Solubility Product · unsaturated or saturated. On the other hand the solubility product is applied...

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Solubility product

A solution which remains in contact with undissolved solute is said to be

saturated. At saturated stage, the quantity of the solute dissolved is

always constant for the given amount of a particular solvent at a definite

temperature.

In case, the solute is an electrolyte, its ionisation occurs in solution and

degree of dissociation depends on the concentration of dissolved

electrolyte at a particular temperature. Thus, in a saturated solution of an

electrolyte two equilibria exist and can be represented as,

SolidAB⇌

Unionised(Dissolved)

AB ⇌ ions

A B

Applying the law of mass action to the ionic equilibrium,

[A ][B ]

K[AB]

Since the solution is saturated, the concentration of unionised molecules

of the electrolyte is constant at a particular temperature, i.e., [AB] K

constant.

Hence, sp[A ][B ] K[AB] KK K (constant)

spK is termed as the solubility product. It is defined as the product of

the concentration of ions in a saturated solution of an electrolyte

at a given temperature.

Consider, in general, the electrolyte of the type x yA B which

dissociates as,

x yA B ⇌ y xxA yB

Applying law of mass action,

y x x y

x y

[A ] [B ]K

[A B ]


When the solution is saturated,

x y[A B ] K (constant) or y x x yx y sp[A ] [B ] K[A B ] KK K

(constant)

Thus, solubility product is defined as the product of concentrations of the

ions raised to a power equal to the number of times the ions occur in the

equation representing the dissociation of the electrolyte at a given

temperature when the solution is saturated.

(1) Difference between solubility product and Ionic product : Both

ionic product and solubility product represent the product of the

concentrations of the ions in the solution. The term ionic product has a

broad meaning since, it is applicable to all types of solutions, may be

unsaturated or saturated.

On the other hand the solubility product is applied only to a saturated

solution in which there exists a dynamic equilibrium between the

undissolved salt and the ions present in solution. Thus the solubility

product is in fact the ionic product for a saturated solution.

(2) Different expressions for solubility products

(i) Electrolyte of the type AB : Its ionisation is represented as,

AB ⇌ A B

Thus, spK [A ][B ]

AgCl ⇌ Ag Cl ; spK [Ag ][Cl ]

4BaSO ⇌ 4Ba SO ; sp 4K [Ba ][SO ]

(ii) Electrolyte of the type AB2 : Its ionisation is represented as,

2AB ⇌ 2A 2B

Thus, 2 2spK [A ][B ]


2PbCl ⇌ 2Pb 2Cl ; 2 2spK [Pb ][Cl ]

2CaF ⇌ 2Ca 2F ; 2 2spK [Ca ][F ]

(iii) Electrolyte of the type A2B : Its ionisation is represented as,

2A B ⇌ 22A B

Thus, 2 2spK [A ] [B ]

2 4Ag CrO ⇌ 242Ag CrO ; 2 2

sp 4K [Ag ] [CrO ]

2H S ⇌ 22H S ; 2 2spK [H ] [S ]

(iv) Electrolyte of the type A2B3 : Its ionisation is represented as,

2 3A B ⇌ 3 22A 3B

Thus, 2 2spK [A ] [B ]

2 3As S ⇌ 3 22As 3S ; 3 2 2 3spK [As ] [S ]

2 3Sb S ⇌ 3 22Sb 3S ; 3 2 2 3spK [Sb ] [S ]

(v) Electrolyte of the type AB3 : Its ionisation is represented as,

3AB ⇌ 3A 3B

Thus, 3 3spK [A ][B ]

3Fe(OH) ⇌ 3Fe 3OH ; 3 3spK [Fe ][OH ]

3AlI ⇌ 3Al 3I ; 3 3spK [Al ][I ]

(3) Criteria of precipitation of an electrolyte : For the precipitation of

an electrolyte, it is necessary that the ionic product (the product of the

ions) must exceed its solubility product. For example, if equal volumes of

30.02M AgNO solution and 2 40.02M K CrO solution are mixed, the

precipitation of 2 4Ag CrO occurs as the ionic product exceeds the solubility

product of 2 4Ag CrO which is 122 10 .


In the resulting solution, 20.02[Ag ] 0.01 1 10 M

2

and 2 24

0.02[CrO ] 0.01 1 10 M

2

Ionic product of 2 22 4 4Ag CrO [Ag ] [CrO ] 2 2 2(1 10 ) (1 10 )

61 10

61 10 is higher than 122 10 and thus precipitation of 2 4Ag CrO

occurs.

(4) Relationship between solubility and solubility product : Salts

like 4 4 2AgI, BaSO , PbSO , PbI etc., are ordinarily considered insoluble but

they do possess some solubility. These are sparingly soluble electrolytes. A

saturated solution of sparingly soluble electrolyte contains a very small

amount of the dissolved electrolyte. It is assumed that whole of the

dissolved electrolyte is present in the form of ions, i.e., it is completely

dissociated.

