1

For insoluble substances like silver bromide (AgBr), the molar solubility canoe quite small. In the case of AgBr, the value is 5.71x10-7 moles per liter. . -··-------···-----

• For very soluble substances (like sodium nitrate, NaN03), this value can be quite high, exceeding 10.0 moles per liter of solution in some cases .

• The molar solubility of a substance is the number of moles that dissolve per liter of solution . •

Since K is always calculated by just multiplving concentrations, it is called a "solubility product" constant - Ksp.

Calculating the K.P from the Molar Solubility

-~

Notice that the Ag2S04 is left out of the expression! Why?

Since this is an equilibrium, we can write an equilibrium expression for the reaction:

Dissolving silver sulfate, Ag2S04, in water When silver sulfate dissolves it dissociates info ions. When the solution is saturated, the following equilibrium exists:

Example; BaS04(S)+-+Ba+ (aq) + S04 2• (aq)

t pure solid

When we write the eq. cons. For heterogeneous equilibrium we ignore the cone. Of pure solids and liquid .therefore,

The _eq. cons.exp. For dissolving of a slightly soluble ionic solid depends only on the molar cone. Of the species in solution

Ksp=[Ba2+][SO/°]

•

The solubility product is the equilibrium constant for the reaction in which a solid salt dissolves , to give its constituent ions in solution

Solid is an ionic compound It is a strong electrolyte---------- completely ionized

++ 2· . Example; BaS04cs)+-+Ba (aq) + S04 (aq)

SOLUBILITY EQT!ILIBRIA Solubility product ksp

Analytical Chemistry

·(, -{ S' -b-Z)-e ,,r~:J~~

~--

Determine the Ksp of calcium fluoride (C~2), given that its molar solubility is 2.14 x 10-4 moles per liter. l-CaF2 (s) <=> Ca2+ (aq) + 2 F- (aq)______ _ .

------- 2

x., = (X) (2X)2

x., = 4X3

x~jK:

2X x CaF2 (S) ~ Ca 2+

(aq) + 2F- (aq)

When AgBr dissolves, it dissociates like this: 1-AgBr (s) <=>Ag+ (aq) +Br" (aq) The Ksp expression is: 2-KsP =[Ag+] [Br-] There is a 1:1 ratio between AgBr and Ag+ and there is a 1:1 ratio between AgBr and Br". This means that, when 5. 71 x 10--7 mole per liter of AgBr dissolves, it produces 5. 71 x 10-7 moles per liter of Ag+ and 5.71x10- 7 moles per liter of Br- in solution. Putting the values into the Ksp expression, we obtain:

. 3-Ksp = (5.71x10--7) (5.71x10-'7) = 3.26 x 10-13

S is molar solubility. The molar solubility of a substance is the number of moles that dissolve per liter of solution _

Determine the K,;p of silver bromide, given that its molar solubility is 5.71x10- 7 nioles per liter.

Ksp=[Ag+][Cr ] =X2

Sor X = ..JK;

x x AgCl (s) +-Ag" (aq) +er (aq)

AB •

calculating the KsP of the substance 1) Write the chemical equation for the substance dissolving and dissociating. 2) Write the Ksp expression. 3) Insert the concentration of each ion and multiply out.

3

The Ksp expression is: x., =- [Ce4j [103-)4 We know the following:·~ ---- .. -·· - - - - -- - -- -~- These is a 1:4 ratio between the concentrations of the cerium(IV) ion and the iodate ion .There is a 1:1 ratio between the molar solubility and the cerium(IV) ion's concentration. Therefore: Ksp = (1.80 x 10-4)(7.20 x 10-4)4

Ksp = 4.84 x 10-17

~ Calculate the x., for Ce(I03)4, given that its molar solubility is 1.80 x 10-4 mol/L

~Ksp X = 4x27

x., =(2X)2 (3X)3

5)A2B3 or A3B2 , Pb3(po4)2 BhS3 (s) !:; 2Bi3+ (aq) +3s2- (aq)

