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Objectives. Define precipitation reactions. Determine the solubility and solubility product of different salts. Compare the solubility of different salts. Define precipitation titrations. Define how to detect end point of argentometric titrations. Precipitation reactions.

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  • Define precipitation reactions.

  • Determine the solubility and solubility product of different salts.

  • Compare the solubility of different salts.

  • Define precipitation titrations.

  • Define how to detect end point of argentometric titrations.

Precipitation reactions
Precipitation reactions

Simple rule for the solubility of salts in water (VIP)

  • Most nitrate (NO3-) salts are soluble.

  • Most salts containing the alkali metal ions (Li+, Na+, K+, Cs+, Rb+) and the ammonium ion (NH4+) are soluble.

  • Most chloride, bromide and iodide salts are soluble. Notable exceptions are salts containing the ions Ag+, Pb2+ and Hg22+.

  • Most sulfate salts are soluble. Notable exceptions are BaSO4, PbSO4, HgSO4 and CaSO4.

  • Most hydroxide salts are only slightly soluble. The important hydroxides are NaOH and KOH. The compounds Ba(OH)2, Sr(OH)2 and Ca(OH)2 are marginally soluble.

  • Most Sulfide (S2-), carbonate (CO32-), chromate (CrO42-) and phosphate (PO43-) salts are only slightly soluble.

Precipitation reactions cont
Precipitation reactions (cont.)

  • Precipitation reaction occurs when water solutions of two different ionic compounds are mixed, an insoluble solid separates out. This precipitate formed is also ionic, as the cation comes from one solution and the anion comes from the other.

  • The precipitation diagram, enables you to determine whether or not a precipitate will form when dilute solutions of two ionic solutes are mixed. (very important figure). (shaded parts are precipitates)

Precipitation reactions cont1
Precipitation reactions (cont.)

  • If a cation in solution 1 mixes with an anion in solution 2 to form an insoluble compound (coloured squares, that compound will precipitate. Cation-anion combinations that lead to the formation of a soluble compound (white squares) will not give a precipitate.

  • Example

  • If solutions of NiCl2 (Ni2+, Cl- ions) and NaOH (Na+, OH- ions) are mixed,

  • A precipitate of Ni(OH)2, an insoluble compound, will form.

  • NaCl, a soluble compound, will not precipitate.

Solubility equilibria and solubility product
Solubility equilibria and solubility product

  • Solubility of a substance, is the amount of substance that dissolves in a given volume of solvent at a given temperature, the solubility is expressed in mol/L.

  • Solubility product constant, Ksp, is the constant for equilibrium expression representing the dissolving of an ionic solid in water.

  • For a salt AaBb, the solubility product can be expressed as,

    Ksp = [A]a[B]b

  • The [AaBb] was not considered in the expression as it is a pure solid.

Solubility equilibria and solubility product cont
Solubility equilibria and solubility product (cont.)

  • It is very important to distinguish between the solubility of a given solid and its solubility product.

  • The solubility product is an equilibrium constant and has only one value for a given solid at a given temperature.

  • Solubility on the other hand, is an equilibrium position that varies according to the amount of solute and the temperature of solution.

Example 1
Example 1

  • Copper(I) bromide has a measured solubility of 2.20x10-4 mol/L at 25°C. calculate its Ksp value.

    CuBr(s) Cu+(aq) + Br-(aq)

    Ksp = [A]a[B]b

    Ksp = [Cu+] [Br-]

    Ksp = [2.20x10-4 mol/L][2.20x10-4 mol/L]

    Ksp = 4.0x10-8 mol2/L2 (the units for Ksp are usually omitted).

Example 2
Example 2

  • Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0x10-15 mol/L at 25°C.

  • Bi2S3(s) 2Bi3+(aq) + 3S2-(aq)

    Notice that, each mol of Bi2S3, produces 2 moles of Bi3+ and 3 moles of S2-, so

  • [Bi3+] = 2 x [Bi2S3] and that of [S2-] = 3 x [Bi2S3]

  • So we can simply say that: (this is valid for all salts)

  • Ksp = [aA]a[bB]b

  • Ksp = [2Bi3+]2.[3S2-]3

  • Ksp = [2x1.0x10-15].[3 x 1.0 x10-15] = 1.1 x 10-73

Example 3
Example 3

The Ksp value for copper(II) iodate Cu(IO3)2, is 1.4x10-7 at 25°C. calculate its solubility at 25°C.

