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www.thechemistryguru.com Solubility Product
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 ]
www.thechemistryguru.com Solubility Product
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 ]
www.thechemistryguru.com Solubility Product
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 .
www.thechemistryguru.com Solubility Product
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
www.thechemistryguru.com Solubility Product
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.
www.thechemistryguru.com Solubility Product
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.
www.thechemistryguru.com Solubility Product
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 .
www.thechemistryguru.com Solubility Product
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
www.thechemistryguru.com Solubility Product
(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.
www.thechemistryguru.com Solubility Product
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 ]
www.thechemistryguru.com Solubility Product
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.
www.thechemistryguru.com Solubility Product
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
www.thechemistryguru.com Solubility Product
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]
www.thechemistryguru.com Solubility Product
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