```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.

Example 16:   The solubility product of silver chloride is 1.5625 × 10-10 at 25°C. Find its solubility in g L-1.
Solution:   Let the solubility of AgCl be 5 mol litre-1.
AgCl  Ag+ + Cl-
Hence, s2 = 1.5625 × 10-10
or       s2 = 1.25 × 10-5 mol L-1
Molecular mass of AgCl = (108 + 35.5) = 143.5
So, Solubility in g litre-1 = mol. mass × s
= 143.5 × 1.25 × 10-5
= 1.79 × 10-3 g L-1
Example 17.   The solubility of PbSO4 in water is 0.038 g L-1 at 25°C. Calculate itssolubility product at the same temperature.
Solution:   Solubility (s) of PbSO4 in mol L-1,
= 0.038/(Mol.mass of PbSO4 )= 0.038/303 = 1.254×10-4
The equilibrium is
PbSO4 ↔ Pb2+ +  SO2-4
s          s
So   Ks - [Pb2+] [SO2-4 ] = s × s = s2
or    Ks = 1.254 × 10-4 × 1.254 x 10-4
= 1.573 × 10-8

Example 18.  The concentration of Ag+ ion in a saturated solution of Ag2CrO4 at 20°C is 1.5 × 10-4 mol L-1. Determine the solubility product of Ag2CrO4 at 20°C.
Solution:  The equilibrium is
Ag2CrO4 ↔ 2Ag+ + CrO2-4
On the basis of this equation, the concentration of CrO2-4  ions will be half of the concentration of Ag+ ions.
Thus, [Ag+] = 1.5 x 10-4 M and [CrO2-4 ]
= 0.75 x 10-4 M
Ks = [Ag+]2[CrO2-4 ] - (1.5 × 10-4)2 (0.75 × 10-4)
= 1.6875 × 10-12

Example 19:   The solubility product of BaSO4 is 1.5 x 10-9. Find out thesolubility in
(i) pure water and
(ii) 0.1 M BaCl2 solution.
Solution:  The equilibrium is
(i)    BaS04  ↔ Ba2+ +  SO-24
Let s be the solubility in mol litre-1; then
Ks = [Ba2+][SO-24 ] = s2
or    1.5 × 10-9 = s2
So   s = 3.87 × 10-5 mol L-1
(ii)   Let s' be the solubility of BaSO4 in 0.1 M BaCl2 solution.
Total Ba2+ ions concentration = (s' + c) mol L-1 and SO-24  ions concentration = s' mol L-1
So   Ks = (s' + c)s' = (s' + 0.1)s'
or     1.5 × 10-9 = (s' + 0.1)s'
or     (s')2 + 0.1s' = 1.5 x 10-9
Neglecting (s')2,
0.1s' = 1.5 × 10-9
or   (s') = 1.5 × 10-8 mol L-1

Example 20:  The solubility of Mg(OH)2 in pure water is 9.57 × 10-3 g L-1. Calculate its solubility in g L-1 in 0.02 M Mg(NO3)2 solution.
Solution:  Solubility of Mg(OH)2 in pure water
= 9.57 × 10-3 g L-1
= (9.57×10-3)/(mol mass) =  mol L-1
= (9.57×10-3)/58 = 1.65×10-4   mol L-1
Further, Mg (OH)2 ↔ Mg2+ + 2OH-
s         2s
Ks = [Mg2+][OH-]2 = s×(2s)2 = 4s3
= 4 × (1.63×10-4)3
= 17.9685 × 10-12

Let s' be solubility of Mg(OH)2 in presence of Mg(NO3)2
[Mg2+] = (s' + c) = (s' + 0.02)
[OH-] = 2s'
So         Ks = (s' + 0.02) (2s')2
17.9685 × 10-12 = 4(s')2(s' + 0.02)
(17.9685×10-12)/4 = (s')3 + 0.02(s')2
[neglecting (s')3]
4.4921 × 10-12 = 0.02 (s')2
or        (s')2 = 4.4921/0.02×10-12
or        s' = 14.9869 × 10-6 mol L-1
Solubility of Mg(OH)2 in g litre-1 = s' × M
= 14.9868 × 10-6 × 58
= 8.69 × 10-4 g L-1 ```
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