giving no sharp end point , therefore small known amount of Mg2+ is added to the Ca2+ solution which can be corrected for it at the end of the titration by using blank titration with the same amount of Mg2+.

We have already explained in the theoretical section that different metal ions react with electron pair donor and form either a coordination compound or complex ions . This reaction generally involves the replacement of the solvent molecules coordinated with the metal ion with other donors . These replacing donors are called ligands and must have at least one pair of unshared electrons ; they can be a charged ion or neutral molecule ; they can also be organic or inorganic species and the complex formed can be ion or neutral like , Cl -, NH3 , H2O , NH2-CH2-COO- etc.

In complex formation , the reaction occurs in steps and step-wise equilibria is formed . Let us take the formation of the complex ML n .

M + L1 ML1 K1 =

M + 2L ML2 K2 =

M + nL MLn Kn =

Each step has its own step-wise stability constant ; the overall stability constant K is related to the wise constants .

K = K1 . k2 . Kn

This relationship is valid only if there no insoluble product formed. One of the most widely used aminopolycarboxylic acid is the Ethylenediaminetetraacetic acid , EDTA. For simplification , we can take H4Y instead of EDTA and Na2H2Y instead of the disodium form . The substance is weak organic acid which undergoes four stability constants .

H4Y H3Y - + H+ K1 = 1.02 X 10 -2

H3Y 3- H2Y 2- + H + K2 = 2.14 X 10 -3

H2Y 2- HY 3- + H + K3 = 6.92 X 10 -7

HY 3- Y 4- + H + K4 = 5.5 X 10 -11

The K values show the ease of the hydrogen ion to be released .The first two values are much greater than the others. The sodium salt is generally used in titrations for determining different metal cations . The important feature of the EDTA titrations is that it combines with all metal ions in ratio of 1:1 and forms stable chelates suitable for reproducible volumetric analysis .

· Types of EDTA titrations:

1. Direct titrations: In this method direction titration is made between the metal ion and standard solution of EDTA . The metal ion solution is usually buffered with NH4+/NH4Cl at pH – 10 .

In certain cases it is very important to prevent any metal hydroxide precipitation by using additional complexing agents like, citrate , tartrate or other organic chelating agent like , 2,2',2"- Nitrilotriethanol (Triethanolamine). The equivalence point is accompanied with change in color of the metal indicator . The end-point can be determined instrumentally by conductometric , amperometric or spectrophotometric methods .

2. Back titration methods: Sometimes it is not possible or difficult to determine the metal by direct titration with EDTA . There are several reasons for this problem ; sometimes the pH necessary for the titration may precipitate the metal , other times undesirable complex may be formed or no suitable indicator is available . In these situations back titration method is applied in which excess of EDTA solution is added to the metal solution and the pH of the resulting solution is adjusted to the desired value , then the excess of the EDTA is titrated with standardized metal ion solution.

To give an example , the reaction between Al3+ and EDTA is very slow , so direct titration is very difficult . The problem is overcome by adding excess amount of EDTA to the Al3+ solution and let the mixture stand still for about 35 minutes . Then the excess EDTA is titrated with standard Zn2+ solution . The Al3+ is calculated from the amount of EDTA consumed.

3. Displacement titration: In this system, metal ions that do not form stable complex (or do not react) with the metal indicator has to be substituted for metal ions which form more stable EDTA complex . Usually this happens in the case of magnesium or zinc complex .

MgY2- + M MY2- + Mg2+

The displaced magnesium is back titrated with standard EDTA.

Metal ion indicators (EDTA Titrations)

The result of EDTA titration depends on precisely locating the end-point ; this is usually achieved by utilizing specific indicators called metal ion indicators. These indicators are organic dyes that chelate with the metal ion in a given range of concentrations, forming specific color . The intensity of the color is discernible within that concentration range (10-6 to 10-7) . The most widely used metal indicator is the Eriochrome Black T

Eriochrome Black T

The sulfonic acid functional group ionizes completely in aqueous solution releasing hydrogen ions but the phenolic group partially ionizes .