The equilibrium for a saturated solution of any sparingly soluble salt may

be expressed as,

x yA B ⇌ y xxA yB

Thus, solubility product, y x x yspK [A ] [B ]

Let '' s moles per litre be the solubility of the salt, then

x yA B ⇌ y xxA yB

xs ys

So, x yspK [xs] [ys] x y x yx y (s)

(i) 1 : 1 type salts : Examples : 4 4AgCl, AgI, BaSO , PbSO , etc.,

AB ⇌ A B


Let solubility of AB be s moles 1litre . So, 2spK [A ][B ] s s s or

spKs

(ii) 1 : 2 or 2 : 1 type salts : Examples :

2 3 2 4 2 2Ag CO , Ag CrO , PbCl , CaF , etc.

2AB ⇌ 2

s 2sA 2B

Let solubility of 2AB be s moles 1litre . So, 2 2 2 3spK [A ][B ] s (2s) 4s

or sp3s K /4

Now, 2A B ⇌ 2

2s s2A B

Let s be the solubility of 2A B. 2 2spK [A ] [B ] 2 3(2s) (s) 4s or

sp3s K /4

(iii) 1:3 types salts : Examples : 3 3 3 3AlI ,Fe(OH) , Cr(OH) , Al(OH) ,

etc.,

3AB ⇌ 3

s 3sA 3B

Let s moles litres–1 be the solubility of 3AB . So,

3 3 3 4spK [A ][B ] s (3s) 27s or sp4s (K /27)

The presence of common ion affects the solubility of a salt. Let AB be a

sparingly soluble salt in solution and A B be added to it. Let s and s be

the solubilities of the salt AB before and after addition of the electrolyte

A B . Let C be the concentration of A B .

Before addition of A B , 2spK s …..(i)

After addition of A B , the concentration, of A and B ions become s and

(s C) , respectively.


So, spK s (s C) …..(ii)

Equating eqs. (i) and (ii),

2s s (s C)

Solubility products of some common sparingly soluble salts at

298 K

Salt Ksp Salt Ksp

AgCl 101.7 10 CdS 283.6 10

2PbCl 51.6 10 HgS 534.1 10

3CaCO 94.7 10 ZnS 222.5 10

4BaSO 101.0 10 PbS 283.4 10

3SrCO 107.0 10 SnS 261.18 10

3Al(OH) 238.5 10 MnS 151.4 10

3Fe(OH) 381.1 10 FeS 181.5 10

2Mg(OH) 111.4 10

(5) Applications of solubility product

(i) In predicting the formation of a precipitate

Case I : When spipK K , then solution is unsaturated in which more

solute can be dissolved. i.e., no precipitation.

Case II : When spipK K , then solution is saturated in which no more

solute can be dissolved.

Case III : When spipK K , then solution is supersaturated and

precipitation takes place.


When the ionic product exceeds the solubility product, the equilibrium

shifts towards left-hand side, i.e., increasing the concentration of

undissociated molecules of the electrolyte. As the solvent can hold a fixed

amount of electrolyte at a definite temperature, the excess of the

electrolyte is thrown out from the solutions as precipitate.

(ii) In predicting the solubility of sparingly soluble salts : Knowing

the solubility product of a sparingly soluble salt at any given temperature,

we can predict its solubility.

(iii) Purification of common salt : HCl gas is circulated through the

saturated solution of common salt. HCl and NaCl dissociate into their

respective ions as,

NaCl ⇌ Na Cl ; HCl ⇌ H Cl

The concentration of Cl ions increases considerably in solution due to

ionisation of HCl . Hence, the ionic product [Na ][Cl ] exceeds the

solubility product of NaCl and therefore pure NaCl precipitates out from

solution.

(iv) Salting out of soap : From the solution, soap is precipitated by the

addition of concentrated solution of NaCl .

Soap

RCOONa ⇌ RCOO Na ; NaCl ⇌ Na Cl

Hence, the ionic product ]][[ NaRCOO exceeds the solubility product of

soap and therefore, soap precipitates out from the solution.

(v) In qualitative analysis : The separation and identification of various

basic radicals into different groups is based upon solubility product

principle and common ion effect.

(a) Precipitation of group first radicals (Pb+2, Ag+ and Hg+2) : The group

reagent is dilute HCl .


sp[Ag ][Cl ] K for AgCl

The first group radicals are precipitated out by dil. HCl , because the ionic

products of the chlorides of these ions exceed their corresponding

solubility products.

(b) Precipitation of group second radicals (Hg+2, Pb+2, Bi+3, Cu+2, Cd+2,

As+3, Sb+3 and Sn+2) : The group reagent is 2H S in presence of dilute HCl .

