a)The Ksp was found to be equal to 13.69
Explanation:
Terminology
Qsp of a dissolving ionic solid — is the solubility product of the concentration of ions in solution.
Ksp however, is the solubility product of the concentration of ions in solution at EQUILIBRIUM with the dissolving ionic solid.
Note that if Qsp > Ksp , the solid at a certain temperature, will precipitate and form solid. That means the equilibrium will shift to the left in order to attain or reach equilibrium (Ksp).
Step-by-step solution:
To solve this:
#./ Substitute the molar solubility of KCl as given into the ion-product equation to find the Ksp of KCl.
#./ Find the total concentration of ionic chloride in each beaker after the addition of HCl. We pay attention to the amount moles present at the beginning and the moles added.
#./ Find the Qsp value to to know if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate.
a) The equation of solubility equilibrium for KCL is thus;
KCL_(s) ---> K+(aq) + Cl- (aq)
The solubility of KCl given is 3.7 M.
Ksp= [K+][Cl-] = (3.7)(3.7) =13.69
The Ksp was found to be equal to 14.
In pure water KCl
Ksp =13.69 KCl =[K+][Cl-]
Let x= molar solubility [K+],/[Cl-] :. × , x
Ksp =13.69 = [K+][Cl-] = (x)(x) = x²
x= √ 13.69 = 3.7 M moles of KCl requires to make 100mL saturated solutio
37M moles/L
The Ksp was found to be equal to 14.
4.0 M HCl = KCl =[K+][Cl-]
Let y= molar solubility :. y, y+4
Ksp =13.69= [K+][Cl-] = (y)(y*+4)
* - rule of thumb
Ksp =13.69= [K+][Cl-] = (y)(y*+4)= y(4)
13.69=4y:. y= 3.42 moles/100mL
y= 34.2moles/L
8 M HCl = KCl =[K+][Cl-]
Let b= molar solubility :. B, b+8
Ksp =13.69= [K+][Cl-] = (b)(b*+8)
* - rule of thumb
Ksp =13.69= [K+][Cl-] = (b)(b*+8)= b(8)
13.69=8b:. b= 1.71 moles/100mL
17.1 moles/L
Therefore in a solution with a common ion, the solubility of the compound reduces dramatically.