H2In- + H2O HIn2- + H3+O K1 = 5 x 10 -7

red blue

HIn2- + H2O In3- + H3+O K2 = 2.8 x 10-12

Most of the EBT metal complexes have generally red color and to observe the color change during the process , the pH must be adjusted to 7 and above so that free indicator color (blue} becomes clear. At the end-point:

MIn + HY3- HIn2- + MY2-

The advantage of the EBT is that it complexes with more than twenty different ions, but only certain number of these metal ions form complexes stable enough to detect the end-point.

Experiment (3-1): Determination of water hardness using EDTA.

Reagents:

A. 0.01 M EDTA (3.8 g of Na2EDTA , dried , purified in 1liter).

B. 6 M HCl

C. 6 M NaOH

D. Methyl red indicator

E. Eriochrome black T (100 mg dissolved in a mixture of 15 ml triethanolamine and 5 ml absolute ethanol).

F. Buffer solution pH-10 (570 ml conc. NH3 + 70 g NH4Cl all diluted to 1liter).

Procedure:

Take accurately measured aliquot of water (not less than 100 ml) , add 0.5 ml of 6 M HCl and boil for few minutes to get rid of CO2 . After cooling the sample add few drops of methyl red and neutralize with NaOH . Add 2 ml of the buffer solution followed by 5 drops of EBT or calmagite indicator . Start titrating the sample with the standard EDTA from the burette until the red color changes to blue . Repeat the process three times and report the result as mgs of CaCO3/liter.

Experiment (3-2a) Determination of magnesium by titrating with EDTA .

Reagents:

1. Standard EDTA (0.01M)

2. Buffer pH-10 (570 ml conc. NH3 + 70 g NH4Cl diluted to liter)

3. Eriochrome Black T indicator

Take prepared sample solution of Mg2+ into 250 ml volumetric flask and dilute to the mark . Take 50 ml of the diluted sample solution into 250 ml conical flask add 2 ml of the buffer solution and 5 drops of the indicator . Titrate with 0.01M Na2H2Y until the color changes from red to blue . Repeat the titration two more times . Estimate the Mg in milligrams/ liter or ppm .

Experiment (3-2b): Determination of Ca by Complexometric back titration

Reagents:

1. EDTA (0.01M , standardized)

2. 0.01M MgSO4 (1.3g MgSO4 . 7H2O in 500 ml distilled water)

3. 6M NaOH

4. 6M HCl

5. Methyl red indicator solution

6. Eriochrome Black T indicator solution (EBT)

Take 25 ml of 0.01M Ca2+ solution into 250 ml conical flasks , add 25 ml distilled water , 2ml buffer solution pH-10 , 5 drops of EBT indicator calamite . Run excess of Na2H2Y from the burette and record the volume . Titrate the excess EDTA with 0.01M MgSO4 solution until the color changes from blue to red . Calculate the amount of Ca2+ in mg/liter .

1 mole of EDTA 1 mole Ca2+

4. Redox – Titrations

We have already seen that oxidation – reduction systems deal with reactions in which loss and gain of electrons are involved . In these reactions , titrimetric procedures can be employed to determine quantitatively one of the reducing or oxidizing agents in a given sample . Like acid – base titrations , we employ indicating system to denote the equivalence point during the titration . The indicating technique as we mentioned before, can be a change in color , redox potential or self – indicating system .

In this section we will briefly discuss about common oxidizing agents in oxidation-reduction titrations such as potassium permanganate , potassium dichromate , cerium (IV) sulfate , and iodine despite that there are many others that fall into this category .

A. Potassium permanganate

This substance is strong oxidizing agent and takes very important role in oxidation-reduction titrations . It is applied as an oxidizing agent in many organic and inorganic reactions in both acidic and basic solutions. In acidic solution , sulfuric acid is usually used since it has no effect upon permanganate in dilute solutions unlike the hydrochloric acid which may reduce the chloride ion and decrease the oxidizing power.