2(Weak electrolyte)

H S ⇌ 22H S ; (Strong electrolyte)

HCl ⇌ H Cl

HCl suppresses ionisation of weakly dissociated SH 2 . With the result only

the ionic product of the sulphides of group-II radicals exceed their

corresponding solubility product and hence only these are precipitated out

leaving the sulphides of III, IV etc. group in solution. Their solubility

products are more.

Group

II

Metal

sulphide

Solubility

product

Group reagent

H2S in acidic

medium

2 3Bi S 721.6 10

HgS 544 10

CuS 441 10

PbS 295 10

CdS 281.4 10

Group

IV

CoS 263 10

H2S in basic

medium

NiS 241.4 10

ZnS 221.0 10

MnS 151.4 10


(c) Precipitation of group third radicals (Fe+3, Al+3 and Cr+3) : The group

reagent is OHNH 4 in presence of ClNH 4 .

4(Weak electrolyte)

NH OH ⇌ 4NH OH ; 4(Strong electrolyte)

NH Cl ⇌ 4NH Cl

Addition of ClNH 4 in group III suppresses the ionisation of OHNH 4 with

the result ionic products of only the group III radicals exceed their

corresponding solubility products and hence other group (IV, V etc)

radicals remain in solution while the III group radicals are precipitated

out as their hydroxides.

(d) Precipitation of group fourth radicals (Co+2, Ni+2, Mn+2 and Zn+2) : The

group reagent is 2H S in presence of 4NH OH.

2H S ⇌ 22H S ; 4NH OH ⇌ 4NH OH ; H OH ⇌ 2H O

OH ions from 4NH OH combine with the H ions from 2H S to form

unionisable water with the result more 2H S is ionised to maintain the

equilibrium (Le-chatelier principle). Thus the concentration of 2S ions

will be high in the solution and thus a stage will be reached when the

ionic product of group IV radicals and sulphide ions exceeds the

corresponding solubility product and thus IV group radicals are

precipitated out.

(e) Precipitation of group fifth radicals (Ba+2, Sr+2, Ca+2) : The group

reagent is ammonium carbonate in presence of 4NH Cl and 4NH OH.

4 2 3(Weak electrolyte)

(NH ) CO ⇌ 24 32NH CO ; 4

(Strong electrolyte)

NH Cl 4NH Cl

Thus due to common ion 4(NH ) , ionisation of 324 )( CONH is suppressed and

thus only the Vth group carbonates having low solubility products will be

precipitated out.


4NH OH is also added during precipitation of the Vth group radicals

because it converts 4 3NH HCO (present in large amount with ammonium

carbonate) into ammonium carbonate otherwise soluble bicarbonates of

group Vth radicals will be formed.

4 3 4 4 2 3 2NH HCO NH OH (NH ) CO H O

(vi) Precipitation of calcium oxalate from calcium acetate : The

solubility product provides an answer why oxalic acid precipitates calcium

oxalate from calcium acetate solution but not from calcium chloride or

calcium nitrate solutions. The reaction between calcium acetate and oxalic

acid can be represented as,

3 2 2 2 4 2 4 3Ca(CH COO) H C O CaC O 2CH COOH

The concentration of 22 4[C O ] ions is sufficient to make the ionic product

2 22 4[Ca ][C O ] greater than the solubility product of calcium oxalate. The

acetic acid formed is a weak acid and this does not affect the ionisation of

oxalic acid. In the case of 2CaCl or 3 2Ca(NO ) , the acids formed are strong

acids which ionise completely in solution. The common H ions suppress

the ionisation of oxalic acid, thereby, decreasing the concentration of

oxalate ions. Under this condition the ionic product does not exceed the

solubility product and thus, precipitation of calcium oxalate does not

occur.

(vii) Calculation of remaining concentration after precipitation :

Sometimes an ion remains after precipitation if it is in excess. Remaining

concentration can be determined,

Example : spleft

K [AB][A ]

[B ]

; sp 22left 2

K [Ca(OH) ][Ca ]

[OH ]


In general sp m nn mm nleft

K [A B ][A ]

[B ]

Percentage precipitation of an ion = Initial conc. - Remaining conc.100

Initial conc.

(viii) Simultaneous Solubility : Solubility of two electrolytes having

common ion; when they are dissolved in the same solution, is called

simultaneous solubility, e.g.,

(a) Solubility of AgBr and AgSCN, when dissolved together.

(b) Solubility of 2CaF and 2SrF , when dissolved together.

(c) Solubility of 2MgF and 2CaF when dissolved together.

Calculation of simultaneous solubility is divided into two cases.

Case I : When the two electrolytes are almost equally strong (having close

solubility product).

e.g., 13spAgBr (K 5 10 ) ; 12

spAgSCN (K 10 )

Here, charge balancing concept is applied.

