Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev

Physical Chemistry

Created by: Asf Institute

Chemistry : Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev

The document Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev is a part of the Chemistry Course Physical Chemistry.
All you need of Chemistry at this link: Chemistry

Solubility (s) and Solubility Product (Ksp)

This is generally used for sparingly soluble salts.

Solubility product (ksp) is an equilibrium constant so like all the equilibrium constants, it depends only on the temperature.

Solubility product: in a saturated solution of a salt, there exists a dynamic equilibrium between the excess of solute and the ions furnished by the part of the solute which has gone in solution.

Solution

AgCl(s) Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev    [Ag+][Cl]

AgCl   Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev      Ag+ + Cl

a-s                    s           s

Ksp = S2

Simple Solubility         Ax By Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev xAy+ + yB–x

Ksp = (xs)x (ys)y = xx.yy (s)x+y

Ksp = xx.yy(s)x+y

 

Factors Affecting Solubility:

  • Effect of common ions on solubility

Because of the presence of common ion the solubility of the salt decreases.

AgCl   Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev    Ag+      +          Cl

-                       s                       s

Added NaCl, which is having common ion as that of sparingly soluble salt, AgCl.

            NaCl    →        Na+      +       Cl

            C                     0                    0

            -                       C         C

So, added electrolyte, will decrease the solubility of AgCl

AgCl Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev Ag+ + Cl

            S'            S'

Ksp = S' (S' + C)

Assumption: C >>> S'

Hence S' + C »C

Ksp = s1 x C

S1 = ksp / C

  • Simultaneous Solubility:

When sparingly soluble salts are added in water simultaneously, there will be two simultaneous equilibria in the solution

AgCl   Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev     Ag+      +     Cl

                        S1+S2          S1                                 Ksp1

AgCl   Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev     Ag+      +     Br

                        S1+S2          S2                                  Ksp2

Ksp1 = (s1 + s2)s1

Ksp2 = (s1 + s2)s2

Adding equation (1) and (2)

K1P1+K1P2 = (S1+s2)

From (1)

Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev

  • Solubility in presence of hydrolysis:

When the ion from sparingly soluble salt, undergo hydrolysis Hence, we have to follow different procedure to calculate solubility of these salts. 

AgCN(s) Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev Ag+ + CN

                        s1         s-x

CN + H2O Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev  HCN + OH

s-x                      x       x

ksp=[Ag+] [CN]                                              … 1

Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev                                      … 2

Multiply (1) and (2)

Ksp  x Kh = [Ag+] [HCN]

Since [HCN] = [OH]

Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev

Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev

Effect of complex formation: 

AgCl(s) Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev  Ag+ (ag) + Cl (ag)

Ag+ + 2NH3Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev [Ag(NH3)2]+ K= stability constant

S          C

-           C-2S                S

Reverse the Reaction,

[Ag(NH3)2]+Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev Ag+ + 2 NH3

S                           -      C-2S

s-x                   x          C-2s+2x

Since, Kd is very very low,

Hence, c – 2s + 2x ≈ C – 2S

Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev

Ksp = x × s                                                                   … 1

Put value of Kd in eq. 1

Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev

S = Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev

Concept of Precipitation: Less the value of S, less the solubility of the salt..Hence, more the extent of precipitation of salt.

AgCl Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev Ag+ + Cl

With the knowledge of solubility product of a sparingly soluble substance, we can predict whether under certain given conditions that substance would be precipitated or not. It may be remembered that a substance gets precipitated when the ionic product i.e. the product of concentrations of its ions present in a solution exceeds the solubility product of the substance,

Ki = [Ag+] [Cl]

If         Kip>Ksp          Precipitation Occurs

If         Kip<Ksp          Unsaturated Solution

If         kip = Ksp         Saturated Solution

Example : 104 Molar Mg(NO3)2 with pH = 9 and if Ksp (Mg(OH)2) = 8.9 × 1012

  1. Will precipitation takes place or not.

  2. At what minimum value of pH precipitation will occur.

a­)  Mg(NO3)2     →    Mg2+ + 2 NO3

            104                -             -

 

              -                  104          10–4

 

            Mg(OH2) Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev  Mg2+ + 20H

 Kip = (10–4) (10–5)2

Ki = 10–14

Ksp of Mg(OH)2 = 8.9 × 10–12

Ksp>Kip

Hence solution is unsaturated. No precipitation will occur.

For Mg (OH)2

            Ksp = [Mg2+][OH]2

            8.9 × 10–12 = 10–4[OH]2

            [OH] = 2.9 × 10–4

pOH = 3.83

pH = 10.47

Hence pH should be greater than 10.47 for the precipitation to occur.

Preferential precipitation of an insoluble salt: 

Silver chloride is insoluble or sparingly soluble, to be more accurate. So is silver iodide. The question arises as to what would happen if potassium iodide solution is added to silver chloride.

AgCl + KI Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev KCl + AgI

Sparingly                     Sparingly
soluble                         soluble

Would the reaction take place towards the right, i.e. would the precipitated of siler chloride change into the precipitate of silver iodide.

For answer, it is necessary to look their respective solubility products.

Solubility product of silver chloride

Ksp (AgCl) = [Ag+] [Cl] = 1.56 × 10–10

And that of silver iodide

Ksp(AgI) = (Ag+)(I) = 0.94 × 10–16

 

Example :  0.1 Molar concentration of Cl ion and 103 molar chromate CrO42 ions are present in a solution. If AgNO3 is added which will precipitate first

            If         Ksp (AgCl)=1012

                        Ksp (Ag2CrO4) 1010

2.  What will be the concentration of Cl ions when Ag2CrO4 begins to precipitate. Also find the percentage of Cl left with respect to original.

Solution           AgCl Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev Ag+ Cl

                          -             S        S

                        Ksp = S2

                        Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev

                        S = 10–6

For Ag2CrO4

                        Ag2CrO4Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev 2Ag+ + CrO42–

                        Ksp = (2S)2S

                        10–10 = 4S3

                        S = (0.25 × 10–10) 1/3

                        S = 2.9 × 10–4

            Since, solubility of AgClis less than solubility of Ag2CrO4.

            Then, obviously AgCl will precipitate first

b)  When Ag2CrO4 begins to precipitate, concentration of CrO42– would be

            10–10 = [Ag+]2[CrO42–]

            10–10 = [Ag+] × 10–3

            [Ag+] = 3.1 × 10–4

For AgCl

            Ksp = [Ag+][Cl]

            10–12 = 2.9 × 10–4 [Cl]

            Cl = 3.4 × 10–9

            Percentage of Cl left = Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev

            Percentage of Cl left = 3.4 × 10–6 %

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Dynamic Test

Content Category

Related Searches

Objective type Questions

,

mock tests for examination

,

Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev

,

past year papers

,

shortcuts and tricks

,

Sample Paper

,

Important questions

,

study material

,

Extra Questions

,

Free

,

Semester Notes

,

Summary

,

pdf

,

Previous Year Questions with Solutions

,

MCQs

,

Exam

,

video lectures

,

practice quizzes

,

Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev

,

ppt

,

Viva Questions

,

Solubility and Solubility product - Ionic Equilibrium Chemistry Notes | EduRev

;