2MnO4- + 16H+ + 10Cl - 2Mn2+ + 5Cl2 + 8H2O

The reduction half-reaction of the permanganate in acidic solution can be written as :

MnO4- + 8H+ + 5e Mn2+ + 4H2O

Since the standard potential is 1.51 volts in acidic solution , it is considered as powerful oxidizing agent. Another advantage of the MnO4- is that there is no need for an indicator in colorless solutions . More important feature of the permanganate is that it can be applied in strong alkaline solutions .

MnO4- + 2H2O + 3e MnO2 + 4OH –

Potassium permanganate cannot be taken as primary standard because it is rarely obtained in pure form since there are always traces of manganese dioxide and remains of organic substances from the distilled water . These substances may reduce the permanganate to manganese dioxide which catalyzes the self-decomposition of the permanganate solution on standing for long time .

MnO4- + 2H2O MnO2 + 3O2 + 4OH –

At the same time any traces of Mn2+ affects the stability of the solution and careful steps has to be followed when preparing the MnO4- solution ; the solution must be covered from unnecessary exposure to bright light. Standardization of the potassium permanganate solution is accomplished by titrating with a standard solution like sodium oxalate in acidic solution .

2MnO4 + 5H2C2O4 + 6H+ 2Mn2+ + 10CO2 + 8H2O

The oxalate solution must be heated to about 90 oC .

· Preparation of 0.1N KMnO4.

Weigh out accurately 3.16g of A.R potassium permanganate into one liter volumetric flask and dissolve it in distilled water by shaking well during the dissolution process. Fill it to the mark and store it in a dark or in diffuse light when is not in use .

B. Potassium dichromate (k2Cr2O7) .

Potassium dichromate is not as powerful oxidizing agent as potassium permanganate , yet it has several advantages over the latter . Potassium dichromate can be obtained in a pure form . It is least affected by direct light , heat and water vapor in the atmosphere . Another advantage is that it is used in acidic solution and quickly reduces to green Cr3+ .

Cr2O72- + 14H+ + 6e 2Cr3+ + 7H2O

as a result it is considered as excellent primary standard . Dichromate solution easily oxidizes the organic compounds in comparison to permanganate solution . The appearance of the green color due to formation of Cr3+ makes the end-point almost impossible to be visually ascertained ; so redox indicator is employed to give clear and distinct color change . Suitable indicators for this titration are A. N-phenyl anthranilic acid (C13H11O2 , 0.1% in 5x10-3N NaOH). N-phenyl anthranilic acid

B. another important indicator is sodium diphenylamine sulfonate (C12H10NNaO3S , 0.2% , w/v , H2O). sodium diphenylamine sulfonate

Preparation of 0.1N K2Cr2O7.

Weight accurately about 4.9 g. of finely powdered and well – dried K2Cr2O7 into 1-liter volumetric flask using suitable funnel to avoid any loss of the salt . Dissolve the salt in distilled water and dilute to the mark . This solution can be standardized with other reducing agent like iron(II) salt .

Preparation of 0.1N Iron(II) sulfate.

This solution can be prepared by accurately weighing 27. 8 g. of ( A.R) FeSO4.7H2O or 39.21g. of (A.R) Iron(II) ammonium sulfate (FeSO4.(NH4)2SO4.6H2O) into 1-liter volumetric flask . Dissolve the salt in distilled water and dilute to the mark.

C. Cerium (IV) Sulfate .

Solution of cerium (IV) sulfate in sulfuric acid is powerful oxidizing agent . It is more powerful than the potassium permanganate and can be a substitute for it in most cases . Cerium (IV) can only be used in acidic solution of 5N and more. There are considerable advantages of Ce(IV) sulfate over other oxidizing agents as a reliable primary standard .

Ce(IV) sulfate solution is indefinitely stable and not affected by exposure to direct light or heat unlike the permanganate solution .

Ce(IV) sulfate solution does not oxidize chloride ions , even in higher concentrations contrary to the situation in potassium permanganate .

