Solubility Product 

 

Table of Content

Different Expressions for Solubility Products Solubility Product of a Weak Electrolyte Common Ion Effect Applications of Solubility Product Significance of Stability Constant Solubility of Metal Hydroxides in Acid Medium Relationship between Solubility and Solubility Product Calculation of Remaining Concentration After Precipitation Simultaneous Solubility Ostwald’s Dilution Law Limitations of Ostwald's Dilution Law Related Resources

If to a given amount of solvent at a particular temperature, a solute is added gradually in increasing amounts, a stage is reached when some of the solute remains undissolved, no matter how long we wait or how vigorously we stir. The solution is then said to be saturated. 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.

Refer to the following video for dissolution of salt in water

Thus, in a saturated solution of an electrolyte two equilibria exist and can be represented as:

AB    AB             

Solid  unionized           ions

  (dissolved)

Applying the law of action to the ionic equilibrium,

 

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’ = K_s (constant)

K_s 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 A_xB_y which is dissociated as:

A_xB_y  xA^y+ + yB^x-

Applying law of mass action,

When the solution is saturated,

[A_xB_y] = K' (constant)

or   A^y+]^x[B^x-]^y = K [A_xB_y] = KK' = K_s (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.

Note:  Solubility product is not the ionic product under all conditions but only when the solution is saturated.
 

• Different Expressions for Solubility Products

Refer to the following video for solubility product

(i) Electrolyte of the type AB:

Its ionisation is represented as:

            AB   A⁺ + B^-

Thus,     K_s = [A⁺][B^-]

AgCl   Ag⁺ + Cl^-;               K_s = [Ag⁺][Cl^-]

BaSO₄  Ba²⁺ + SO^-2₄ ;     K_s = [Ba²⁺][SO^-2₄]

(ii)  Electrolyte of the type AB₂:

Its ionisation is represented as:

            AB₂  A²⁺ + 2B^-

Thus,     K_s = [A²⁺][B^-]²

PbCl₂  Pb²⁺ + 2Cl^-;          K_s = [Pb²⁺][Cl^-]²

CaF₂  Ca²⁺ + 2F^-;            K_s = [Ca²⁺][F^-]²

(iii)  Electrolyte of the type A₂B:

Its ionisation is represented as:

            A₂B ↔  2A²⁺ + B^2-

Thus,     K_s = [A⁺]²[B^2-]

Ag₂CrO₄  2Ag⁺ + CrO^-2₄;    K_s = [Ag⁺]²[CrO^-2₄]

H₂S   2H⁺ + S2^-;                    K_s = [H⁺]²[S^2-]

(iv)   Electrolyte of the type A₂B₃:

Its ionisation is represented as:

            A₂B₃  2A³⁺ + 3B^2-

Thus,     K_s = [A³⁺]²[B^2-]³

As₂S₃  2As³⁺ + 3S^2-;

K_s = [As³⁺]²[S^2-]³

Sb₂S₃  2Sb³⁺ + 3S2^-;

K_s = [Sb³⁺]²[S^2-]³

(v)   Electrolyte of the type AB₃:

Its ionisation is represented as:

           AB₃  A³⁺ + 3B^-

Thus,     K_s = [A³⁺][B^2-]³

Fe(OH)₃    Fe³⁺ + 3OH^-;     K_s = [Fe³⁺][OH^-]³

AH₃    Al³⁺ + 3l^-;      K_s = [Al³⁺][I^-]³
 

• Solubility Product of a Weak Electrolyte

Let degree of ionization of weak electrolyte A_mB_n be 'α'.

                        A_mB_n  mAⁿ⁺ + nB^m-

t = 0                  s          0          0

t_eq                    s-sα     msα      nsα

                        K_sp = [Aⁿ⁺]^m[B^m-]ⁿ

                        = [msα ]^m[nsα]ⁿ

                        K_sp = m^mnⁿ (sα)^m+n

Criteria of precipitation of an electrolyte:

A very useful conclusion is derived from the solubility product concept. No precipitation of the electrolyte occurs if the ionic product is less than the solubility product, i.e., the solution has not reached the saturation stage.

Case I:    When K_ip<K_sp, then solution is unsaturated in which more solute can be dissolved.

