Answer:
a) The ion-product constant Ksp was determined to be 13.69
Explanation:
Terminology
The Qsp for an ionic solid dissolving reflects the solubility product of ions in solution.
Ksp, in contrast, defines the equilibrium state solubility product of these ions in solution when in balance with the dissolving solid.
It’s important to note that if Qsp exceeds Ksp at a particular temperature, precipitation will occur, causing the equilibrium to shift left to maintain balance (Ksp).
Steps to Solve:
To calculate this:
1. Replace the molar solubility of KCl into the ion-product equation to identify the Ksp of KCl.
2. Assess the total concentration of potassium chloride ions in each beaker after HCl has been added, taking into consideration initial moles and those added.
3. Calculate Qsp to determine whether it surpasses Ksp. If Qsp is below Ksp, no precipitation occurs.
Thus, the equilibrium equation for KCl can be expressed as:
KCL_(s) ---> K+(aq) + Cl- (aq)
The provided KCl solubility is 3.7 M.
Ksp= [K+][Cl-] = (3.7)(3.7) =13.69
Therefore, the Ksp was found to be equal to 13.69.
In pure water, KCl
Ksp =13.69 KCl =[K+][Cl-]
Let x represent the molar solubility [K+]/[Cl-]:. x, x
Ksp =13.69 = [K+][Cl-] = (x)(x) = x²
x= √ 13.69 = 3.7 M moles of KCl necessary for a 100mL saturated solution
37M moles/L
The Ksp was determined to be equal to 13.69.
4.0 M HCl = KCl =[K+][Cl-]
Let y signify the molar solubility:. y, y+4
Ksp =13.69= [K+][Cl-] = (y)(y*+4)
* - as a general guideline
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 denote the molar solubility:. B, b+8
Ksp =13.69= [K+][Cl-] = (b)(b*+8)
* - as a general guideline
Ksp =13.69= [K+][Cl-] = (b)(b*+8)= b(8)
13.69=8b:. b= 1.71 moles/100mL
17.1 moles/L
Thus, in a solution containing a common ion, the compound's solubility significantly decreases.
