Calculate the molar solubility of gold(I) chloride (Ksp= 2.0 x 10^-13) in .10 M NaCN. The complex ion formed is [Au(CN)2]- with Kf= 2.0 x 10^38. Ignore any other competing equilibrium systems.
I'm having trouble writing an equation to start off with.
First, I had written
AuCl + NaCN <--> [Au(CN)2]- + NaCl (s)
But I can't seem to balance it correctly.
I think I know how to do the problem if I have just one reactant, but I'm not sure what to do with two. Any help would be appreciated.
The equation is not balanced because the CN- is not balanced. It can be calanced this way.
AuCl + 2NaCN ==> Na[Au(CN)2] + NaCl. In aqueous solution, the sodium salt of the complex ion will ionize into Na^+ and Au(CN)2^- while the NaCl will ionize into Na^+ and Cl^-.
To balance the equation, we need to consider the charges of the ions involved.
AuCl + 2NaCN → Na[Au(CN)2] + NaCl
Here, we see that the complex ion [Au(CN)2]- has a charge of -1, so we need 2 CN- ligands to balance the charge of the Au3+ ion.
Next, we have Na+ ions on both sides of the equation, so we can eliminate them.
AuCl + 2CN- → [Au(CN)2]- + Cl-
The equation is now balanced in terms of charges.
Now, let's consider the solubility equilibrium. The molar solubility of a compound is given by the concentration of the compound when it reaches equilibrium with its ions in solution.
The solubility product constant, Ksp, is the equilibrium constant for the dissolution of a sparingly soluble compound in water.
For gold(I) chloride (AuCl), the equilibrium expression would be:
Ksp = [Au+][Cl-]
Since AuCl is a sparingly soluble compound, we can assume that a negligible amount of it will dissociate into its ions in solution. Therefore, the concentration of Au+ can be approximated as equal to the molar solubility of AuCl.
For the complex ion [Au(CN)2]-, the equilibrium constant Kf is related to the stability constant of the complex. In this case, Kf is given as 2.0 x 10^38.
The complex ion formation can be represented as:
[Au(CN)2]- ⇌ Au+ + 2CN-
We can use the stability constant Kf to determine the concentration of Au+ and CN-.
Now, let's set up an ICE table to determine the concentrations at equilibrium:
Initial: AuCl: x Cl-: 0
Final: AuCl: x Cl-: x
Change: AuCl: 0 Cl-: x
Since Cl- comes from the dissociation of AuCl, its concentration will be equal to the concentration of AuCl, which is x.
Now, for the complex ion formation:
Initial: [Au(CN)2]-: 0 CN-: 2x
Final: [Au(CN)2]-: x CN-: x
Change: [Au(CN)2]-: x CN-: -x
The CN- concentration will decrease by x, and the [Au(CN)2]- concentration will increase by x.
From the ICE table, we can substitute the concentrations into the equilibrium expression for the complex ion formation:
Kf = [Au+][CN-]^2 / [Au(CN)2]-
Kf = (x)(x)^2 / x
Simplifying:
Kf = x^2
Now, we need to determine the value of x and substitute it back into the solubility expression for AuCl to find the molar solubility.
Ksp = [Au+][Cl-]
Ksp = (x)(x)
Ksp = x^2
We know that Ksp = 2.0 x 10^-13, so we can solve for x:
2.0 x 10^-13 = x^2
Taking the square root of both sides:
x = √(2.0 x 10^-13)
x ≈ 1.41 x 10^-7 M
Therefore, the molar solubility of gold(I) chloride in 0.10 M NaCN is approximately 1.41 x 10^-7 M.