Solubility Equilibria and Precipitation Titrations

SOLUBILITY EQUILIBRIA AND PRECIPITATION TITRATIONS

1© L. Lapitan 2011

Brought to you by USTLeaksCopyright 2013

2013/3/6

SOLUBILITY EQUILIBRIA AND PRECIPITATION TITRATIONS

2© L. Lapitan 2011

Required Reading: Ch 3 and 11(Fundamentals of Analytical Chemistry : An Introduction By Skoog, West, Holler ,Crouch, and Chen – 8th ed.)`

2013/3/6

3

Solubility Equilibria• MmXx (s) m Mn+ (aq) + x Xy- (aq)

• The equilibrium is established when we have a saturated

solution of ions forming the solid and solid is dissociatingto form the ions in solution. The rates of these processes

must be equal.

© L. Lapitan 20112013/3/6

4

Solubility Equilibria• For a dissolution process, we give the

equilibrium constant expression the name solubility product (constant) Ksp. For

• MmXx (s) m Mn+ (aq) + x Xy- (aq)• Ksp = [Mn+]m [Xy-]x

• refer to a specific balanced equation(by definition this balanced equation is one mole of solid becoming aqueous ions)

• at a specific temperature.© L. Lapitan 20112013/3/6

5

Problem

• Write the expressions of Ksp of:a) AgClb) PbI2

c) Ca3(PO4)2

d) Cr(OH)3

[ ][ ][ ][ ][ ] [ ][ ][ ]33

sp

234

32sp

22sp

sp

OHCrK

POCaK

IPbK

ClAgK

−+

−+

−+

−+

=

=

=

=

© L. Lapitan 20112013/3/6

6

Problem1. If a saturated solution of BaSO4 is prepared by

dissolving solid BaSO4 in water, and [Ba2+] = 1.05 x 10-5

mol⋅L-1, what is the Ksp for BaSO4?

[ ][ ][ ] [ ]

( )10-

sp

25-sp

5-24

22sp

24

2sp

24

2 4

10 x 1.1 K

10 x 1.05 K

M 10 x 1.05 SO Ba x if xK

SOBaK

(aq)SO (aq) Ba (s)BaSO

=

=

====

=

+

−+

−+

−+→←

2. Calculate the Ksp for Ca3(PO4)2 (FW = 310.2) if the solubility is 8.1x10-4 g/L [Ans. 1.3x10-26]

© L. Lapitan 20112013/3/6

7

Molar solubility

• If we know the Ksp value for a solid, we can calculate the molar solubility, which is the number of moles of the solid that can dissolve in a given amount of solvent before the solution becomes saturated.

• The molar solubility leads to the solubility (by using the molar mass) which is the mass of the solid that can dissolve in a given amount of solvent before the solution becomes saturated.

© L. Lapitan 20112013/3/6

8

Problem

2. A handbook lists the aqueous solubility of lithiumphosphate (Li3PO4) as 0.034 g per 100 mL at 18 °C.What is the Ksp of lithium phosphate at 18 °C? Themolar mass of Li3PO4 is 115.794 g⋅mol-1. [Answer: Ksp =2.0 x 10-9]

1. Lead iodide, PbI2 is a dense, golden yellow “insoluble”solid used in bronzing and in organometal work, requiringgolden color. Calculate the molar solubility of lead iodidein pure water at 25oC given its Ksp = 7.1 x 10-9

© L. Lapitan 20112013/3/6

Solubility Product Constant• Ksp indicates how soluble an ionic compound

is in water at a certain temperature

2013/3/6

Equilibrium Reaction Ksp @ 25oC

AgBr(s) = Ag+(aq) + Br-(aq) 7.7 x 10-13

AgCl(s) = Ag+(aq) + Cl-(aq) 1.6 x 10-10

AgI(s) = Ag+(aq) + I-(aq) 8.3 x 10-17

• The higher the Ksp, the more soluble the compound is.

AgCl > AgBr > AgI

© L. Lapitan 2011 9

Note: the solutes are above are f the same type (MX or MX2 or M2X…) therefore their molar solubilities can be related in the same way as their Kspvalues)

• If solutes are not of the same type, you’ll have to calculate the each molar solubility and compare the results.

