The molar solubility of gold(I) chloride is 4.510-7 mol/L. Calculate the solubility constant of this
compound.
Well, let's dive into the world of solubility constants, shall we? It's like solving a mystery, but with chemistry!
So, we know that the molar solubility of gold(I) chloride is 4.5 x 10-7 mol/L. To find the solubility constant (Ksp), we need to set up an equilibrium expression using this molar solubility.
Now, gold(I) chloride can dissociate into gold(I) ions (Au+) and chloride ions (Cl-). The equation would look like this:
AuCl ↔ Au+ + Cl-
Since the stoichiometric coefficient of AuCl is 1, we can say that the concentration of Au+ and Cl- would also be 4.5 x 10-7 mol/L.
Now, let's put these concentrations into our equilibrium expression:
Ksp = [Au+][Cl-]
Ksp = (4.5 x 10-7)(4.5 x 10-7)
Calculating that, we get a result of 2.025 x 10-13 mol2/L2.
So there you have it, the solubility constant (Ksp) of gold(I) chloride is approximately 2.025 x 10-13 mol2/L2. It's a tiny number, but it describes the equilibrium between the dissolved and undissolved forms of the compound. Pretty fascinating, right?
To calculate the solubility constant (Ksp) of gold(I) chloride, we need to use the molar solubility given. The molar solubility represents the concentration of the ions in a saturated solution.
The balanced equation for the dissociation of gold(I) chloride is:
AuCl <=> Au+ + Cl-
The solubility product expression for this reaction is:
Ksp = [Au+][Cl-]
Since the molar solubility of gold(I) chloride is 4.5 x 10-7 mol/L, we can use this value to represent the concentration of both Au+ and Cl- ions in the equation.
Therefore, the solubility constant (Ksp) of gold(I) chloride is:
Ksp = (4.5 x 10^-7)(4.5 x 10^-7) = 2.025 x 10^-13
To calculate the solubility constant of a compound, we need to use the molar solubility. The solubility constant, also known as the solubility product constant (Ksp), is a measure of the equilibrium between a sparingly soluble compound and its dissolved ions in a solution.
In this case, the molar solubility of gold(I) chloride is given as 4.5x10^-7 mol/L.
The general reaction equation for the dissolution of a compound can be written as:
AuCl(s) ⇌ Au+ + Cl-
The solubility product constant expression for this reaction can be written as:
Ksp = [Au+][Cl-]
Since the molar solubility of gold(I) chloride is the same as the concentration of the ions in the equilibrium state, we can substitute the values into the Ksp expression:
Ksp = (4.5x10^-7)(4.5x10^-7)
Ksp = 2.025x10^-13
Therefore, the solubility constant of gold(I) chloride is 2.025x10^-13.
.......................AuCl(s) ----> Au^+ + Cl^-
I.......................solid...............0.........0
C.....................solid-x.............x..........x
E......................solid................x..........x
Ksp = (Au^+)(Cl^-) = (x)(x)
The problem tells you that x = 4.5E-7
Substitute that number and solve for Ksp.
Post your work if you get stuck.