JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Solubility Product

In a saturated solution of sparingly soluble electrolyte two equilibria exist and can be represented as, \[\underset{\text{Solid}}{\mathop{AB}}\,\] ? \[\underset{\begin{smallmatrix}

 \text{Unionised} \\

 \text{(Dissolved)}

\end{smallmatrix}}{\mathop{AB}}\,\]? \[\underset{\begin{smallmatrix}

  \\

 i\text{ons}

\end{smallmatrix}}{\mathop{{{A}^{+}}+{{B}^{-}}}}\,\]

Applying the law of mass action,\[\frac{[{{A}^{+}}][{{B}^{-}}]}{[AB]}=K\]

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,  \[[{{A}^{+}}][{{B}^{-}}]=K[AB]=K{K}'={{K}_{sp}}\] (constant)

\[{{K}_{sp}}\] 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 \[{{A}_{x}}{{B}_{y}}\] which dissociates as, \[{{A}_{x}}{{B}_{y}}\]? \[x{{A}^{y+}}+y{{B}^{x-}}\]

Applying law of mass action, \[\frac{{{[{{A}^{y+}}]}^{x}}{{[{{B}^{x-}}]}^{y}}}{[{{A}_{x}}{{B}_{y}}]}=K\]

When the solution is saturated,  \[[{{A}_{x}}{{B}_{y}}]={K}'\] (constant) or

\[{{[{{A}^{y+}}]}^{x}}{{[{{B}^{x-}}]}^{y}}=K[{{A}_{x}}{{B}_{y}}]=K{K}'={{K}_{sp}}\] (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, either unsaturated or saturated and varies accordingly.

On the other hand, the term 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 at a constant temperature.

(2) Different expression for solubility products

(i) Electrolyte of type AB (1 : 1 type salt) e.g., \[AgCl,\ BaS{{O}_{4}}\]

\[AgCl\]? \[\underset{x}{\mathop{A{{g}^{+}}}}\,+\underset{x}{\mathop{C{{l}^{-}}}}\,\]

\[{{K}_{sp}}=[A{{g}^{+}}][C{{l}^{-}}]\] ; \[{{K}_{sp}}={{x}^{2}}\]; \[x=\sqrt{{{K}_{sp}}}\]

 (ii) Electrolytes of type\[A{{B}_{2}}\](1:2 type salt) e.g.,\[PbC{{l}_{2}},\ Ca{{F}_{2}}\]

\[PbC{{l}_{2}}\] ? \[\underset{x}{\mathop{P{{b}^{2+}}}}\,+\underset{2x}{\mathop{2C{{l}^{-}}}}\,\]

\[{{K}_{sp}}=[P{{b}^{2+}}]\,{{[C{{l}^{-}}]}^{2}}\];\[{{K}_{sp}}=[x]\ {{[2x]}^{2}}\];\[{{K}_{sp}}=4{{x}^{3}}\] 

\[x=3\sqrt{{{K}_{sp}}/4}\]

(iii) Electrolyte of type A₂B (2 : 1 type salt) e.g.,\[A{{g}_{2}}Cr{{O}_{4}},\ {{H}_{2}}S\]

\[A{{g}_{2}}Cr{{O}_{4}}\] ? \[\underset{2x}{\mathop{2A{{g}^{+}}}}\,+\underset{x}{\mathop{CrO_{4}^{2-}}}\,\]

\[{{K}_{sp}}={{[A{{g}^{+}}]}^{2}}\ [CrO_{4}^{2-}]\];\[{{K}_{sp}}={{[2x]}^{2}}\ [x]\];\[{{K}_{sp}}=4{{x}^{3}}\]\[x=\sqrt[3]{\frac{{{K}_{sp}}}{4}}\]

(iv) Electrolyte of type \[{{A}_{2}}{{B}_{3}}\](2 : 3 type salt)

 e.g., \[A{{s}_{2}}{{S}_{3}},\ S{{b}_{2}}{{S}_{3}}\]

\[A{{s}_{2}}{{S}_{3}}\]? \[\underset{2x}{\mathop{2A{{s}^{3+}}}}\,+\underset{3x}{\mathop{3{{S}^{2-}}}}\,\]

\[{{K}_{sp}}={{[A{{s}^{3+}}]}^{2}}{{[{{S}^{2-}}]}^{3}}\] ; \[{{K}_{sp}}={{[2x]}^{2}}{{[3x]}^{3}}\]; \[{{K}_{sp}}=4{{x}^{2}}\times 27{{x}^{3}}\]

\[{{K}_{sp}}=108{{x}^{5}}\] ; \[x=\sqrt[5]{\frac{{{K}_{sp}}}{108}}\]