2X 3X

Ksp=27 X4

x = 4{KsP ...}27

Ksp =(X) (3X)3 X 3X

4)AB3 Fe(OH)3 (s) !:; Fe3+ (aq) +30ff (aq)

x = 3!ffe

Ksp =(2X)2 (X) Ksp= 4 X3

x 2X

3)A2B ------------Ag2Cr04 Ag2S04 (s) ~ 2Ag+ (aq) + S042-(aq)

2+, - 2 2-Ksp = [Ca J [F ] . 3-Ksp = (2.14 x 10-4) ( 2x2.14 X 10-4)2 = 3.92 X 10-11

( 4.28 x 1 o-4)2

4

ion product, Qsp. If the solution is not saturated, no precipitate will form. In this case, the product is called the ion product,

Qsp· . - ~- ---- --~----

[Ca2'] [C0:.1 -.

[Li"]2[C031

Agt.l(s) =Ag+(aq)+Cr(aq) (Ag"] [Cl1

. . .

Equilibrium . Expressbnfor K,p and C2,p

Mn (II) hydroxide, Mn(OH)2 Nickel sulfide, NiS Silver chromate, Ag2Cr04

Zinc carbonate, ZnC03

Calcium fluoride, CaF2

Solubility Products, Ksp1 and Ion Products Q~P formations of precipitates are chemical equilibria phenomena" and we usually write these heterogeneous equilibrium in the following manner, and call the equilibrium constants solubility products, Ksr,:

Then write the Ksp expression for the salt. • For each salt below, write a balanced equation showing its dissociation in water. •

Writing solubility product expressions ...

example, what is the concentration ofHg2 2• in equilibrium with 0.1 O M er in a solution of KCI containing excess, un dissolved Hg2Cl2(s)?

)- Calculate the Ksp for Mg3(P04)2, given that its molar solubility is 3 .57 x 1 o-6 mol/L The Ksp expression is: Ksp = [Ivig2+]3 [P043-]2

We know the following: These is a 3:1 ratio between the concentration of the magnesium ion and the molar solubility of the magnesium phosphate .There is a 2:1 ratio between the concentration of the phosphate ion and the molar solubility of the magnesium phosphate. · · Therefore: Kso = (1.071X10-5)3 (7.14 X 10-6)2 Ks~= 6.26 x 10-26 .

5

This will happen only if Qsp > Ksp in our mixture.

If a precipitate forms, it means the solubility equilibrium has shifted BACKWARDS.

Ksp = 2 x 10·16 = [Pb2l[Cr0l]

PbCr04 (s) ~ Pb2+ (aq) + CrO,/- (aq)

Step 1: Is a sparingly soluble salt formed? We can see that a double replacement reaction can occur and produce PbCr04. Since this salt has a very small Ksp, it may precipitate from the mixture. The solubility equilibrium is:

Q = [Pb2+] [SO/·]= (0.0016) (0.004) = 6.4X10 ·G> Ksp = 6.3x10 -7

Q>Ksp -ti. ppt will form

[Pb2+] 0.1 x8.0x10 .3 0.0016 M 0.1+0.4

[SOl·] = 0.4 xS.OxlO -3 0.004 M 0.1+0.4

REMEMBE.Rr TAKE INTO ACCOUNT OF CONCENTRATION OF IONS IN MIXTURE (DllUTION UPON MIXING)

PbS04{S)~ Pb2•(aq)+SOl'(aq)

Will a precipitate form when 0.10 L of 8.0 x 10·3 M Pb(N03)2 is added to 0,40 L of 5.0 x Ht3 M tia2SQ4? (K1P = 6.3 x 10·7)

Predicting Precipitation (Ksp vs Q)

Qsp > Ksp Oversaturate solution Qsp = Ksp Saturate solution Qsp < Ksp Unsaturated solution • Is the product of the concentrations of the ions at any moment in time. When the ion product is equal to the solubility product for the salt ,the system is at equilibrium . . ~ ..

THE ION PRODUCT The role of the ion product Qsp in solubility calculation

6

1. Common Ion Effect 2. pH of solution 3. Formation of Complexes

Factors affecting solubility

This question deals with the concept of ion product, Qsp· H Qsp = [Pb2+] [Br"]2 < Ksp the solution is unsaturated.