Ex. AgCl and Ag2CrO4

2013/3/6 © L. Lapitan 2011 10

Solubility Product Constant

Which has the greater molar solubility: AgCl with Ksp = 1.8 x 10-10 or

Ag2CrO4 with Ksp = 1.1 x 10-12?

Answer: The molar solubility of AgCl is 1.3 x 10-5 M while the molar solubility of Ag2CrO4 is 6.5 x 10-5 M. Silver chromate has a higher molar solubility.

The Common-ion Effect• MmXx (s)m Mn+ (aq) + x Xy- (aq)

• If we have dissolved a solid in pure water and we add to this solution another solution containing one of the common ions, then Le Chatalier’s Principle tells us what will happen:

• The presence of the common-ion in the added solution will force the dissolution reaction to the left, meaning more solid

will form!

11© L. Lapitan 20112013/3/6

12

The Common-ion Effect

© L. Lapitan 20112013/3/6

13

MmXx (s)m Mn+ (aq) + x Xy- (aq)

• If instead of dissolving a solid in pure water we try and dissolve it into a solution that already contains one of the common ions, then Le Chatalier’sPrinciple tells us what will happen:

• The presence of the common-ionalready in solution will force the dissolution reaction to the left, meaning less solid will dissolve than would dissolve in pure water!

The Common-ion Effect

© L. Lapitan 20112013/3/6

14

Problem

2. Calculate the molar solubility of MgF2 (Ksp = 7.4 x 10-11) in pure water and in 0.10 mol⋅L-1 MgCl2 at 25 °C. [Answer: The molar solubility is 2.6 x 10-4 M in pure water and 1.4 x 10-5 M in 0.10 M magnesium chloride.]

1. What is the molar solubility of PbI2 in 0.1 M KI solution? Ksp = 7.1 x 10-9

© L. Lapitan 20112013/3/6

15

Criteria for Precipitation and its Completeness• Can we predict if a solid will form if we mix

two solutions of different ions?• Consider the mixing of two different

solutions, one with Ca2+ ions and one with F-

ions. A formation of solid is the dissolution reaction in reverse, so we can express the reaction using the dissolution equation

CaF2 (s) Ca2+ (aq) + 2 F- (aq) Ksp = [Ca2+ ][F-]2

© L. Lapitan 20112013/3/6

16

• When we mix the solutions (BE CAREFUL –mixing ALWAYS changes the concentrations of both our ions!)

• the system is most likely not at equilibrium.

• Like in other equilibrum problems, we can use a reaction quotient Qsp (often called the ion product) to tell us in which direction the system must go to reach equilibrium

Qsp = [Ca2+ ][F-]2

Criteria for Precipitation and its Completeness

© L. Lapitan 20112013/3/6

17

If Qsp > Ksp, the solution is supersaturated, so the system is not at equilibrium. The concentration of the ions is greater than it would be at equilibrium, and so the reaction wants to shift from ions towards the solid.

We expect precipitation to occur!

If Qsp = Ksp, the solution is saturated, and the system is at equilibrium.

No precipitation occurs!


© L. Lapitan 20112013/3/6

18

•If Qsp < Ksp, the solution is unsaturated, so the system is not at equilibrium. The concentration of the ions is less than it would be at equilibrium, and so the reaction wants to shift from solid towards the ions.

No precipitation can occur!


© L. Lapitan 20112013/3/6

19

Problem

2. Will a precipitate form when 0.150 L of 0.10 mol⋅L-1

Pb(NO3)2 and 0.100 L of 0.20 mol⋅L-1 NaCl are mixed?Ksp of PbCl2 is 1.2 x 10-5 [Answer: Qsp = 3.8 x 10-4 > Ksp so precipitation should occur]

1. Three drops of 0.20 M KI solution is added to 100.0 mLof 0.0101 M Pb(NO3)2. Will a precipitate of lead iodide form? (Assume 1 drop = 0.05 mL)

© L. Lapitan 20112013/3/6

20

Complete Precipitation

• Generally we treat precipitation as complete if 99.9% of the original ion concentration has been lost to the precipitate.

