Ionic Equilibrium : 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,

\[\underset{Solid}{\mathop{AB}}\,\rightleftharpoons \underbrace{\underset{\begin{smallmatrix} Unionised \\ (Dissolved)\end{smallmatrix}}{\mathop{AB}}\,\rightleftharpoons \underset{ions}{\mathop{{{A}^{+}}+{{B}^{-}}}}\,}_{{}}

Applying the law of mass action to the ionic equilibrium,

\[\frac{[{{A}^{+}}][{{B}^{-}}]}{[AB]}=K

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, [A+][B–] = K[AB] = KK’ = Ksp (constant)

Ksp 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 AxBy which dissociates as,

\[{{A}_{x}}{{B}_{y}}\rightleftharpoons x{{A}^{y+}}+y{{B}^{x-}}  Applying law of mass action,

\[\frac{{{[{{A}^{y+}}]}^{x}}{{[{{B}^{x-}}]}^{y}}}{[{{A}_{x}}{{B}_{y}}]}=K

When the solution is saturated,

[AxBy] = K’ (constant) or [Ay+]x[Bx–]y = K[AxBy] = KK’ = Ksp (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 board 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,         Ksp = [A+][B–]

                      AgCl ⇌ Ag+ + Cl– ; Ksp = [Ag+][Cl–]

                      BaSO4 ⇌ Ba++ + SO4– ; Ksp = [Ba++][SO4–]

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

                      AB2 ⇌ A2+ + 2B–

Thus,         Ksp = [A2+][B–]2

                      PbCl2 ⇌ Pb2+ + 2Cl– ; Ksp = [Pb2+][Cl–]2

                      CaF2 ⇌ Ca2+ + 2F– ; Ksp = [Ca2+][F–]2

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

                      A2B ⇌ 2A+ + B2–

Thus,         Ksp = [A+][B2–]

                      Ag2CrO4 ⇌ 2Ag+ + CrO42– ; Ksp = [Ag+]2[CrO42–]

                      H2S ⇌ 2H+ + S2– ; Ksp = [H+]2[S2–]                           

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

                      A2B3 ⇌ 2A3+ + 3B2–

Thus,         Ksp = [A+]2[B2–]

                      As2S3 ⇌ 2A3+ + 3S2– ; Ksp = [As3+]2[S2–]3

                      Sb2S3 ⇌ 2Sb3+ + 3S2– ; Ksp = [Sb3+]2[S2–]3

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

                      AB3 ⇌ A3+ + 3B–

Thus,         Ksp = [A3+][B–]3

                      Fe(OH)3 ⇌ Fe3+ + 3OH– ; Ksp = [Fe3+][OH–]3

                      AlI3 ⇌ Al3+ + 3I– ; Ksp = [Al3+][I–]3

(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 0.02M AgNO3 solution and 0.02M K2CrO4 solution are mixed, the precipitation of Ag2CrO4 occurs as the ionic product exceeds the solubility product of Ag2CrO4 which is 2 × 10–12.

In the resulting solution, [Ag+] = \[\frac{0.02}{2} = 0.01 = 1 × 10–2 M and [CrO42–] = \[\frac{0.02}{2} = 0.01 = 1 × 10–2 M and Ionic product of Ag2CrO4 = [Ag+]2[CrO42–] = (1 × 10–2)2(1 × 10–2) = 1 × 10–6

1 × 10–6 is higher than 2 × 10–12 and thus precipitation of Ag2CrO4 occurs.

(4)     Relationship between solubility and solubility product: Salts like AgI, BaSO4, PbSO4, PbSO4, PbI2 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,

                                AxBy ⇌ xAy+ + yBx–

Thus, solubility product,     Ksp = [Ay+]x[Bx–]y

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

                                AxBy ⇌ xAy+ + yBx–

So,              Ksp = [xs]x[ys]y = xx × yy (s) x+y

(i)      1 : 1 type salts : Examples : AgCl, AgI, BaSO4, PbSO4, etc.,

                             AB ⇌ A+ + B–

Let solubility of AB be s moles litre–1. So, Ksp = [A+][B–] = s × s = s2 or $s=\sqrt{{{K}_{sp}}}$ 

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

Ag2CO3, AgCrO4, PbCl2, CaF2, etc.

\[A{{B}_{2}}\rightleftharpoons A_{s}^{2+}+\underset{2s}{\mathop{2{{B}^{-}}}}\,

Let solubility of AB2 be s moles litre–1. So,

Ksp = [A2+][B–]2 = s × (2s)2 = 4s3 or  \[s=3\sqrt{{{K}_{sp}}/4}

Now,   \[{{A}_{2}}B\rightleftharpoons 2A_{2s}^{+}+\underset{s}{\mathop{{{B}^{2-}}}}\,

Let  be the solubility of A2B. Ksp = [A+]2[B2–] = (2s)2(s) = 4s3 or  \[s=3\sqrt{{{K}_{sp}}/4}

(iii)    1:3 types salts : Examples : AlI3, Fe(OH)3, Cr(OH)3, Al(OH)3 etc.,

                         \[A{{B}_{3}}\rightleftharpoons A_{s}^{3+}+\underset{3s}{\mathop{3{{B}^{-}}}}\,

Let  moles litres–1 be the solubility of AB3. So, 

Ksp = [A3+][B–]3 = s × (3s)3 = 27s4 or \[s=4\sqrt{({{K}_{sp}}/27)}

The presence of common ion affects the solubility of a salt. Let AB be a sparingly soluble salt in solution and AB 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  be the concentration of A’B.

Before addition of A’B,        Ksp = s2               …..(i)

After addition of A’B, the concentration, of  A+ and B– ions become  and (s’ +C), respectively.