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

Case III:  When K_ip > K_sp, 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 solution as precipitate.

Thus, for the precipitation of an electrolyte, it is neces­sary that the ionic product must exceed its solubility product For example, if equal volumes of 0.02 M AgN0₃solution and 0.02 M K₂Cr0₄ solution are mixed, the precipita­tion of Ag₂Cr0₄ occurs as the ionic product exceeds the solubility product of Ag₂Cr0₄ which is 2 × 10^-12

In the resulting solution,

[Ag⁺] = 0.02/2 = 0.01 = 1*10^-2 M

and  [CrO^2-₄ ]= 0.02/2 = 0.01 = 1*10^-2 M

Ionic product of Ag₂CrO₄ = [Ag⁺]² [CrO^2-₄]

                                     = (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.
 

• Common Ion Effect

 Let AB to the weak electrolyte. Considering its dissociation,

AB  A⁺ + B^-

and applying law of mass action we have

 K = [A⁺][B^-]/[AB]

The equilibrium constant, K, has a definite value at any given temperature. If now another electrolyte furnishing the A⁺ and B^- ions be added to the above solution. It will increases the concentration of either A⁺ ions or B^- ions (whichever has been added) and in order that K may remain constant, the concentration of AB must increases, i.e., the equilibrium will shift to the left hand side.

In other words, the degree of dissociation of an electrolyte (weak) is suppressed by the addition of another electrolyte (strong) containing a common ion. This is termed as common ion effect. Acetic acid is a weak electrolyte and its ionization is suppressed in presence of a strong acid (H⁺ ion as common ion) or a strong salt like sodium acetate (acetate ion as common ion). Similarly, the addition of NH₄Cl or NaOH to NH₄OH solution will suppress the dissociation of NH₄OH due to common ion either NH^-₄ or OH^-.

As a result of common ion effect, the concentration of the ion not in common in two electrolytes, is decreased. The use of this phenomenon is made in qualitative analysis to adjust concentration of S^2- ions in second group and OH^- ion concentration in third group of analysis.

 

• Applications of Solubility Product

(i)  Purification of common salt

Natural common salt consists of many insoluble and soluble impurities. Saturated solution of common salt is prepared and insoluble impurities are filtered off. Hydrogen chloride gas (HCl) is circulated through the saturated solution. 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 ionization HCl. Hence, the ionic product [Na⁺][Cl^-] exceeds the solubility product of sodium chloride and, therefore, pure sodium chloride precipitates out from solution.

(ii)  Salting out of soap

Soap is a sodium salt of higher acids. From the solution, soap is precipitated by the addition of concentrated solution of sodium chloride. Soap and sodium chloride are present in the form of ions.

C_nH_2n+1 COONa ↔  C_nH_2n+1 COO^- + Na⁺

             Soap

NaCl ↔  Na⁺ + Cl^-

Thus, the concentration of Na+ ions increases considerably on addition of NaCl solution. Hence, the ionic product [C_nH_2n+1COO^-] [Na⁺] exceeds the solubility product of soap and, therefore, soap precipitates out from the solution.

(iii)  Manufacture of sodium bicarbonate (baking soda):

 In Solvay's soda process. CO₂ gas is passed through ammonical brine to precipitate out NaHCO₃.

                NH₄OH + CO₂ → NH₄HCO₃

            NH₄HCO₃ + NaCl → NaHCO₃ + NH₄Cl

    NaHCO₃ is precipitated first because of its lower solubility product as compared to those of NH₄Cl, NH₃HCO₃ and NaCl.

    Thus, baking soda (NaHCO₃) can be quantifiably estimated.

(iv) Application of solubility product in quantitative analysis

1. Estimation of barium as barium sulphate:

H₂SO₄ as precipitating agent is added to the aqueous solution of BaCl₂.

BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

Precipitation of BaSO₄ takes place when its ionic product exceeds solubility product. H₂SO₄ is added in slight excess to ensure complete precipitation. Large excess of H₂SO₄ is harmful for complex formation.

2. Estimation of silver as silver chloride:

NaCl solution is added to the silver nitrate solution, slight excess of NaCl is added to ensure complete precipitation.