• For example, if our initial [Pb2+] is 0.10 M, then precipitation by adding I- is completewhen our solution contains a [Pb2+] less than 1 x 10-4 M.

© L. Lapitan 20112013/3/6

1. The first step in a commercial process in whichmagnesium is obtained from sea water involvesprecipitating Mg2+ as Mg(OH)2. the magnesium ionconcentration in sea water is about 0.059 M. If seawater sample is treated so its [OH-] is maintained at0.20 x 10-3 M.(a) what will be the [Mg2+ ]remaining in the solutionwhen precipitation stops Ksp = 1.8 x 10-11

(b) can we say that precipitation of Mg(OH)2 iscomplete under these conditions?

© L. Lapitan 2011 21

Problem

2013/3/6

22

Problem2. A typical Ca2+ concentration in seawater is 0.010 M.

Will the precipitation of Ca(OH)2 be complete from a seawater sample in which [OH-] is maintained at 0.040 M? Ksp of Ca(OH)2 is 5.5 x 10-6

[Answer: Since the final [Ca2+] is 3.4 x 10-3 M, which is 34 % of 0.010 M, the precipitation is not complete.]

3. What [OH-] should be maintained in a solution if, after precipitation of Mg2+ as solid magnesium hydroxide, the remaining [Mg2+] is to be at a level of 1µg⋅L-1? Molar mass Mg is 24.305 g⋅mol-1 ,Ksp of Mg(OH)2 is 1.8 x 10-11 [Answer: [OH-] needed is 1.6 x 10-2 M]

© L. Lapitan 20112013/3/6

Separation of Ions by Fractional Precipitation• The process by which two or more aqueous

substances in a solution are separatedthrough the addition of a common ion, takingadvantage of their different concentrationneeds in order to form a precipitate.

2013/3/6 © L. Lapitan 2011 23

24

• If we have a solution with both CrO42-

ions and Br- ions • and add a large amount of Ag+ ions at

once, • then both Ag2CrO4 and AgBr will

precipitate• in our container at the same time.

© L. Lapitan 20112013/3/6

Separation of Ions by Fractional Precipitation

25

• If we slowly add the Ag+ solution instead the solid with the significantly lower molar solubility (AgBr in this case – do the calculations to check this for yourself)

• will precipitate first• and consume the added Ag+

preferentially.

© L. Lapitan 20112013/3/6


26

• In other words, • the concentration of Ag+

• CAN NOT become large enough• to precipitate Ag2CrO4

• until the AgBr• precipitation is complete.

© L. Lapitan 20112013/3/6


27

Fractional precipitation

© L. Lapitan 20112013/3/6

• Sodium Iodide (NaI) crystals are slowly added to a solution that is 0.100 M Pb(NO3)2 and 0.100 M AgNO3. Which will precipitate first?Ksp AgI = 8.31 x 10-17 Ksp PbI2 = 7.1 x10-9

2013/3/6 © L. Lapitan 2011 28

Problem

Which Will Precipitate First?1. CaCO3

(Ksp: 8.7x10-9)2. CaF2

(Ksp: 4.0 x 10-11)3. Ca3(PO4)2

(Ksp: 1.2 x 10-26)

2013/3/6 © L. Lapitan 2011 29

Problem

Problem1. Solid NaI is slowly added to a solution that is 0.010 M

in Cu+ and 0.010 M Ag+.(a) Which compound will begin to precipitate first?(b) Calculate [Ag+] when CuI just begins to precipitate.(c) What % of Ag+ remains in solution at this point?

• AgI Ksp = 8.3 x10-17

• CuI Ksp = 1.00 x 10-12

2013/3/6 © L. Lapitan 2011 30

[Answer a) AgI; b) 1.6 x 10-17 M; 1.6 x 10-3%]

31

Problem1. AgNO3 is slowly added to a solution with [Cl-]

= 0.115 M and [Br-] = 0.264 M. What percent of the Br- remains unprecipitated at the point at which AgCl (s) begins to precipitate?