So,                       Ksp = s’(s’ + C)                        …..(ii)

Equating eqs. (i) and (ii),

                                s2 = s’ (s’ + C)

Solubility products of some common sparingly soluble salts at 298 K

Salt Ksp Salt Ksp
AgCl 1.7 × 10–10 CdS 3.6 × 10–28
PbCl2 1.6 × 10–5 HgS 4.1 × 10–53
CaCO3 4.7 × 10–9 ZnS 2.5 × 10–22
BaSO4 1.0 × 10–10 PbS 3.4 × 10–28
SrCO3 7.0 × 10–10 SnS 1.18 × 10–26
Al(OH)3 8.5 × 10–23 MnS 1.4 × 10–15
Fe(OH)3 1.1 × 10–38 FeS 1.5 × 10–18
Mg(OH)2 1.4 × 10–11

 

(5)     Applications of solubility product

(i)      In predicting the formation of a precipitate

Case I : When Kip < Ksp, then solution is unsaturated in which more solute can be dissolved. i.e., no precipitation.

Case II : When Kip = Ksp, then solution is saturated in which no more solute can be dissolved.

Case III : When Kip < Ksp, 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 .

\[\underset{\text{Soap}}{\mathop{RCOONa}}\,\rightleftharpoons RCO{{O}^{-}}+N{{a}^{+}}\,\,;\,\,NaCl\rightleftharpoons N{{a}^{+}}C{{l}^{-}}

Hence, the ionic product [RCOO–][Na+] 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.

                      [Ag+][Cl–] > Ksp 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 H2S in presence of dilute HCl.

\[\underset{Weak\,\,electrolyte}{\mathop{{{H}_{2}}S}}\,\rightleftharpoons 2{{H}^{+}}+{{S}^{-2}}\,\,;\,\,\underset{(Strong\,\,electrolyte)}{\mathop{HCl}}\,\rightleftharpoons {{H}^{+}}+C{{l}^{-}}

HCl suppresses ionisation of weakly dissociated H2S. 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
Bi2S3 1.6 × 10–72
HgS 4 × 10–54
CuS 1 × 10–44
PbS 5 × 10–29
CdS 1.4 × 10–28
 

 

Group IV

 CoS 3 × 10–26 H2S in basic medium
NiS 1.4 × 10–24
ZnS 1.0   10–22
MnS 1.4 × 10–15

 

(c)      Precipitation of group third radicals (Fe+3, Al+3 and Cr+3) : The group reagent is NH4OH in presence of NH4Cl.

\[\underset{\text{(Weak}\text{electrolyte)}}{\mathop{N{{H}_{4}}OH}}\,\rightleftharpoons NH_{4}^{+}+O{{H}^{-}}\,\,;\,\,\,\underset{(Strong\,electrolyte)}{\mathop{N{{H}_{4}}Cl}}\,\,\,\rightleftharpoons \,\,NH_{4}^{+}+C{{l}^{-}}   

Addition of NH4Cl in group III suppresses the ionisation of NH4OH 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 H2S in presence of NH4OH.

          H2S ⇌ 2H+ + S–2 ; NH4OH ⇌ NH4+ + OH– ; H+ + OH– ⇌ H2O

OH– ions from NH4OH combine with the H+ ions from H2S to form unionisable water with the result more H2S is ionised to maintain the equilibrium (Le-chatelier principle). Thus the concentration of S–2 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 NH4Cl and NH4OH.

\[\underset{\text{(Weak}\text{electrolyte)}}{\mathop{{{(N{{H}_{4}})}_{2}}C{{O}_{3}}}}\,\rightleftharpoons 2NH_{4}^{+}+CO_{3}^{-2}\,\,;\,\,\,N{{H}_{4}}Cl\,\,\to \,\,NH_{4}^{+}+C{{l}^{-}}

Thus due to common ion \[(NH_{4}^{+}), ionisation of  ${{(N{{H}_{4}})}_{2}}C{{O}_{3} is suppressed and thus only the Vth group carbonates having low solubility products will be precipitated out.

NH4OH is also added during precipitation of the Vth group radicals because it converts NH4HCO3 (present in large amount with ammonium carbonate) into ammonium carbonate otherwise soluble bicarbonates of group Vth radicals will be formed.

             NH4HCO3 + NH4OH → (NH4)2 CO3 + H2O

(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,

            Ca(CH3COO)2 + H2C2O4 → CaC2O4 + 2CH3COOH                 

The concentration of [C2O42–] ions is sufficient to make the ionic product [Ca2+][C2O42–] 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 CaCl2 or Ca(NO3)2, 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 :  \[{{[{{A}^{+}}]}_{left}}=\frac{{{K}_{sp}}[AB]}{[{{B}^{-}}]}  ;  \[{{[C{{a}^{2+}}]}_{left}}=\frac{{{K}_{sp}}[Ca{{(OH)}_{2}}]}{{{[O{{H}^{-}}]}^{2}}}        

In general   \[[{{A}^{n+}}]_{left}^{m}=\frac{{{K}_{sp}}[{{A}_{m}}{{B}_{n}}]}{{{[{{B}^{m-}}]}^{n}}}

Percentage precipitation of an ion = \[\left[ \frac{\text{Initial conc}\text{. }-\text{ Remaining conc}\text{.}}{\text{Initial conc}\text{.}} \right]\times 100

(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 CaF2 and SrF2, when dissolved together.

(c)      Solubility of MgF2 and CaF2 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., AgBr (Ksp = 5 × 10–13) ; AgSCN (Ksp = 10–12)

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., CaF (Ksp = 3.4 × 10–11) ; SrF2 (Ksp = 2.9 × 10–9)

Most of fluoride ions come of stronger electrolyte.