In order to find the solubility problem, there is a need of understanding of dissociation reaction and solubility. Any reagent is consumed up to a certain limit to reach equilibrium of reaction. The limit is called solubility limit.
Dissociation reaction of AgCl in water
At equilibrium, AgCl is formed one mole of Ag+, Cl- each.
So, the solubility of AgCl is either equal to conce of Ag+, or Cl-
solubility of AgCl = [Ag+] = [Cl-]
In order to find solubility product, use the equation as
Ksp = [A]c[B]d
Compare with the reaction type with standard equation AB ? cA + dB,
So that we get, a= 1, b=1
And thus
Ksp = [Ag+][Cl-]
since [Ag+] = [Cl-]
Ksp = [Ag+][Ag+]= 1.26 x 10-5¬( given in question)
Thus [Ag+]2= (1.26 x 10-5)
[Ag+] = 1.6 x 10-10 M
BaF2
BaF2 in water is dissociated is
BaF2 (s) ? Ba+ (aq) + 2 F- (aq)
Since the mole of Ba+, F- are not equal, so solubility of BaF2 is equal to Ba+
Solubility = [Ba+] = 3.15 x 10-3 M
&[F-] = 2 [Ba+]
Solubility product
Ksp = [Ba+][F-]2
Ksp = ([Ba+])(2 [Ba+])2
Ksp = 4[Ba+]3
Ksp = XXXXXXXXXXx 10-3 M)3
Document Preview: In order to find the solubility problem, there is a need of understanding of dissociation reaction and solubility. Any reagent is consumed up to a certain limit to reach equilibrium of reaction. The limit is called solubility limit. Dissociation reaction of AgCl in water At equilibrium, AgCl is formed one mole of Ag+, Cl- each.
So, the solubility of AgCl is either equal to conce of Ag+, or Cl- solubility of AgCl = [Ag+] = [Cl-] In order to find solubility product, use the equation as
Ksp = [A]c[B]d Compare with the reaction type with standard equation AB ? cA + dB,
So that we get, a= 1, b=1
And thus Ksp = [Ag+][Cl-] since [Ag+] = [Cl-] Ksp = [Ag+][Ag+]= 1.26 x 10-5 ( given in question) Thus [Ag+]2= (1.26 x 10-5) [Ag+] = 1.6 x 10-10 M BaF2 BaF2 in water is dissociated is BaF2 (s) ? Ba+ (aq) + 2 F- (aq) Since the mole of Ba+, F- are not equal, so solubility of BaF2 is equal to Ba+
Solubility = [Ba+] = 3.15 x 10-3 M &[F-] = 2 [Ba+] Solubility product Ksp = [Ba+][F-]2 Ksp = ([Ba+])(2 [Ba+])2 Ksp = 4[Ba+]3 Ksp = XXXXXXXXXXx 10-3 M)3 Ksp = 124 x 10-9 Ksp =1.24 x 10-6
Answer: The solubility product of AgCl is 1.6 x 10-10. The solubility product of BaF2 is 2 x 10-6.