(v) Electrolyte of type \[A{{B}_{3}}\](1 : 3 type salt)

 e.g.,\[AlC{{l}_{3}},\ Fe{{(OH)}_{3}}\]

\[AlC{{l}_{3}}\]? \[\underset{x}{\mathop{A{{l}^{+++}}}}\,+\underset{3x}{\mathop{3C{{l}^{-}}}}\,\]

\[{{K}_{sp}}=[A{{l}^{+3}}][3C{{l}^{-}}]\] ; \[{{K}_{sp}}=[x]\ {{[3x]}^{3}}\]

\[{{K}_{sp}}=27{{x}^{4}}\]; \[x=\sqrt[4]{\frac{{{K}_{sp}}}{27}}\].

(3) Criteria of precipitation of an electrolyte : When Ionic product of an electrolyte is greater than its solubility product, precipitation occurs.

(4) Applications of solubility product

(i) In predicting the formation of a precipitate

Case I : When\[{{K}_{ip}}<{{K}_{sp}}\], then solution is unsaturated in which more solute can be dissolved. i.e., no precipitation.

Case II : When \[{{K}_{ip}}={{K}_{sp}}\], then solution is saturated in which no more solute can be dissolved but no ppt. is fomed.

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 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,

\[NaCl\] ? \[N{{a}^{+}}+C{{l}^{-}}\]; \[HCl\]?\[{{H}^{+}}+C{{l}^{-}}\]

The concentration of \[C{{l}^{-}}\] ions increases considerably in solution due to ionisation of \[HCl\]and due to common ion effect, dissociation of NaCl is decreased. Hence, the ionic product \[[N{{a}^{+}}][C{{l}^{-}}]\] exceeds the solubility product of \[NaCl\] and therefore pure \[NaCl\] precipitates out from the solution.

(iv) Salting out of soap : From the solution, soap is precipitated  by the addition of concentrated solution of \[NaCl\].

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

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⁺ , Hg⁺²) The group reagent is dilute HCl. \[[A{{g}^{+}}][C{{l}^{-}}]>{{K}_{sp}}\] for AgCl.

(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 \[{{H}_{2}}S\] in presence of dilute \[HCl\]. \[[P{{b}^{+2}}][{{S}^{-2}}]>{{K}_{sp}}\] for \[PbS\].

(c) Precipitation of group third radicals (Fe⁺³, Al⁺³ and Cr⁺³) The group reagent is \[N{{H}_{4}}OH\] in presence of \[N{{H}_{4}}Cl\].

\[[F{{e}^{+3}}]{{[O{{H}^{-}}]}^{3}}>{{K}_{sp}}\]

(d) Precipitation of group fourth radicals (Co⁺², Ni⁺², Mn⁺² and Zn⁺²) : The group reagent is \[{{H}_{2}}S\] in presence of \[N{{H}_{4}}OH\].

\[[C{{o}^{+2}}][{{S}^{-2}}]>{{K}_{sp}}\]

(e) Precipitation of group fifth radicals (Ba⁺², Sr⁺², Ca⁺²)  The group reagent is ammonium carbonate in presence of \[N{{H}_{4}}Cl\] and \[N{{H}_{4}}OH\]. \[[B{{a}^{+2}}]\ [CO_{3}^{-2}]>{{K}_{sp}}\]

(vi) 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  \[[{{A}^{n+}}]_{left}^{m}=\frac{{{K}_{sp}}[{{A}_{m}}{{B}_{n}}]}{{{[{{B}^{m-}}]}^{n}}}\]

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

(vii) Calculation of simultaneous solubility : Solubility of two electrolytes having common ion; when they are dissolved in the same solution, is called simultaneous solubility.

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., \[AgBr\ ({{K}_{sp}}=5\times {{10}^{-13}})\]; \[AgSCN\ ({{K}_{sp}}={{10}^{-12}})\]

Here, charge balancing concept is applied.

Charge of \[A{{g}^{+}}\]=   Charge of \[B{{r}^{-}}\]+  Charge of \[SC{{N}^{-}}\]

\[[A{{g}^{+}}]\]              =        \[[B{{r}^{-}}]\]                     +              \[[SC{{N}^{-}}]\]

\[(a+b)\]           =          \[a\]                           \[b\]

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

 e.g., \[Ca{{F}_{2}}\,({{K}_{sp}}=3.4\times {{10}^{-11}})\]; \[Sr{{F}_{2}}\ \ ({{K}_{sp}}=2.9\times {{10}^{-9}})\]

Most of fluoride ions come of stronger electrolyte.