Ksp values AgCl = 1.8 x 10-10

AgBr = 5.0 x 10-13

Answer: 0.12 % of Br- remains.

© L. Lapitan 20112013/3/6

Precipitation Titrations• Titrations with silver nitrate are sometimes

called argentometric titrations.

10-32

Calculate the pAg of the solution during the titration of 50.00 mL of 0.0500 M NaCl with 0.1000 M AgNO₃ after the addition of the following volumes of reagent: (a) 0.00 mL, (b) 24.50 mL,(c) 25.00 mL, (d) 25.50 mL.

Example

32© L. Lapitan 20112013/3/6

(a) Because no AgNO₃ has been added, [Ag⁺] 0 and pAg is indeterminate. (b) At 24.5 mL, [Ag⁺] is very small and cannot be computed from stoichiometric considerations, but [Cl¯] can be obtained readily.

10-3333© L. Lapitan 20112013/3/6

(c) This volume corresponds to the equivalence point where [Ag⁺]=[Cl¯] and

10-34

(d) At this volume of the titrant, there is excess of the Ag+ ions

34© L. Lapitan 20112013/3/6

The Shapes of Titration Curves

10-35

•Effect of titrant concentration on precipitation titration curves. Note the increased sharpness of the break for the more concentrated solution, A.

Curve A shows 50.00 mL of 0.0500 M NaCl with 0.1000 M AgNO3, Curve B shows 50.00 mL of 0.00500 M NaCl with 0.01000 M AgNO3 .

35© L. Lapitan 20112013/3/6

•Effect of reaction completeness on precipitation titration curves.Note that smaller values of Ksp give much sharper breaks at the end point.

10-36

The Shapes of Titration Curves

For each curve, 50.00 mL of a 0.0500 M solution of the anion was titrated with 0.1000 M AgNO3 36© L. Lapitan 20112013/3/6

Precipitation Titrations: Indicators• Three types of end points are encountered in

titrations with silver nitrate:(1) chemical, (2) potentiometric, and (3) amperometric.

• The Mohr Method: Formation of a Colored Precipitate.

10-3737© L. Lapitan 20112013/3/6

Precipitation Titrations: Indicators

• the high chromate ion concentration imparts such an intense yellow color to the solution.

• lower concentrations of chromate ion are generally used, and, as a consequence, excess silver nitrate is required before precipitation begins. An additional excess of the reagent must also be added to produce enough silver chromate to be seen.

10-38

The Mohr Method: Formation of a Colored Precipitate

38© L. Lapitan 20112013/3/6

• The Mohr titration must be carried out at a pH of 7 to 10 because chromate ion is the conjugate base of the weak chromic acid. Consequently, in acidic solutions, where the pH is less than 7, the chromate ion concentration is too low to produce the precipitate.

• Titration at pH < 7 : Formation of HCrO4-

• Titration at pH > 10 : Formation of Ag hydroxide ppt

10-39

Precipitation Titrations: IndicatorsThe Mohr Method: Formation of a Colored Precipitate

39© L. Lapitan 20112013/3/6

• Fluorescein is a typical adsorption indicator• In the early stages of the titration of chloride

ion with silver nitrate, the colloidal silver chloride particles are negatively charged because of adsorption of excess chloride ions. The dye anions are repelled from this surface by electrostatic repulsion.

10-40

Precipitation Titrations: IndicatorsThe Fajans Method: Adsorption Indicators

40© L. Lapitan 20112013/3/6

• Beyond the equivalence point, the silver chloride particles strongly adsorb silver ions and thereby acquire a positive charge. Fluoresceinate anions are now attracted into the counter-ion layer.

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

10-41

Precipitation Titrations: IndicatorsThe Fajans Method: Adsorption Indicators

41© L. Lapitan 20112013/3/6

• Silver ions are titrated with standard solution of thiocyante ion:

• Iron (III) serves as the indicator

• The titration must be carried out in acidic solution to prevent precipitation of iron(III) as the hydrated oxide.

10-42

Precipitation Titrations: IndicatorsThe Volhard Method: Forming a Colored

Complex

42© L. Lapitan 20112013/3/6

• The most important application of the Volhardmethod is for the indirect determination of halide ions.

