# JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Solubility Product

Solubility Product

Category : JEE Main & Advanced

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.

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