  • Ksp = [Cu2+]. [2IO3-]2

  • Let the solubility be x then

  • Ksp = [x].[2x]2

  • Ksp= 4x3

  • X= = 3.3x10-3 mol/L

Example 4
Example 4

  • Calculate the solubility of Ba(IO3)2 in a solution prepared by mixing 200 ml of 0.01 M Ba(NO3)2 with 100 ml of 0.1 M NaIO3. (Ksp= 1.57x10-9)

  • In this case we have two sources of barium and the share of Ba(NO3)2 as a source of Ba2+ ions is much higher than that due to the solubility of Ba(IO3)2, so we consider the concentration of Ba2+ ions as that of the Ba(NO3)2 .

Example 4 cont
Example 4 (cont.)

  • First we have to calculate the No. of moles of each ion, no. of moles = conc. x volume

  • no. of mmol Ba2+ = 200 x 0.01 = 2.00 mmol

  • no. of mmol IO3- = 100x0.1 = 10.0 mmol

  • Each mmol of Ba2+ needs 2 IO3-.

  • Excess mmol of NaIO3 = 10.0 - 2(2.00) = 6.00

  • [IO3-] = 6.00/300 = 0.02 M (300 = total volume)

  • Ksp = [Ba2+][2IO3-]2

  • [Ba2+] = 1.57x10-9 / (0.02)2 = 3.93x10-6M

Relative solubilities
Relative solubilities

A salt’s Ksp value gives us information about its solubility.

However, we must be careful in using Ksp values to predict the relative solubilities of a group of salts as there are two possible cases:

  • 1. The salts being compared produce the same number of ions. For example consider

    AgI Ksp = 1.5x10-16

    CuI Ksp = 5.0x10-12

    CaSO4 Ksp = 6.1x10-5

    Each of these solids dissolves to produce two ions

    Salt cation + anion

     We can compare the solubilities by comparing the Ksp values. CaSO4(s) > CuI(s) > AgI(s)

Relative solubilities cont
Relative solubilities (cont.)

2. The salts being compared produce different numbers of ions. For example consider

CuS Ksp = 8.5 x 10-45

Ag2S Ksp = 1.6x10-49

Bi2S3 Ksp = 1.1x10-73

Because these salts produce different number of ions when they dissolve, the Ksp cannot be compared directly to determine relative solubilities. First we have to calculate the solubilities,

CuS Solubility = 9.2x10-23

Ag2S Solubility= 3.4x10-17

Bi2S3 Solubility = 1.0x10-15

from which we get

(More Soluble) Bi2S3 > Ag2S > CuS (less soluble)

Precipitation titrations
Precipitation titrations

  • Precipitation titrations, is based on reactions that yields ionic compounds of limited solubility, precipitates.

  • The slow rate of formation of most precipitates limits the number of precipitating agents that can be used in titrations.

Precipitation titrations cont
Precipitation titrations (cont.)

  • Silver nitrate is considered as a very important precipitating agent that can be used for the determination of the halogens, the halogen like anions, mercaptans, fatty acids and several divalent inorganic anions.

  • Titrations with silver nitrate are

    sometimes called

    argentometric titrations.

Factors affecting the sharpness of end point of titration
Factors affecting the sharpness of end point of titration

A) Effect of analyte and reagent concentration.

  • The end point is much sharp on using high concentrations of both reactants, compare curves below.

B effect of solubility product
B) Effect of solubility product

From the figure:

  • It is clear that the change in pAg at the equivalence point becomes higher as the solubility product become smaller, (the precipitate is less soluble).

  • Note that ions forming precipitates with solubility products much larger than about 10-10 do not yield satisfactory end points.

Detecting end point of titration
Detecting end point of titration

  • The end point can be determined chemically by two ways:

  • No indicator method

    Disappearance or appearance of a precipitate.

  • Indicator method

    Formation of a coloured precipitate, formation of a coloured complex or using adsorption indicators.

No indicator method
No indicator method

  • a) Disappearance of precipitate.

  • Example, titration of silver ion with cyanide ion.

  • From the first drop of addition of cyanide, silver cyanide is formed and precipitation is continued until all silver is precipitated as silver cyanide.

  • Adding more cyanide will form the soluble argentocyanide complex.

  • Ag+ + CN- AgCN

  • AgCN + CN- [Ag(CN)2]-

  • The process is continued until all silver cyanide is dissolved and the end point is the disappearance of last trace turbidity.

No indicator method cont
No indicator method (cont.)

  • b) Appearance of precipitate

  • Example (Liebeg’s method for determining cyanide ion with silver ion, the reverse of the previous method).