AgNO₃ + NaCl →  AgCl + NaBO₃

Again, precipitation of AgCl takes place when ionic product of AgCl exceeds its solubility product.

3. In a similar manner. We estimate lead as lead chromate, calcium as calcium oxalate, etc.

(i)  Precipitation of the sulphides of group II and IV

 Hydrogen sulphide is a weak electrolyte and is used for the precipitation of various sulphides of group II and IV in quantitative analysis.

It ionizes to a small extent in water:

 H₂S ↔ 2H⁺ + S^2-

Applying law of mass action,

K = ([H⁺ ]² [S^2- ])/[H₂ S]

The concentrations of S2^- ions can be decreased by increasing concentration of H⁺ions and it can be increased by decreasing concentration of H⁺ ions. In group II, lower concentration of sulphide ions is required as the solubility products of the sulphides of group II are low while higher concentration of sulphide ions is required in group IV as the solubility products of the suphides of group IV are high. The values of solubility products of various sulphides are given below.

    

           

The concentration of S2^- ions in group II is lowered by maintaining acidic medium. In the presence of HCl, the ionization of H₂S is suppressed due to common ion effect. The concentration is so adjusted that only ionic products of the sulphides of group II exceed their solubility products and, therefore, get precipitated. However, CdS has somewhat higher value. For its precipitation, dilution of the solution is done which increases ionization of H₂S and thereby increasing concentration of S2^- ions.

In group IV, higher concentration of S2^- ions is needed. This is done by changing the medium for acidic to alkaline. Ammonium hydroxide is added, the OH^- ions furnished by NH₄OH remove H⁺ ions from solution in the form of water molecules as,

                          H⁺ + OH^- ↔  H₂O

More of the ionization of H₂S occurs and, thus, concentration of S2^- ions increases. It becomes so high that ionic products of the sulphides of group IV exceed their solubility products add they get precipitated.

(ii)  Precipitation of III group hydroxides

When NH₄OH is added in presence of NH₄Cl then precipitation of II group hydroxides takes place, i.e., Al(OH)₃, Fe(OH)₃ and Cr(OH)₃ are precipitated. Solubility product of III group hydroxides is less than those of higher group hydroxides.

 NH₄Cl  →  NH⁺₄ + Cl^-

NH₄OH  ↔  NH⁺₄  + OH^-

NH⁺₄  ion furnished by NH₄Cl lowers the ionisatin of NH₄OH and hence the concentration of hydroxide ion OH^-. At low concentration of hydroxide ion only III group hydroxides precipitate.

(v) Fractional Precipitation

It is a technique of separating two or more ions from a solution by adding a reagent that precipitates first one ion and then the second.

Let us suppose 0.1 M Ba²⁺ and 0.1 M Sr²⁺ in aqueous solution. K₂CrO₄ is added as precipitating agent. K_sp BaCrO₄ is 1.2 × 10^-10 and K_sp SrCrO₄ is 3.5 × 10^-5.

[CrO^2-₄]concentration required to precipitate BaCrO₄

        = K_sp/[Ba²⁺] = 1.2 * 10^-10/0.1 = 1.2 * 10^-9

BaCrO₄ will precipitate first because it requires low concentration of CrO^2-₄  ions. On addition of chromate ions, BaCrO₄ starts precipitating when chromate ion concentration reaches 1.2 × 10^-9 M. When CrO^2-₄  ion concentration reaches upto 3.5 × 10^-4 M, then SrCrO₄ also starts precipitating.

Remaining concentration of Ba²⁺ when SrCrO₄ starts precipitation.

=(K_sp BaCrO₄)/[CrO₄^2-] = (1.2×10^-10)/(3.5×10^-4)= 3.4×10^-7 M

% remaining concentration = (3.4 * 10^-7)/0.1 * 100

                                              = 0.00034%

(vi) Stability Constant  

Let us consider dissociation of the ion FeBr⁺.

 FeBr⁺ ↔   Fe²⁺ + Br^-

Dissociation constant for above equilibria may be given as

K_d = [Fe²⁺ ][Br^- ]/[FeBr⁺ ]

Reciprocal of dissociation constant is called stability constant.