• A measured excess of standard silver nitrate solution is added to the sample, and the excess silver ion is determined by back-titration with a standard thiocyanate solution.

10-43


Complex

43© L. Lapitan 20112013/3/6

• a distinct advantage over other titrimetric methods of halide analysis because such ions as carbonate, oxalate, and arsenate (which form slightly soluble silver salts in neutral media but not in acidic media) do not interfere.

• Silver chloride is more soluble than silver thiocyanate.

10-44


Complex

44© L. Lapitan 20112013/3/6

• This reaction causes the end point to fade• This error can be circumvented by filtering the

silver chloride before undertaking the back-titration.

10-45


Complex

45© L. Lapitan 20112013/3/6

Indicators for Precipitation Titrations: Summary

1. Volhard Method: Indicator -> Ferric Alum Endpoint : bloody red color Rxn: Ag+ + SCN- = AgSCN(s)

2. Mohr Method: Indicator -> Chromate (CrO42-)

Endpoint: brick red color Rxn:2Ag+ + CrO4

2- = Ag2CrO4 (s)

3. Fajans Method: Indicator -> FluoroceinEndpoint: AgX:Ag+::Fl- (adsorbed pink color)(Fl- symbolizes the anion of a weak acid)

46© L. Lapitan 20112013/3/6

Blank Titration• In Blank determinations, all steps for the

analysis are performed in the absence of the sample.

• Blank determinations reveal errors due to interfering contaminants from vessels or reagents employed in analysis

• Blanks also allow the analyst to correct titration data for the volume of reagent needed to cause an indicator to change color at an endpoint

© L. Lapitan 2011 472013/3/6

Applications of Precipitation Titrations1) To an aqueous solution containing solution of a

1.6000 g sample consisting of a mixture of CaBr2⋅H2O(FW = 308.0) and inert matter is added 52.00 mL of0.200 M AgNO3. The excess Ag+ requires 4.0 mL of0.1000 M KCNS for the precipitation of AgCNS. Whatis the percentages of Br and of inert matter in thesample?

48© L. Lapitan 20112013/3/6

2) A 0.5131-g sample containing KBr is dissolved in 50 mL of distilled water. Titrating with 0.04614 M AgNO3requires 25.13 mL to reach the Mohr end point. A blank titration requires 0.65 mL to reach the same end point. Report the %w/w KBr in the sample.

Applications of Precipitation Titrations

49© L. Lapitan 20112013/3/6

3) A mixture containing only KCl and NaBr is analyzed bythe Mohr method. A 0.3172-g sample is dissolved in50 mL of water and titrated to the Ag2CrO4 endpoint, requiring 36.85 mL of 0.1120 M AgNO3. A blanktitration requires 0.71 mL of titrant to reach thesame end point. Report the %w/w KCl and NaBr inthe sample.


50© L. Lapitan 20112013/3/6

4) The phosphorus in a 4.258-g sample of aplant food was converted to PO4

3- andprecipitated as Ag3PO4 through the addition of50.00 mL of 0.0820 M AgNO3. The excesssilver nitrate was then back titrated with 4.06mL of 0.0625 M KSCN. Express the results ofthis analysis in terms of % P2O5.

2013/3/6 © L. Lapitan 2011 51


A 2.414 g sample containing KCl, K2SO4, and inert materialswas dissolved in sufficient water to give 250.00 mL ofsolution. A Mohr Titration of the 50.00 mL aliquot required41.36 mL of 0.05818 M AgNO3. A second 50.00 mL aliquotwas treated with 40.00 mL of 0.1083 mL M NaB(C6H5)4. Thereaction is:

NaB(C6H5)4 + K+ →KB(C6H5)4 (s) + Na+

The solid was filtered, redissolved in acetone and titratedwith 49.98 mL of the AgNO3 solution according to thereaction:

KB(C6H5)4 + Ag+ → AgB(C6H5)4(s) + K+

Calculate the percentage of KCl and K2SO4 in the sample.

2013/3/6 © L. Lapitan 2011 52


Documents

Solubility Equilibria and Precipitation Titrations