[0.012] [0.024]2 = 6.9x10"6 < 4x10"5 (Ksp)•

Chemical analysis gave [Pb2+] = 0.012 M, and [Br"] = 0.024 M in a solution. From a table, you find Ksp for PbBr2 has a

value of 4x10·5 • Is the solution saturated, oversaturated or unsaturated?

Either way, no ppte will form!

Note: If Qsp = Ksp, the mixture is saturated If Qsp < Ksp, the solution is unsaturated

the two solutions are mixed! Since Qsp >> Ksp, a precipitate will form when

Step 4: Compare Qsp to Ksp.

Qsp = 1.6 x 104

Qsp .; [Pb2j[CrO/] = (0.0080 M)(0.020 M)

Step 3: Calculate Qsp for the mixture.

{Pb2·•]= C,V1 = (Oc024M)(15 ml) =O~OOBOM Pb2+ V2 (45mL}

CjV1 (0.030M)(20 ml) 2 [CrO.e.2°]= y2 (4Sml) 0.020M Cr04 -

Step 2: find the concentrations of the ions that form the sparingly soluble salt. Since we are mixing twosolutlons in this example, the concentrations of the Pb1+ and Crol'-will be diluted. We have to do a dilution calculation!

Which of the following substances will be more soluble in acidic solution than in basic solution: (a) 8

COMPARE WITH 1. __.,J. x=Solubillty=L'Z xl T' M S=[Mg2+] = 0.18 M

What is the solubility ofMg(OH)2 in a less alkaline solution buffered at pH 9?pOH = 5 Mg(OH)2 (s) ~ Mg2+ (aq) + 20ff (aq)

[Off] = 1.0 x 10·5 M Ksp = [Mg2+] [Off]2 Ksp = 1.8 X 10"11 (25 °C) = [Mg2+](1.0 X 10"5)2

2.

Ksp = [Mg2l[Off]2 = (x)(2x)2 x=Solubility = 1.651 x 10·4 mol/L = 1. 7 xl 0-4 M

[Off]= 2x = 3.302x10-4 M pOH = ~ log(3.302x 10-4) = 3.48 pH = 10.5 pH of Solution

2. pH of Solution Mg(OH)2 (s) ~ Mg2+ (aq) + 20ff (aq)

Ksp = 1.8 x 10·11 (25 °C)Calculate the molar solubility of Mg(OH)2 in pure water. What is the pH of the resulting solution?

{X+YJ Ksp ( [H30 +] [H. 3. o.·.+ ]2 + - - - -)· S::; xx yY 1 + K + K -

a a Ka-:

poly basic add HaA

2 s c· [H3.0+]) S ·=.Ksp 1+ · Ka General rule

Salt B)(A, Of

In generalrule;the solubtlltv ofsllghtlv sqluble salts containing basic anions increases as [H I increases (as PH is lowered)

As a result the solubility eq. of CaF2 is shifted to right. CaF2 (s) ~ Ca2+ (aq) + 2F- {aq)

1 addition of H30+

F-+ H+ HF Over all Ca F2 (s) +2H+ (aq) ~ Ca2+ (aq} + 2HF (aq)

9

For example, AgCl, whose Ksp = 1.8 10-10, will dissolve in the presence of aqueous ammonia because of the interaction between Ag+ and the Lewis base NH3,

This process can be viewed as the sum of two reactions, the solubility equilibrium of AgCl

and the Lewis acid-base interaction between Ag+ and NH3:

Lei.visatld + Lewis base ::-""" adduct · Electron pair Electron pair

acceptor donor

Lewis bases other than water can also interact with metal ions, particularly with transition-metal ions. Such interactions can have a strong effect on the solubility of a metal salt.

3-Formation of Complex Ions

A characteristic property of metal ions is their ability to act as Lewis acids,or electron-pair acceptors, toward water molecules, which act as Lewis bases, or electron-pair donors.

