Q. 674.1( 7 Votes )

Determine the sol

Answer :

To determine the solubility, we apply the formula

Ksp = [A+][B-]


Where Ksp is the solubility product which is equal to the product of ionic concentrations in a saturated solution.


a) Silver chromate


ionization of silver chromate:


Ag2CrO4 2Ag+ + CrO42-


As we know that,


Ksp = [A+] [B-]


Where A and B are the ions dissolved


In the above reaction,


[A+] = [Ag+]


[B-] = [CrO42-]


Ksp = [Ag+]2 + [CrO42-]


As Ksp of Ag2CrO4 =1.1 × 10-12 (given)


Let ‘s’ be the solubility of Ag2CrO4


[Ag+] = 2s


[CrO42-] =s


1.1 × 10–12 = (2s)2 s


1.1 × 10-12 = 4(s)3


0.275 × 10–12 = s3


s =


s = 0.65 × 10-4


Thus, Molarity of Ag+ = 2s= 2× 0.65× 10-4 = 1.30× 10-4 M


Molarity of CrO42- = s = 0.65 × 10-4 M


b) Barium chromate


ionization of barium chromate:


BaCrO4 Ba2+ + CrO42-


As we know that,


Ksp = [A+] [B-]


Where A and B are the ions dissolved


In the above reaction,


[A+] = [Ba2+]


[B-] = [CrO42-]


Ksp = [Ba2+] + [CrO42-]


As Ksp of BaCrO4 =1.2 × 10-10 (given)


Let ‘s’ be the solubility of BaCrO4


[Ba2+] =s


[CrO42-] =s


1.2 × 10–10 = (s× s)


1.2 × 10-10 = (s)2


s =


s = 1.09 × 10-5


Thus, Molarity of Ba2+ and CrO42- = s= 1.09× 10-5M


c) Ferric hydroxide


ionization of ferric hydroxide


Fe(OH)3 Fe3+ + 3OH-


As we know that,


Ksp = [A+] [B-]


Where A and B are the ions dissolved


In the above reaction,


[A+] = [Fe3+]


[B-] = [OH-]


Ksp = [Fe3+] + [OH-]


As Ksp of Fe(OH)3 =1.0 × 10-38 (given)


Let ‘s’ be the solubility of Fe(OH)3


[Fe3+] =s


[3OH-] = 3s


1.0× 10–38 = (s)(3s)3


1.0 × 10-38 = (27s)4


0.37 × 10–38 = s3


s =


s = 1.39 × 10-10


Thus, Molarity of Fe3+ = s= 1.39× 10-10 M


Molarity of OH- = 3s = 4.17 × 10-10M


d) Lead chloride


ionization of lead chloride:


PbCl2 Pb2+ + 2Cl-


As we know that,


Ksp = [A+] [B-]


Where A and B are the ions dissolved


In the above reaction,


[A+] = [Pb2+]


[B-] = [Cl-]


Ksp = [Pb2+] + [Cl-]2


As Ksp of PbCl2 =1.6 × 10-5 (given)


Let ‘s’ be the solubility of PbCl2


[Pb2+] =s


[Cl-] =2 s


1.6 × 10–5 = (s)(2s)2


1.6 × 10-5 = (4s)3


0.4 × 10–5 = s3


s =


s = 1.58 × 10-2


Thus, Molarity of Pb2+ = s= 1.58× 10-2 M


Molarity of Cl- = 2s = 3.16 × 10-2 M


e) Mercurous Iodide


ionization of mercurous iodide:


Hg2I2 Hg22+ + 2I-


As we know that,


Ksp = [A+] [B-]


Where A and B are the ions dissolved


In the above reaction,


[A+] = [Hg22+]


[B-] = [I-]2


Ksp = [Hg22+] + [I-]2


As Ksp of Hg2I2 = 4.5 × 10–29


Let ‘s’ be the solubility of Hg2I2


[Hg22+] =s


[I-] 2s


4.5 × 10–29 = (s)(2s)2


4.5 × 10–29 = (4s)3


4.5 × 10–29 = s3


s =


s = 2.24× 10-10


Thus, Molarity of Hg22+ = s= 2.24× 10-10 M


Molarity of I- = 2s = 4.48 × 10-10 M


Note: Ksp increases with increase in temperature.


In a saturated solution, Ksp = [A+][B-]


In an unsaturated solution of AB, Ksp> [A+][B-] means more solute can be dissolved.


In a super saturated solution of AB, Ksp< [A+][B-] means precipitation will start to occur


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

What is the minimNCERT - Chemistry Part-I

Equal volumes of NCERT - Chemistry Part-I

What is the maximNCERT - Chemistry Part-I

The solubility prNCERT - Chemistry Part-I