K_s = [FeBr⁺ ]/([Fe²⁺] [Br^- ])

Let us consider the formation of complex K₂Cd(CN)₄, Complex ion is Cd(CN₄^2-)  where oxidation state of central metal Cd²⁺ is (2+). Complexing process proceeds in four steps as

Here          Ks = K₁K₂K₃K₄.
 

• Significance of Stability Constant

Greater will be the value of stability constant more stable will be the complex.

Note :       

(a) If on addition of a common ion in a salt solution (sparingly soluble), formation of complex ion takes place, then ionization increases, i.e., equilibrium shifts towards right hand direction to maintain the value of K_sp constant. It means, addition of common ion in the case of complex formation increases the solubility of the sparingly soluble salt which is against the concept of common ion effect.

(b) When we add an electrolyte to another electrolyte solution having no common ion, then ionization of the later increases.

(c) For a given electrolyte solubility product is always constant at a particular temperature.
 

• Solubility of Metal Hydroxides in Acid Medium

 H⁺ ion furnished by the medium effects the solubility of metal hydroxide, say M(OH)², because of neutralization of OH^- ion by H⁺ ion.

           M(OH)²  ↔  M²⁺ + 2OH^-

           K_sp of M(OH)² = [M2⁺][OH^-]²

           [M²⁺] =  K_sp/[OH^-]²

           [H⁺] [OH^-] = K_w = 10^-14

           [OH^-]² = 10^-28/[H⁺]²

           From Eqs. (i) and (ii), we have

           [M²⁺] = K_sp[H⁺]²/10^-28
 

• Relationship between Solubility and Solubility Product

Salts like Agl, BaS0₄, PbS0₄, Pbl₂, 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:

A_xB_y ↔ xA^y+ + yB^x-

Thus, solubility product, K_s = [A^y+]^x[B^x-]^y

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

          AxBy ↔  xA^y+ + yB^x-

                       xs       ys

So                   K_s = [xs]x[ys]y

                           = x^x.y^y(s)^x+y

(i)  1:1 type salts:

Examples: AgCl, Agl, BaSO₄, PbSO₄, etc.

Binary electrolyte: AB ↔  A⁺ + B^-

                                     s      s

Let solubility of AB be s mol litre^-1.

So               K_s = [A⁺][B^-] = s × s = s²

or                s = √H_s

(ii)  1:2 or 2:1 type salts:

Examples: Ag₂CO₃, Ag₂CrO₄, PbCl₂, CaF₂, etc.

Ternary electrolyte:

AB₂ ↔ A²⁺ + 2B^-

           s       2s

Let solubility of AB₂ be s mol litre^-1.

So          K_s = [A2⁺][B^-]² = s × (2s)² = 4s³

or           s = 3√K_s/4

              A₂B ↔ 2A⁺ + B^2-

                         s      s

      Let s be the solubility of A₂B.

                  K_s = [A⁺]²[B^2-]

                      = (2s)²(s) = 4s³

        or          s = 3√K_s/4

(iii)  1: 3 type salts:

       Examples: All3, Fe(OH)₃, Cr(OH)₃, Al(OH)₃, etc.

       Quaternary electrolyte:  AB₃ ↔ A3⁺ + 3B^-

         Let s mol litre^-1 be the solubility of AB₃.

                    K_s = [A³⁺][B^-]³ = s × (3s)³ = 27s⁴

       or             s  = 4√K_s/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, K_s=s²                      ... (i) 

After addition of A'B, the concentration of A⁺ and B^- ions become s' and (s' + c), respectively.

So    K_s = s'(s' + c)                                 .... (ii)

Equating Eqs. (i) and (ii),

s² = s'(s' +c)

• Calculation of Remaining Concentration After Precipita­tion

Sometimes an ion remains after precipitation if it is in excess. Remaining concentration can be determined, e.g..

(i)    [A⁺]_left = K_sp [AB]/[B^-]

(ii)    [Ca²⁺]_left =  Ksp[Ca(OH)₂]/[OH^-]²

(iii)    [Aⁿ⁺]^m_left = K_sp[A_mB_n]/[B^m-]ⁿ

Percentage precipitation of an ion

= [Initial conc. -Left conc./Initial conc.] × 100
 

• Simultaneous Solubility

Solubility of two electrolytes having common ion; when they are dissolved in the same solution, is called simultaneous solubility, e.g.,

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

(ii)    Solubility of CaF₂ and SrF₂, when dissolved together.