(c) The solubility of BaS04 is largely unaffected by changes in pH because so,': is a rather weak base and thus has little tendency to combine with a proton. However, BaS04 is slightly more soluble in strongly acidic solutions.

(d) The solubility of AgCl is unaffected by changes in pH because er is the anion of a strong acid and therefore has negligible basicity.

Overall: CaCO:)(s) + 2IP(aq) _,. ci"'(aq)' + COi{g) + HP(l)

CO}~{aq) + 2H'"(aq} · · lli~(!Uf) H~~(aq} ~C°'i{g) + HP(/)

Ni(Oli}i($)-- .....•. •Ni:t-+·cllq) + 20H-(aq) 20H-·(aq) + 2H +(aq) ~ 2HP(l)

Overall~ Ni(OH)2(s) + 21-I"'(aq) ~Nl.,.(itq) + 21{.z(Xl)

Ni(OH)z(s) (b) CaC03(s) (c) BaS04 (s) (d) AgCI (s)? SOLUTION (a) We can conclude that Ni(OH)2(s) will be more soluble in acidic solution because of the basicity of

011; the It ion reacts with the OH~ ion, forming water: (b) Similarly, CaC03(s) dissolves in acid solutions because co,': is a basic anion:

10

Ag+c·· ···) f '»...TU~( ·~---...: A~TU_\...tt ·\ . .. "'I · · . '·"" ~ aq1 ~ at>\' u. "JIZ \aq1 ·

• For a Lewis base such as N H3 to increase the solubility of a metal salt, it must be able to interact more strongly with the metal ion than water does.

• The N H3 must displace solvating H20 molecules in order to form Ag(NH3)/:

Such an equilibrium constant is called a formation constant, K1

The stability of a complex ion in aqueous solution can be measured by Kr

For example, the equilibrium constant for formation of Ag{NH3)/ is L7 x107

. t~)i+] ..... 1" = [Ag+j[Nitji = 1.1 x 101

The presence of NH3 drives the top reaction, the solubHity equilibrium of AgCt, to the right as Ag+(aq) ls removed to form Ag{NH3)i.+. a metal ion and the Lewis bases bonded to it, such as

Ag(NH3)i+, is called a complex ion

Qv:erall: AgCl{s) + 2NH3(ttq) ......,.._-:-"' Ag(NH:J1 + (aq) + Cl" (aq

AgC~($) ~Ag+ {aq) + er (aq) Ag i:(aq) + 2Nll:J(aq) .··· ·. . Ag(Nli,.t)i +(aq)

11

FROM stepwise f ormatlon constants

tPt1+1"" K,[Pn2+n-1 ... ,1,<lx rn2>f7•9x 10~3x1.ox 10~3,

• 7.9 X 10-it M l Phb(aq}J "" ~2l Ph2+ nr J1 .. i.J X 10-~ M

[J>b1;1- ~.,f Ph2~1rr1s = 6.6 x 10-s M [Pbll"' J "" ~J[ Pb2+ ]( .r y'· eo 2.4 X 10-10 M

b) If, instead. we; take. [I" J = LQ M, lhcn analogous compuutions sl1qw tha!

. - - -------(Pb2.,, J • 7.9 x 1Ct9. M--- lPbI_;J • 6.6 X IO-~ M I Pbl+ l * 1l) X 10'"1 M t.Pbii ... ,I st 2.4 x 10-4 M

lPbh(aq}j"'" 1.1.X10-5 M

Fin9 the wnccnlratior\$ of Pbl "~ Pbl2(aq), Phl i 1< ;iud Pbl}- in a solution saturated "ith Pblz(.S) and conlai.nfog diwl\·ed r with a (.1JOCCrtltillk1n of (a) 0.001 0 M: and (b) L-0 M.