There are no variable oxidation states when reacting with reducing agents so that the equivalent weight and molar weight are the same .

Ce4+ + 1e Ce3+

These properties besides others makes the Ce(IV) solution preferable substitute for permanganate solution . different Ce(IV) salts are commercially available in high standard of purity making it suitable for preparing excellent primary standard solution .

Preparation of 0.1N Ce(IV) sulfate.

66 g of ammonium ceric sulfate [Ce(SO4)2 . 2(NH4)2SO4. 2H2O] is accurately weighed and transferred into a mixture of 30 ml of conc. H2SO4 and 30 ml of distilled water . Heat gently to dissolve , cool the solution and filter it through fine – porosity sintered glass crucible ; dilute to liter and mix thoroughly . The solution can be standardized with 0.1N sodium oxalate or with ammonium Fe (II) sulfate using Ferroin ( 1,10-phenanthroline Fe (II) sulfate ) as indicator .

D. Iodimetry titration.

Titration analysis relating iodine solutions is called Iodometric titrations . Saturated aqueous solution of iodine is about 1 x 10-2M at room temperature , so to achieve higher concentrations of iodine solution we have to introduce external iodide ions to form soluble tri-iodide complex .

I2(s) + I - I3- K = 7.1 x 10 2

This external iodide ions comes from potassium iodide salt , so in iodimetry titrations we have to employ concentrated iodide solution .

I2 + 2S2O32- 2I - + S4O62- Or

I3- + 2S2O32- 3I - + S4O62-

Iodine is weak oxidizing agent and employed in the titrations of certain strong reducing agents . Iodine solution is not stable solution and needs re-standardization from time to time ; one of the reasons for instability is that iodine crystals are volatile (sublime) so there may be certain loss during the preparation of the solution. Also , iodine slowly attacks the stoppers other than glass . Another reason may be a change in the solution concentration due to possible air oxidation .

4I - + O2 2I2 + 2H2O

The reaction is catalyzed by light , heat and acidic traces . The titration reaction is sufficient to be self-indicating . Aqueous solution of iodine has yellowish brown color and disappearance of this color during the titration can serve as the end-point of the titration. Although the system can serve as self – indicator , starch solution is used to give sharp end – point . Starch gives an intense blue color with I2 even in very low concentrations .

Preparation of starch:

Take about 2g of soluble starch into about one liter of water . Boil the mixture until clear solution is obtained . Cool and store in stoppered bottle .

Preparation of 0.1N iodine solution

Weigh 20 g of potassium iodide (free of iodate) , into 40ml of distilled water in a glass stoppered volumetric flask . Weigh out about 12.7g of iodine (use rough balance to avoid iodine vapor) and transfer it to potassium iodide solution . Stopper the glass and shake well until the iodine is dissolved and fill to the mark .

Experiment (4-1) Determination of Iron(II) by titrating with K2Cr2O7 solution .

Requirements :

A. Titration apparatus

B. 0.1N K2Cr2O7 (A.R)

C. 0.1N Iron(II) sulfate or ammonium Iron(II) sulfate .

D. Sodium diphenylamine sulfonate (1% solution in 0.005N NaOH ).

E. Ortho-phosphoric acid (concentrated).

Titrate 25.0ml of the acidified Iron(II) sulfate solution with standard 0.1N potassium dichromate solution using sodium diphenylamine sulfonate solution (0.5ml) and 2.5ml concentrated phosphoric acid . Continue adding the dichromate solution slowly with constant stirring until the green color changes into a grey-green . Then begin to add the dichromate drop wise until the first blue – violet appears permanently . Repeat the process three times , report the result as usual and calculate the amount of Iron .

1 mol of K2Cr2O7 6mol Fe .

Experiment (4-2). Estimation of hydrogen peroxide by titrating against potassium permanganate.

