Respuesta :
Answer:
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.
Answer:
(a) 13.69
(b) i beaker 1: 0g
ii beaker 2: 0g
Explanation:
a. The solubility equilibrium equation for KCl is
KCl(s) ⇄ K⁺(aq) + Cl⁻(aq)
3.7M KCl contains equal moles of K ions and Cl ions
therefore, the ion-product expression is written thus
Ksp = [K⁺][Cl⁻]
= [3.7][3.7]
= 13.69
b. from the first two beakers containing 100 mL and 3.7M KCl
moles of K⁺ = moles of Cl⁻ = moles of KCl = 3.7moles in 1L
if 3.7M Implies 3.7 moles in 1L or 1000 mL or 1000 cm³
how many moles will be contained in 100 mL
this is calculated as follows
3.7moles/Liter * 100 mL
[tex]\frac{3.7 moles KCl}{1000 mL} * 100 ml = 0.37moles KCl[/tex]
= 0.37moles K⁺ = 0.37moles Cl⁻
4.0 M HCl, contains
[tex]\frac{4 moles HCl}{1000 mL} *100mL = 0.4moles HCl = 0.4 moles H = 0.4moles Cl[/tex] in 100mL
8.0M HCl, contains
[tex]\frac{8moles HCL}{1000mL} *100mL=0.8mole HCl=0.8molesH=0.8molesCl[/tex] in 100mL
now, in the first beaker 100 mL of 4M HCl is added to 100 mL of 3.7M KCl
total moles of Cl⁻ (0.4 + 0.37) moles = 0.77 moles
total moles of K⁺ remains 0.37 moles
total volume of solution = (100mL + 100mL) = 200mL/1000mL = 0.2L
total moles of Cl⁻ per Liter = 0.77moles/0.2L = 3.85M Cl⁻
total moles of K⁺ per Liter = 0.37moles/0.2L = 1.85M K⁺
Qsp must be greater or equal to Ksp for Precipitation to occur, that is
Qsp ≥ Ksp
Qsp = [K][Cl] = [1.85][3.85] = 7.12 this is less than 13.69(Ksp)
hence no KCl will precipitate in the first beaker
since there is no precipitate, there is therefore no need for calculating the mass precipitated
and the answer is 0g
(bii) now, in the second beaker 100 mL of 8M HCl is added to 100 mL of 3.7M KCl
total moles of Cl⁻ (0.8 + 0.37) moles = 1.17 moles
total moles of K⁺ remains 0.37 moles
total volume of solution = (100mL + 100mL) = 200mL/1000mL = 0.2L
total moles of Cl⁻ per Liter = 1.17moles/0.2L = 5.85M Cl⁻
total moles of K⁺ per Liter = 0.37moles/0.2L = 1.85M K⁺
Qsp must be greater or equal to Ksp for Precipitation to occur, that is
Qsp ≥ Ksp
Qsp = [K][Cl] = [1.85][5.85] = 10.82 this is less than 13.69(Ksp)
hence no KCl will precipitate also in the second beaker
since there is no precipitate, there is therefore no need fo calculating the mass precipitated
and the answer is 0g