(a) Front K~p for Reaction

Jn generaITheoverall,orcumulative,formation constants are denoted ~i:

M. + nX: JJ.. . MX~ ~11- - [MX"JllMHXl~

Kt""' [Pbl+:VIPbHJ[rJ =LO x lo':? ~1=lPhlz(aq}lt[Pb~+HrJz==1.4 x u}t

~J = -[Pbl;JtIPQ?:+ ni,- J1 = sJ x J(il

~-.f =-· I Pblj'"" Jtf Pb2 +-JI I"" J4 "" 3,:Q X IO~

Pb~+. 1- ~ Pbl+ . + -~ PbH '>I- ~ Phl , . ·1· . + - ,• ...---- ' . z._..,,_1q

.Pn~+ + sr ~ PM; .Pb. 2 + _- ·"l"" ~ Pbl2- .. - + ... ~ .t

inthe·presenceofexcessl-highconcentrationso.fl-ca.usesolidPbl2.todissclve.We explain this by the formation of esertesot comptex ions

· ..

Pbl2(s) ~- Pb2+ + 2r

EXAMPLE; If Pb2+ and 1- only reacted toform solid Pb 12:, then the solubility of Pb h would alwaysbeveryiow

·~---·---

12

Because~ is quite Iarge, we begin with the assumption that essentially all of the Ag+ is converted to Ag(NH3)2+,

in accordance with Equarton Ag+ (aq) + l:NH3(aq) • =- Ag(Nl~:J)2 -r (4q)

Thus, [Ag+] will be small at equilibrium. tf:[Ag+] is n.010 M initiaU~ then [Ag(NH3J2 +] wiU be 0.010 M

. Let the concentration ofAg+ at. equilibrium hex. Then, at equilibrium ·

• EXERCISE • Calculate the concentration of Ag+ present in

solution at equil.ibrium when concentrated ammonia is added to a 0.010 M solution of AgN03 to give an equilibrium concentration of [NH3] = 0.20 M, Neglect the small volume change that occurs on addition of NH3.'

k s - SQ (1 +k1[L]+K1K2[LJ 2+--)

Where k is formation or stability constants.

S= ~ksp (l+k1[L]+K1K2[L] 2+------) Or

The solubility of slightly soluble salt which form a complex calculated

13

v a:1·· a·· x 10-10 l\sp .. I . . ..

~ Calculate the molar solubility of AgC! i.n 12 M NH3•

KspAgCl = 1.8x10~10 ·Kt Ag(NHa)i = 1.7x107

Ag Cl ( s) <= Ag+{aij + Cl-(aq). Arfa<tl+ 2NH3(aq) -~ Ag(NH~l(aq) Ke1.7x 107 AgCl(s) + 2NH3{aq) ¢:;:;; Ag(NH3)i+ + Cl-(aq) K= KspX Kr

=3.06 x10·3

K=[Cd(NH3}42+J[C03U= x2 = 6.76x10-s [NH3]4 (1.S-4*}4

x= 11.H x 10-2 Mi

Calculate the molar solubility of CdC03 in 1.5 M NHa. Note that Cd2+ forms the Cd(NHal/+ complex ion for which Kt is 1.3 x 107• Ksp for CdCO:i is 5.2x10-12•

CdC03(s) ¢:::== -G&!+-fa:EI1 + CQ32-(aq) Ksp=S.2x10-12 td~faf.tl+ 4NH3(aq) ¢:::== Cd(NH3)42+(aq) Ke1.3 x 107 CdC03(sJ + 4NH3 (aq) ¢:::== Cd(NH3)42+(aq)+ C032-(aq) K= Kspx Kr

=(5.2x10-12)(1.3x107) =6.76x10-5

Such an equilibrium constant is called a formation constant, K1

The stability of a complex ion in aqueous solution can be measured by Kr

For example, the equilibrium constant for formation of Ag{NH3)i+ is 1.7 x107

x. I~ 1~··1n7 ., ;:: .--~· _,_-~ ;:;::;'' ·-" x: \T (Ag L - .

~ ·-

14

4)Temperature The sofubility of slightly soluble salt increased as the result of increasing the temperature of the solution. increasing the temperature, increase the cone. Of ions in the soln. and Ksp increase.

x= !6.6x10-1 Mi

K=[Ag(NH3}i+] [Cl]= x2 = 3.06x10-3 [NH3]2 -. (12-~