Permanganate solution reacts quantitatively with acidified hydrogen peroxide solution according to the following equation :

2MnO4- + 5H2O2 + 6H+ 2Mn2+ + 5O2 + 8H2O

Procedure:

Take 25ml of 12% hydrogen peroxide solution into 500ml volumetric flask and dilute it to the mark and mix thoroughly . Transfer 25ml of this solution into the titrating flask , dilute it with 200ml distilled water and 20ml of 20% H2SO4 Titrate with standard 01N KMnO4 until the permanent faint pink appears . Repeat the process until you get within the range of 0.1ml difference . Report the data as usual and estimate the percent of hydrogen peroxide(w/v) in the original solution.

Experiment (4-3): Determination of NO2- by titrating with KMnO4.

Permanganate solution readily reacts with simple nitrite salt in acidic solution according to the following ionic equation :

2MnO4- + 5NO2- + 6H+ 2Mn2++ 5NO3- + 3H2O

Procedure:

Take accurate weight of ordinary potassium nitrite (1-1.05g) , dissolve in a cold water and dilute it to 250ml in a volumetric flask , and shake well transfer 25ml of 0.1N KMnO4 into 500ml conical flask , add 300ml of 0.8N sulfuric acid and heat up to 40C0. In this titration , nitrite solution must be in the burette to avoid the lost of some nitrous oxide liberated otherwise . Begin the titration slowly with constant stirring until the permanganate is just decolorized . Calculate the amount of nitrite salt.

Experiment (4-4): Determination of copper in CuSO4 by Iodometric titration.

Requirements:

A. Titration apparatus .

B. 0.1N CuSO4.5H2O (3g in 250ml)

C. 0.1N iodine solution

D. Starch solution indicator

E. KI salt

F. 0.1N sodium thiosulfate solution (6.25g in 250ml)

Procedure

Take 50 ml of the prepared copper sulfate solution into 250ml titration flask ; add 1g of potassium iodide salt (or 10ml of a 10% KI solution) , shake the mixture a bit and titrate liberated iodine against 0.1N sodium thiosulfate solution using starch indicator .

2CuSO4 + 4KI 2CuI + I2 + K2SO4

2Cu2+ I2 S2O32-

Qualitative Analysis

Solubility Equilibria and fractional precipitation.

In qualitative analysis , formation of a precipitate is an important technique to separate mixtures of ions and identify them systematically ; so the study of precipitation reactions are an essential part of qualitative inorganic analysis . A precipitate can be in crystalline or colloidal form and can be separated from the rest of the solution constituents by filtration or centrifuging . To understand the precipitation process , we must have extensive knowledge of the solubility of the different inorganic salts . This topic has been discussed quantitatively in precipitation titrations .The most important physical factors that affect the solubility of a substance are the temperature , concentration , and the nature of the solvent. In general the solubility of a substance increases with increase in temperature and can serve as the bases of separation between the ions in a solution . Typical example is the separation of lead and silver in group I cations. Solubility also depends on the concentration and the nature of ions in the solution . The presence of common ion in the solution considerably decreases the solubility of the precipitate . As we already explained this phenomena is known as common ion effect .

The effect of pH on precipitate depends on the nature of the precipitate . For example in basic precipitate ,

M(OH)2 M 2+(aq) + 2OH –(aq)

Increasing the pH (supplying more OH-) will shift the equilibrium to the left according to the Le Chatelier principle and decrease the solubility of the precipitate. The reverse is true if the pH is decreased and the equilibrium shifts to the right increasing the solubility of the precipitate. Another example is the insoluble CaCO3 ,

CaCO3 + H2O Ca2+(aq) + HCO3-(aq) + OH –(aq)

Decreasing the pH shifts the equilibrium to the right resulting the dissolution of the precipitate .