(iii)   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 closesolubility product), e.g.,

AgBr(K_sp = 5 x 10^-13);    AgSCN (K_sp = 10^-12)

(See Example 23)

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.,

CaF₂ (K_sp = 3.4 x 10^-11);    SrF₂ (K_sp = 2.9 x 10^-9)

Most of fluoride ions come of stronger electrolyte.
 

• Ostwald’s Dilution Law

According to Arrhenius theory of electrolyte dissociation, the molecules of an electrolyte in solution are constantly splitting up into ions and the ions are constantly reuniting to form unionized molecules. Therefore, a dynamic equilibrium exists between ions and unionized molecules of the electrolyte in solution. It was pointed out by Ostwald that like chemical equilibrium, law of mass action van be applied to such systems also.

Consider a binary electrolyte AB which dissociates into A⁺ and B- ions and the equilibrium state is represented by the equation:

                         AB   A⁺ + B^-

Initially t = 0       C         0     0

At equilibrium     C(1-α)   Cα    Cα

So, dissociation constant may be given as

K = [A⁺][B^-]/[AB] = (Cα * Cα)/C(1-α) =   Cα² /(1-α)                                    

For very weak electrolytes,

α <<< 1,  (1 - α ) = 1

 .·.                    K = Cα²

                      α = √K/C 

Concentration of any ion = Cα = √CK .

From equation (ii) it is a clear that degree of ionization increases on dilution.

Thus, degree of dissociation of a weak electrolyte is proportional to the square root of dilution.
 

Limitations of Ostwald's Dilution Law

The law holds good only for weak electrolytes and fails completely in the case of strong electrolytes. The value of 'α' is determined by conductivity measurements by applying the formula Λ/Λ_∞. The value of 'α' determined at various dilutions of an electrolyte when substituted in Eq. (i) gives a constant value of K only in the case of weak electrolytes like CH₃COOH, NH₄OH, etc. the cause of failure of Ostwald's dilution law in the case of strong electrolytes is due to the following factors"

• The law is based on the fact that only a portion of the electrolyte is dissociated into ions at ordinary dilution and completely at infinite dilution. Strong electrolytes are almost completely ionized at all dilutions and Λ/Λ_∞ does not give accurate value of 'α'.

• When concentration of the ions is very high, the presence of charges on the ions appreciably effects the equilibrium. Hence, law of mass action its simple form cannot be strictly applied in the case of string electrolytes.

Example 1:  The ionization constant of HCN is 4 × 10^-10. Calculate the concentration of hydrogen ions in 0.2 M solution of HCN containing 1 mol L^-1 of KCN?

Solution:  

The dissociation of HCN is represented as

 HCN ↔ H⁺ + CN^-

Applying law of mass action,

 K_a =   ([H⁺ ][CN^-])/[HCN] or [H⁺ ] (K_a [HCN])/[CN^- ]

In presence of strong electrolyte, the total CN^- concentration comes from KCN which undergoes complete dissociation. It is further assumed that dissociation of HCN is very-very small and the concentration of HCN can be taken as the concentration of undissociated HCN.

Thus, [HCN] = 0.2 M and [CN^-] = 1M

Putting these values in the expression

[H⁺] = (K_a [HCN])/([CN^- ]) = (4×10^-10×0.2)/1 = 8×10^-11 mol L^-1

[Note: When KCN is not present, the [H+] concentration is equal to √CK i.e., √(0.2*4*10^-10) = 8.94 *10^-8mol L^-1 . This shows that concentration of H⁺ ions fails considerably when KCN is added to HCN solution.

Example 2:  When 0.100 mole of ammonia, NH₃, is dissolved in sufficient water to make 1.0 L solution, the solution is found to have a hydroxide ion concentration of 1.34 × 10^-3 M. Calculate K_b for ammonia.

