Chemistry SOLUBILITY OF SPARINGLY SOLUBLE SALTS AND SOLUBILITY PRODUCTS
Please Wait... While Loading Full Video

SOLUBILITY OF SPARINGLY SOLUBLE SALTS AND SOLUBILITY PRODUCTS

Solubility and Solubility Product :

When a salt is dissolved in water, it dissolves as the salt,

`AgCl(s)` `->` ` AgCl(aq)` and then the dissolved salt dissociates to give the ions

`AgCl (aq) -> Ag^(+) (aq) + Cl^(-)(aq)`.

After a certain amount of the salt has been dissolved, the solution would become saturated with the salt. Now, if some more of the salt is dissolved, the salt would dissolve to give ions while at the same time the ions would precipitate to give back the salt. This amount (moles) of the salt that has made the solution saturated per liter of solution is called the solubility of the salt.

`AgCl (s) ⇋ Ag ^(+) (aq) + Cl^(-) (aq)` `K = ([Ag^(+)][Cl^(-)])/([AgCl])`

Since the concentration of `AgCl`, which is a solid, is constant, `K[AgCl] = [Ag^(+)] [Cl^(-)] = K_C`

This `K_c` is given the symbol, `K_(sp)`. This means that a solution cannot have the product of the concentration of `Ag^(+)` and `Cl^(-)` to be more than `K_(sp)` (called solubility product).

`A_x B_y(s) ⇋ x A^(y+)( aq) + yB^(x-) (aq)`

`text( xs ys)`

For saturated solution (at equilibrium),


`K_(eq) = ([A^(y+)]^x xx [B^(x-)]^y)/[A_xB_y]`

`K_(eq)[A_xB_y] = [A^(y+)]^x [B^(x-)]^y`

`K_(sp) =[A^(y+)]^x [B^(x-)]^y` (since `A_xB_y` is a pure solid)

`K_(sp) = (xs)^(x)(ys)^(y)`

`K_(sp) = x^(x) xx y^(y) xx s^(x+y)`

Calculation of Solubilities of Salts :

i) `text(Solubilities of AgCl (salt of a strong acid and strong base) in water :)`

`AgCl` would dissolve in water as,

`AgCl (s) -> Ag^(+) (aq) + Cl^(-) (aq)`

At saturation point

`AgCl (s) = Ag^(+)(aq) + Cl^(-) (aq)`

If the solubility of the salt is `x` moles/L

`[Ag^(+)] = xM`, `[Cl^(-)] = xM` `=>` `x^2 = K_(sp)`, `x = sqrt(K_(sp))`


ii) `text(Solubility of AgCl in a solution that is having)` `0.1` `M` `text(in)` `AgNO_3` :

`AgCl` would dissolve and finally reach saturation.

`AgCl(s) ⇋ Ag^(+)(aq) + Cl^(-) (aq)`

The `K_(sp)` of `AgCl` is approximately `10^(-10)`. If `AgCl` were to be dissolved in water (pure), its solubility would have been `10^(-5)M` (previous section). In the presence of `0.1` `M` `AgNO_3` its solubility will decrease due to common ion effect. This means that `[Ag^(+)]` from `AgCl` would be less than `10^(-5) M.` Hence, we can ignore the contribution of `Ag^(+)` from `AgCl`.

If the solubility of `AgCl` is `x` moles/L in the presence of `0.1` `M` `AgNO_3`, then

`[Ag^(+)] =` `0.1``M`, [`Cl^(-)`] = `x` `M` `=>` `x = K_(sp)/0.1 = 10^(-9)` moles/L

iii) `text(Solubility of)` `CH_3COOAg` `text((salt of weak acid and strong base) in water :)`

`CH_3COOAg` dissolves and reaches saturation. Since it is a salt of weak acid and strong base, it would hydrolyse. If the solubility of the salt is `x` moles/L, then

`CH_3COOAg(s) ⇋ CH_3COO^(-)(aq) + Ag^(+)(aq)`

At eqb : `x` `-` `y` `x`

`CH_3COO^(-) (aq) + H_2O ⇋ CH_3COOH (aq) + H^(+) (aq)`

At eqb : `x` `-` `y` `y` `y`

Where `y` is the amount of `CH_3COO^(-)` ion that is hydrolysed.

`(x - y) x = K_(sp)`

`y^2/(x-y) = K_w/K_a`

Knowing the values of `K_(sp)` and `K_a`, solubility of the salt can be calculated.

iv) `text(Solubility of)` `CH_3COOAg` `text((salt of a weak acid and strong base)` `text(in an acid buffer of pH= 4 (assuming that the buffer does not have any common ion by))` `CH_3COOAg) :`

`CH_3COOAg` would dissolve and reach equilibrium. It would then be hydrolysed. If the solubility of the salt is `x` `M` in this solution, then

At eqb : `CH_3COOAg (s) ⇋ CH_3COO^(-) (aq) + Ag^(+) (aq)`

`x` `-` `y` ` x`

`CH_3COO^(-) (aq) + H^(+) ⇋ CH_3COOH (aq)`

`x` `-` `y` `10^(-4)` `y`

Since the solution is a buffer, the `pH` will be maintained.