Solubility Product

This topic has been discussed to certain level in precipitation titration . We have seen that in saturated solution an equilibrium is established between the precipitate and its dissolved ions . We know that AgCl is sparingly soluble in water and at the saturation point the solution attains equilibrium so that:

AgCl(s) Ag +(aq) + Cl –(aq)

We call this equilibrium heterogeneous equilibrium due to the presence of solid phase in the solution . The equilibrium constant , Ksp , for the system is expressed as :

Ksp = , Ksp is the solubility product

This equilibrium constant can be stated as

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

The reason for deleting [AgCl] is already clarified in precipitation titrations and chemical equilibrium where many examples has been given . If dissolution of a solid occurs and without reaching an equilibrium , then we think of reaction quotient to explain the system ; in this case we speak of ion product quotient , (Q) , to predict whether a precipitate can form or not . Q is expressed the same way as equilibrium but the system is not at equilibrium. If we add silver ions to chloride solution , and the system is not yet at equilibrium , then ion product will be :

Q = [Ag+]Q [Cl -]Q

The value of Q in relation to Ksp will be in one of the following cases:

A. Q Ksp , [Ag+]Q [Cl -]Q 1.6 x 10-10 , then the system is not yet saturated and the precipitation cannot occur .

B. Q = Ksp , [Ag+]Q [Cl -]Q = 1.6 x 10-10 , then the system is at equilibrium and saturated solution is formed .

C. Q Ksp , [Ag+]Q [Cl -]Q 1.6 x 10-10 , the system is supersaturated and will tend to precipitate until it reaches the Ksp .

Predicting the occurrence of precipitation.

In a given system we can predict whether a precipitate takes place or not when we add two solutions together . This process has a practical application in an industrial or laboratory preparations .For example , we can manipulate the concentrations involved in a given reaction in such a way that the ion product exceed the solubility product constant in order to create the desired precipitate .

Let us take an example of river water in which the chloride concentration is 10 -5M that is [Cl -] = 10 -5 . If we add 0.2M AgNO3 solution to a sample of the river water will there be a precipitate or not ?

Solution:

We know that the Ksp of [Ag+][Cl -] = 1.6 x 10 -10 Q = [2 x 10 -1][10 x 10 -5] = 2 x 10 -6 In this case Q Ksp , and the system tend to precipitate If on the other hand , we take the distilled water in which [Cl -] = 10 -11 instead of river water , then Q = [2 x 10 -1][10 x 10 -11] = 2 x 10 -12 since Q Ksp the precipitation will not occur.

Fractional Precipitation

In qualitative analysis , we need sometimes to separate ions in a solution so that we can remove some of them and leave others behind in the solution . This process is achieved through precipitation and the method is called fractional precipitation. We have to employ the solubility product to know which of two ions can be precipitated first ; this has great analytical importance in both qualitative and quantitative analysis . Estimation of halides by Mohr's method can be taken as an example .

In this method, silver nitrate solution is titrated against chloride solution employing potassium chromate as an indicator. Both silver chloride and silver chromate are insoluble in water .

Ksp = [Ag+][Cl -] = 1.6 x 10-10 , Ksp = [Ag+]2[CrO42-] = 2.4 x 10-12

For silver chloride , [Ag+] =

For silver chromate , [Ag+] = , so that =

= 1 : 1.1 x 10 8

The result shows that potassium chromate is more soluble than silver chloride and more chromate ions are always present in the solution so any addition of silver nitrate to the solution containing both chloride and chromate ions , the chloride ions precipitate first and continue to precipitate until the concentration of Cl - decreases to the ratio expressed above. If sodium chloride solution of 0.1 N is titrated with nitrate solution containing 2.0 x 10 -3 K2CrO4 , the concentration of Cl - at which the CrO42- to precipitate can be calculated ,

= so that [Cl -] = = 4.3 x 10 -6

This must be the concentration of Cl - so that precipitation of CrO42- can take place.

Chemical tests on simple inorganic salts.

In this practical , we will begin with the tests of certain simple inorganic salts. since suitable laboratories or enough facilities are rarely available or non-existent in the developing countries , simple , save and clear chemical tests have been outlined. Simple inorganic salts usually consist of two parts ; negatively charged part called acidic radical (anion) and the positively charged part called basic radical (cation) .Thus the salt is electrically neutral (equal positive and negative charges) . Before we begin the tests , the student should have a clear scheme of how to report the results . One of the ways to lay out the activity is to make the following table on the notebook which can designed with help of the instructor. Modest example is given below.