Charge of Ag = Charge of Br + Charge of SCN

[Ag ] = [Br ] + [SCN ]

(a b) = a b

Case II : When solubility products of two electrolytes are not close, i.e.,

they are not equally strong.

e.g., 11sp2CaF (K 3.4 10 ) ; 9

sp2SrF (K 2.9 10 )

Most of fluoride ions come of stronger electrolyte.


Example 1 : If the concentration of lead iodide in its saturated

solution at Co25 be 32 10 moles per litre, then its solubility product is

(a) 64 10 (b) 128 10 (c) 96 10 (d) 932 10

Solution: (d) 2S

PbI ⇌ S 2S

Pb 2I

3 3 3spK 4S 4 [2 10 ] 932 10

Example 2 : How many grams of 2 4CaC O (molecular weight = 128)

on dissolving in distilled water will give one litre saturated solution.

9 2 2sp 2 4[K (CaC O )] 2.5 10 mol l

(a) 0.0064 g (b) 0.01280 g (c) 0.0128 g (d) 1.2800 g

Solution: (a) 9ss K 2.5 10 55 10 M

Weight of 5 12 4CaC O 5 10 128 gL 0.0064 g

Example 3 : In a saturated solution of calcium phosphate, the

concentration of 34PO ions is 73.3 10 M . The spK of 3 4 2Ca (PO ) will be

(a) 311.32 10 (b) 321.32 10 (c) 331.32 10 (d) 351.32 10

Solution: (b) 2 33 4 2 4Ca (PO ) 3Ca 2PO

2 3 74

3 3[Ca ] [PO ] (3.3 10 )M

2 2 74.95 10 M

2 3 3 2sp 4K [Ca ] .[PO ] 7 3 7 2(4.95 10 ) (3.3 10 ) 321.32 10

Examples based on Solubility product and simultaneous

solubility


Example 4 : If the solubility product spK of a sparingly soluble salt

2MX at o25 C is 111.0 10 , the solubility of the salt in mole litre–1 at this

temperature will be

(a) 142.46 10 (b) 41.36 10 (c) 72.60 10 (d) 101.20 10

Solution: (b) 2MX ⇌2

2

(S) (2S)

M 2X

3spK 4S

spKS 3

4

111 103

4

41.36 10

Example 5 : A solution has 20.05M Mg and 30.05M NH . Calculate

the concentration of 4NH Cl required to prevent the formation of 2Mg(OH)

in this solution. spK of 122Mg(OH) 9.0 10 and ionisation constant of

53NH 1.8 10

(a) 6.7 M (b) 0.67 M (c) 0.067 M (d) 0.0067 M

Solution: (c) The maximum concentration of [OH ] ions that will

precipitate 2Mg(OH) is calculated by applying the equation,

2 2spK [Mg ][OH ]

12

sp2 102

K 9.0 10[OH ] 1.8 10

0.05[Mg ]

or 5[OH ] 1.34 10 M

3NH is present in solution in the form of OHNH 4

3 20.05

NH H O ⇌ 40.05

NH OH ⇌ 4NH OH

The ionisation of OHNH 4 is suppressed by the addition of ClNH 4 (Strong

electrolyte)

3 4

4NH NH OH

4

[NH ][OH ]K K

[NH OH]


Whole of the concentration of 4

NH ions is provided by 4NH Cl

4 4NH OH

4

K [NH OH][NH ]

[OH ]

5

5

1.8 10 0.050.067M

1.34 10

;

i.e., 4[NH Cl] 0.067M .

Example 6 : Calculate simultaneous solubility of AgCNS and AgBr in a

solution of water. 12spK AgCNS 1 10 , 13

spK AgBr 5 10

(a) 7 18.16 10 mol L (b) 6 17.39 10 mol L

(c) 12 153.2 10 mol L (d) 22 190 10 mol L

Solution: (a) Let the solubility of AgCNS and AgBr is water be a

and b respectively.

AgCNS ⇌ aa

Ag CNS ; AgBr ⇌ bb

Ag Br

[Ag ] a b, [CNS ] a and [Br ] b

spK AgCNS [Ag ][CNS ] a(a b)

121 10 a(a b) …..(i)

spK AgBr [Ag ][Br ] b(a b)

135 10 b(a b) …..(ii)

Dividing equation (i) by (ii),

12

13

1 10 ab5 10

;

a2

b or a 2b

Putting the value of a in equation (i),

2 126b 1 10 ; 2 121b 10

6 ; 7 1b 4.08 10 mol L

7a 2 4.08 10 7 18.16 10 mol L

Solubility Product · unsaturated or saturated. On the other hand the solubility product is applied...

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Transcript of Solubility Product · unsaturated or saturated. On the other hand the solubility product is applied...