Solubility Product

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 dissolutionof 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’ = Ks (constant)

Ks 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 is dissociated as:

AxBy  xAy+ + yBx-

Applying law of mass action,



When the solution is saturated,

[AxBy] = K' (constant)

or   Ay+]x[Bx-]y = K [AxBy] = KK' = Ks (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 sulubility product

​

(i) Electrolyte of the type AB:

Its ionisation is represented as:

AB   A+ + B-

Thus,     Ks = [A+][B-]

AgCl   Ag+ + Cl-;               Ks = [Ag+][Cl-]

BaSO4  Ba2+ + SO-24 ;     Ks = [Ba2+][SO-24]

(ii)  Electrolyte of the type AB2:

Its ionisation is represented as:

AB2  A2+ + 2B-

Thus,     Ks = [A2+][B-]2

PbCl2  Pb2+ + 2Cl-;          Ks = [Pb2+][Cl-]2

CaF2  Ca2+ + 2F-;            Ks = [Ca2+][F-]2

(iii)  Electrolyte of the type A2B:

Its ionisation is represented as:

A2B ↔  2A2+ + B2-

Thus,     Ks = [A+]2[B2-]

Ag2CrO4  2Ag+ + CrO-24;    Ks = [Ag+]2[CrO-24]

H2S   2H+ + S2-;                    Ks = [H+]2[S2-]

(iv)   Electrolyte of the type A2B3:

Its ionisation is represented as:

A2B3  2A3+ + 3B2-

Thus,     Ks = [A3+]2[B2-]3

As2S3  2As3+ + 3S2-;

Ks = [As3+]2[S2-]3

Sb2S3  2Sb3+ + 3S2-;

Ks = [Sb3+]2[S2-]3

(v)   Electrolyte of the type AB3:

Its ionisation is represented as:

AB3  A3+ + 3B-

Thus,     Ks = [A3+][B2-]3

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

AH3    Al3+ + 3l-;      Ks = [Al3+][I-]3

Solubility product of a weak electrolyte

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

AmBn  mAn+ + nBm-

t = 0                  s          0          0

teq                    s-sα     msα      nsα

Ksp = [An+]m[Bm-]n

= [msα ]m[nsα]n

Ksp = mmnn (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 thesolubility product, i.e., the solution has not reached the saturation stage.

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

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 solution as precipitate.

Thus, for the precipitation of an electrolyte, it is neces­sary that the ionic productmust 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  [CrO2-4 ]= 0.02/2 = 0.01 = 1*10-2 M

Ionic product of Ag2CrO4 = [Ag+]2 [CrO2-4]

= (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 S2- 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.

CnH2n+1 COONa ↔  CnH2n+1 COO- + Na+

Soap

NaCl ↔  Na+ + Cl-

Thus, the concentration of Na+ ions increases considerably on addition of NaCl solution. Hence, the ionic product [CnH2n+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+ + S2-

Applying law of mass action,

K = ([H+ ]2 [S2- ])/[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. Ksp BaCrO4 is 1.2 × 10-10 and Ksp SrCrO4 is 3.5 × 10-5.

[CrO2-4]concentration required to precipitate BaCrO4

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

BaCrO4 will precipitate first because it requires low concentration of CrO2-4  ions. On addition of chromate ions, BaCrO4 starts precipitating when chromate ion concentration reaches 1.2 × 10-9 M. When CrO2-4  ion concentration reaches upto 3.5 × 10-4 M, then SrCrO4 also starts precipitating.

Remaining concentration of Ba2+ when SrCrO4 starts precipitation.

=(Ksp BaCrO4)/[CrO42-] = (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

Kd = [Fe2+ ][Br- ]/[FeBr+ ]

Reciprocal of dissociation constant is called stability constant.

Ks = [FeBr+ ]/([Fe2+] [Br- ])

Let us consider the formation of complex K2Cd(CN)4, Complex ion is Cd(CN42-)  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 Ksp 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.

(vii)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-

Ksp of M(OH)2 = [M2+][OH-]2

[M2+] =  Ksp/[OH-]2

[H+] [OH-] = Kw = 10-14

[OH-]2 = 10-28/[H+]2

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

[M2+] = Ksp[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:

AxBy ↔ xAy+ + yBx-

Thus, solubility product, Ks = [Ay+]x[Bx-]y

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

AxBy ↔  xAy+ + yBx-

xs       ys

So                   Ks = [xs]x[ys]y

= xx.yy(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               Ks = [A+][B-] = s × s = s2

or                s = √Hs

(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          Ks = [A2+][B-]2 = s × (2s)2 = 4s3

or           s = 3√Ks/4

A2B ↔ 2A+ + B2-

s      s

Let s be the solubility of A2B.

Ks = [A+]2[B2-]

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

or          s = 3√Ks/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.

Ks = [A3+][B-]3 = s × (3s)3 = 27s4

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

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

So    Ks = 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 = Ksp [AB]/[B-]

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

(iii)    [An+]mleft = Ksp[AmBn]/[Bm-]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(Ksp = 5 x 10-13);    AgSCN (Ksp = 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 (Ksp = 3.4 x 10-11);    SrF2 (Ksp = 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,

Ka =   ([H+ ][CN-])/[HCN] or [H+ ] (Ka [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+] = (Ka [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 Kb 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

Kb = [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

Ka = [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

NHiC            ↔          H+     +     NiC-

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

Ka =  (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 30o 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 Ka.

Degree of dissociation at 0.66 M concentration = 0.0145.

Applying α = √Kα/C

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

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

α1 = √Kα/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 S2- and HS- ions in solution. Given  and  for H2S are 10-7 and 1.3 × 10-13 respectively.

Solution:  H2S ↔  H+ + HS-

Kα1=[H+ ][HS- ]/[H2 S]            ....... (i)

Further HS- = H+ + S2-

Kα2=[H+ ][HS2- ]/[H2 S]             ....... (ii)

Multiplying both the equations

Kα1×Kα2=([H+ ]2 [HS2- ])/[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.

[S2-]=(Kα1× Kα2[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

You can also refer to following links

IIT JEE Syllabus of Chemistry

Reference books of Physical Chemistry for IIT JEE


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