We will begin our tests with acid radicals . These acid radicals or anions can be classified into three main subgroups , on the base of their reaction with mineral acids.

1. Hydrochloric acid group:

This group reacts with dilute HCl liberating different gases.

1. Carbonates : most of the carbonates are insoluble in water except carbonates of ammonium and alkali metals . All carbonates react with dilute HCl liberating gas CO2. Both solid and solution of these salts can be used .

Procedure:

Approximately 0.5 g or 5ml sodium carbonate solution is taken . Add few milliliters of 4M HCl . Observe the effervescence and evolution of CO2. Let this gas pass through clear solution of lime water and notice the turbidity formed .

2HCl + CO3- - CO2 + H2O + 2Cl –

CO2 + Ca(OH)2 CaCO3 + H2O

Lime water

2. Bicarbonates:

Similar tests can be performed , NaHCO3 + HCl CO2 + NaCl + H2O There are certain chemical tests to distinguish between the carbonates and bicarbonates:

Take two test tubes one with Na2CO3 solution and other with NaHCO3 solution. To each one add few drops of magnesium sulfate (MgSO4 . 7H2O) solution notice the difference .

Mg2 + HCO3- Mg(HCO3)2 , soluble salt , no precipitate

Mg 2++ CO3- - MgCO3 white precipitate .

Heat the test-tube that contains the bicarbonate solution to boiling and notice the precipitation .

Mg(HCO3)2 MgCO3 + H2O + CO2

3. Sulfites : Sulfites react with dilute HCl evolving sulfur dioxide gas (SO2). The evolution of the gas is slow at the room temperature but evolves more rapidly on gentle heating.

A. Add approximately one ml. of 4M HCl to a test-tube containing few ml. of 0.5M Na2SO3.7H2O solution . Notice the evolution of suffocating odor of SO2 , which can be identified by exposing to a filter paper wetted with acidified sodium dichromate . Green color is formed on the paper due to formation of Cr3+ ions.

SO2 + Cr2O7 2- + 2H+ 2Cr 3+ + 3SO4 2- + H2O

Reddish-orange green

The gas can also be identified by exposing filter paper with iodate(IO3-) and starch solution in which case blue color is formed due to formation of iodine.

B. With barium chloride, the sulfites form white precipitate of barium sulfite soluble in dilute HCl . The precipitate reacts with mineral acid liberating SO2 gas .

SO32- + Ba2+ BaSO3

BaSO3 + H+ Ba2+ + SO2 + H2O

C. With acidified (dilute H2SO4) potassium permanganate solution , the violet color of MnO4- disappears owing to reduction of Mn7+ to Mn2+ .

SO32- + 2MnO4- + 6H+ Mn2+ + 5SO42- + 3H2O

Violet colorless

D. With potassium dichromate solution acidified with dilute H2SO4 ; green color is formed.

SO32 + Cr2O72- + 8H+ 2Cr3+ + 3SO42- + 4H2O

4. Thiosulfates (S2O32-) , to run the following tests use approximately 0.5M Na2S2O3 solution .

a) With dilute HCl: Turbid solution is formed due to precipitation of element sulfur . By gentle heating , SO2 gas is liberated which is identified by exposing a filter paper moistened with acidified potassium dichromate (green color is formed) .

Na2S2O3 + HCl S + SO2 + NaCl

b) With iodine solution ; S2O32- reacts with iodine solution in which case immediate de-colorization of iodine takes place and tetra-thionate is formed

2S2O32-+ I2 2I - + S4O62-

c) With silver nitrate solution; white precipitate of Ag2S2O3 is formed which dissolves in excess of thiosulfate forming a complex ion of [Ag(S2O3)2]3- . The white precipitate turns to black silver sulfide on standing .