    In this case, the solution contains excess of cyanide ions which precipitates as Silver cyanide when silver is added. Titration is continued until all the cyanide is transferred to soluble argentocyanide complex. The first excess of silver ion added will form the insoluble silver argentocyanide complex,

    AgCN + KCN K[Ag(CN)2]

    K[Ag(CN)2] + Ag+ Ag[Ag(CN)2] + K+

    The end point is detected by the first appearance of turbidity. Most silver salts are soluble in silver cyanide Cl-, I-, Br-, SCN-.

Indicator methods a formation of coloured precipitate mohr s method
Indicator methodsa- Formation of coloured precipitate (Mohr’s method)

  • Sodium chromate can serve as an indicator for argentometric determination of chloride, bromide and cyanide ions by reacting with silver ion to form a brick red colour of Ag2CrO4 at the equivalence point.

  • Titration reaction

    Ag+ + X- AgX(s) white

  • Indicator reaction

    2Ag+ + CrO42- Ag2CrO4(s) red

Mohr s method cont
(Mohr’s method) (cont.)

  • The solubility of silver chromate is several times greater than that of silver chloride or bromide. Thus silver chloride will precipitate first and then chromate will precipitate after.

  • This requires a high concentration of chromate (yellow colour) which will obscures the red color of the silver chromate.

  • So we have to use a lower concentration of chromate thus for precipitation of silver chromate excess amount of silver will be needed which is a positive error in the results.

Mohr s method cont1
(Mohr’s method) (cont.)

  • A good way is to make a correction by standardizing silver nitrate against a primary standard solution of sodium chloride using the same conditions.

  • Mohr titration must be carried out at neutral or slightly alkaline pH 7-9.

  • In acidicmedium, the equilibrium is shifted towards formation of dichromate. Silver dichromate is considerable soluble than chromate so again over consumption of silver nitrate occurs

  • 2CrO42- + 2H+ Cr2O72- + H2O

  • If the medium is strongly alkaline, silver precipitates as the oxide.

B adsorption indicators fajan s method
b- Adsorption indicators: Fajan’s method

  • Adsorption indicator is an organic compound that tends to be absorbed onto the surface of the solid in precipitation titration. The adsorption occurs near the equivalence point.

  • Fluorescein is a typical adsorption indicator useful for the titration of chloride ion with silver nitrate.

    In aqueous solution, (neural or slightly alkaline 7-9 while strongly acidic medium hinders its dissociation), fluorescein partially dissociates into hydronium ions and negatively charged fluoresceinate ions that are yellow green. The fluoresceinate ion forms an intensely red colour with silver ions.

Fajan s method cont
Fajan’s method (cont.)

  • In the early stages of titration, the colloidal silver chloride are negatively charged because of adsorption of excess chloride ions. The indicator anions are repelled and the solution impart a yellow green colour.

  • Beyond the equivalence point, the silver chloride strongly adsorb silver ions and acquires a positive charge thus the indicator anions are attracted and surround the colloidal silver precipitate particles.

  • The net result is the appearance of red color of silver fluorescinate in the surface layer of the solution surrounding the solid.

  • This colour is due to adsorption and not to precipitation.

C formation of a coloured complex the volhard method
c- Formation of a coloured complexThe Volhard method

  • In Volhard’s method, a standard solution of thiocyante is used as titrant for silver ion or mercuric ions using ferric as indicator.

  • The method can be applied for indirect determination of halides, cyanide, arsenate, phosphate and oxalate which form insoluble silver salts in acid medium.

  • This method is indirect as excess of standard silver nitrate solution is added to the halide solution and this excess is then determined by thiocyante as follows:

  • Ag + SCN- AgSCN

  • Ferric ion serves as indicator by forming a red colour with excess thiocayante

  • Fe3+ + 2SCN- [Fe(SCN)6]2+

The volhard method cont
The Volhard method (Cont.)

  • Notice that,

  • The titration must be carried out in acid medium to prevent precipitation of ferric hydroxide.

  • The indicator concentration is not critical but at high concentrations of iron its yellow colour may harden detecting the end point.

  • When determining chloride, the precipitate of AgCl must be filtered or nitrobenzene is used to coat the precipitate before titration. WHY?

The volhard method cont1
The Volhard method (Cont.)

  • Silver chloride is more soluble than silver thiocyante thus at the end point the reaction

  • AgCl(s) + SCN- AgSCN(s) + Cl-

  • As a consequence,when titrating with thiocyante first all excess silver ions are precipitated then the first excess of thiocyante will displace chloride in silver chloride precipitate to form the more insoluble silver thiocyante. This consumes a excess amount of titrant (Thiocyante).