`(x - y) xx C = K_(sp)`

`y'/(10^(-4) xx(x' -y')) = 1/K_a`

Since in presence of basic buffer, the degree of hydrolysis will be suppressed by already existing `OH^(-)` ions, therefore the approximated formula which can be used is

`x^2 = K_(sp) (CH_3COOAg)` (neglecting `y`)

Knowing `K_(sp)` and `K_(a)`, the solubility can be calculated.

v) `text(Solubility of)` `CH_3COOAg` `text(in a buffer solution of pH= 9 :)`

Following the same logic as give in the earlier section,

`CH_3COOAg (s) ⇋ CH_3COO^(-) (aq) + Ag^(+) (aq)`

At eqb : `x` `-` `y` `x`

`CH_3COO^(-)(aq) + H_2O ⇋ CH_3COOH + OH^(-)`

At eqb : `x` `-` `y` `y` `10^(-5)`

Where `x` `M` is the solubility of the salt and `y` the extent to which it is hydrolysed.

(`x - y`) `x` `= K_(sp)`

`(y^n xx 10^(-5))/(x^n -y^n) = K_w/K_a`

Knowing, `K_(sp)` and `K_(a)`, the solubility can be calculated.

vi) `text(Solubility of)` `AgCl` `text(in an aqueous solution containing)` `NH_3` :

Let the amount of `NH_3` initially be `a` `M`. If the solubility of the salt is `x` moles/L, then

`AgCl (s) ⇋ Ag^(+) (aq) + Cl^(-) (aq)`

At eqb : `x` `-` `y` `x`

`Ag^(+) (aq) + 2NH_3 (aq) ⇋ Ag(NH_3)_2^(+) (aq)`

`x` `-` `y` `a` `-` `2y` `y`

Where `y` is the amount of `Ag^(+)` which has reacted with `NH_3`.

`(x -y) x= K_(sp)`


`y/((x-y)(a-2y)^2) = K_f` (formation constant of `Ag(NH_3)_2 ^(+))`

Knowing `K_(sp)` and `K_f`, the solubility can be calculated.

Precipitation of Salts :

It should be noted that if the ionic product is more than `K_sp`, the equilibrium will shift in backward direction and some amount of salt is thrown out of solution (precipitation) and finally an equilibrium is developed in such a way that the ionic product becomes equal to solubility product.

If ionic product is less than `K_(sp)`, the solution is unsaturated and more of the salt can be dissolved. If ionic product is equal to `K_(sp)`, the solution is saturated and if ionic product is more than `K_(sp)`, the solution is called super saturated.

From the solubility product principle, we may infer that,

- When the ionic product of a salt in solution is equal to its solubility product, the solution is saturated and the undissociated salt remains in equilibrium with its ions in the solution.

If Ionic Product `= K_(sp)`; a precipitate will not form and the solution is satuated in that salt. (or we can say that solution is at a critical stage, when precipitation just begins, but actually has not occurred yet in real sense).

- When the ionic product of a salt in solution is less than the solubility product, the solution is unsaturated and the solution contains only ions and no undissociated salt.

If Ionic Product `< K_(sp)`; a precipitate will not be formed and the solution will be unsaturated

- When the ionic product is greater than the solubility product, the excess ions in solution combine and gets precipitated. So when the product of the concentrations of the constituent ions raised to appropriate powers exceeds the solubility product of the salt, the salt will be precipitated.

If ionic Product `(I P) > K_(sp)`; precipitation takes place till `I.P.` equals `K_(sp)`.

Let us have a solution containing more than one ion capable of forming a precipitate with another ion, which is added to the system. The added ion will selectively form precipitate with any one of the ions present in the solution. This process of selectively or prefentially precipitating an ion from a solution of more than one ion is called selective precipitation. The selective precipitation of ions from a solution in the form of a salt, (which is partially soluble) can be done by adding precipitating agent drop by drop.

If the stoichiometry of the precipitated salts is same, then the salt with minimum solubility product (and hence also the minimum solubility) will precipitate first, followed by the salt of next higher solubility and so on. For example, in a solution containing `Cl^(-)`, `Br^(-)` and `I^(-)` ions, when `Ag^(+)` ions are added, then out of the three, the least soluble silver salt is precipitated first. If the addition of `Ag^(+)` ions is continued further, eventually a stage will be reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on.

 
SiteLock