Solution:          

NH₃  +   H₂O  ↔  NH⁺₄

At equilibrium (0.100 -1.34 ×10^-3) M      1.34×10^-3 M

                         = 0.09866 M    +    OH^-

                           1.34 × 10^-3 M

 K_b = [NH₄⁺ ][OH^- ]/[NK₃ ] = (1.34×10^-3 × 1.34×10^-3)/0.09866

                         =1.8199×10^-5

 Example 3:   Ka for HA is 4.9 ×10^-8. After making the necessary approximation, calculate for its decimolar solution

(a)   % dissociation

(b)   H⁺ ion concentration.

Solution:  For a weak electrolyte.

α = √K/C = √((4.9×10^-8)/0.1)

            = 7 × 10^-4

% dissociation = 100 × α = 100 × 7 × 10^-4

            = 7 ×10^-2

HA       ↔         H⁺     +      A^-

C(-1-α)           Cα                    Cα

[H⁺] = C × α = 0.1 ×7 × 10^-4 = 7 × 10^-5 mol L^-1

Example 4:  Nicotinic acid (Ka = 1.4 ×10-5) is represented by the formula HNiC. Calculate its per cent dissociation in a solution which contains 0.10 mole of nicotine acid per 2 litre of solution.

Solution:  Initial concentration of the nicotinic acid = 0.10/2 = 0.05molL-1

                           HNiC    ↔      H⁺ + NiC^-

Equilibrium conc. (0.05-x)      x      x

As x is very small, (0.05 - x), can be taken as 0.05

K_a = [H⁺ ][NiC^- ]/[HNiC] = (x × x)/0.05

or    x² = (0.05) × (1.4 × 10^-5)

or    x = 0.83 × 10^-3 mol L^-1

% dissociation = (0.83×10^-3)/0.05×100 = 1.66

Alternative method:   Let α be the degree of dissociation

                             NiHC            ↔          H⁺     +     NiC^-

At equilibrium 0.05 (1 -α)       0.05α        0.05α

K_a =  (0.05α×0.05α)/0.05(1-α)

As α is very small, (1-α) →  1.

So,   1.4 × 10^-5 = 0.05 α²

or    α√((1.4×10^-5)/0.05=1.67×10^-2 )

Per cent dissociation = 100 × α = 100 × 1.67 ×10^-2  = 1.67

Example 5:  At 30^o C the degree of dissociation of 0.066 M HA is 0.0145. What would be the degree of dissociation of 0.02 M solution of the acid at the same temperature?

Solution:  Let the ionization constant of the acid be K_a.

Degree of dissociation at 0.66 M concentration = 0.0145.

Applying α = √K_α/C

0.0145 =  √K_α/0.066                       ..... (i)

Let the degree of dissociation of the acid at 0.02 M concentration be α₁.

α₁ = √K_α/0.02                               ...... (ii)

Dividing Eq. (ii) by Eq. (i)

α₁/0.0145=√(0.066/0.02)=1.8166

or   α₁ = 0.0145 × 1.8166 = 0.0263

Example 6:  A solution contains 0.1 M H₂S and 0.3 M HCl. Calculate the concentration of S^2- and HS^- ions in solution. Given  and  for H₂S are 10^-7 and 1.3 × 10^-13 respectively.

Solution:  H₂S ↔  H+ + HS-

K_α1=[H⁺ ][HS^- ]/[H₂ S]            ....... (i)

Further HS^- = H⁺ + S^2-

K_α2=[H⁺ ][HS^2- ]/[H₂ S]             ....... (ii)

Multiplying both the equations

K_α1×K_α2=([H⁺ ]² [HS^2- ])/[H₂S]

 Due to common ion, the ionization of H₂S is suppressed and the [H⁺] in solution is due to the presence of 0.3 M HCl.

[S^2-]=(K_α1× K_α2[H₂S])/[H⁺]²=(1.0×10^-7×1.3×10^-13×(0.1))/(0.3)²

Putting the value of [S2^-] in Eq. (ii)

1.3×10^-13 = (0.3×1.44×10^-20)/([HS^-])

or [HS^-] (0.3×1.44×10^-20)/(1.3×10^-13)= 3.3×10^-8 M
 

Related Resources

• Reference books of Physical Chemistry for IIT JEE

• IIT JEE Syllabus of Chemistry 

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