CALCULATE THE SOLUBILITY OF CuBr in 0.69 M NH3
...CuBr(s) + 4NH3 ==> Cu(NH3)4^2+ + Br^-
I....solid....0.69.....0............0
C....solid.....-4x.....x............x
E....solic....0.69-x...x............x
What we usually do is to say that this reaction goes essentially to completion since Kf for the complex is such a large number of 1.1E13 (you should look this up in your text/notes and use what it shows). So we redo the above ICE chart but work backwards (from right to left an start with 0.69 M for the complex.
---------------------------------
I.....0.......0......0.172....0.172
C.....x.......4x..... -x........-x
E.....x.......4x.....0.172-x..0.172-x
Substitute E line into K expression and solve for x.
To calculate the solubility of CuBr in 0.69 M NH3, we can use the concept of complex ion formation and the equilibrium constant.
The dissolution of CuBr in water can be represented by the following equation:
CuBr(s) ⇌ Cu2+(aq) + 2Br-(aq) (Equation 1)
In the presence of NH3, ammonia can form a complex ion with the copper ion:
Cu2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4](2+)(aq) (Equation 2)
The equilibrium constant (Kf) for the complex ion formation is given as 2.1 x 10^13.
We can write the combined equation for the dissolution of CuBr in NH3 as:
CuBr(s) + 4NH3(aq) ⇌ [Cu(NH3)4](2+)(aq) + 2Br-(aq) (Combined Equation)
Let's assume that 'x' Moles of CuBr dissolves in NH3. Then, the concentration of CuBr(aq) is (0.69 - x) M, [Cu(NH3)4](2+)(aq) is 'x' M, and [Br-] is 2x M.
Using the concentrations and the equilibrium constant, we can write the following expression for the equilibrium constant (Kc):
Kc = ([Cu(NH3)4](2+)(aq) * [Br-](aq)^2) / ([CuBr](aq)) * ( [NH3](aq))^4)
Substituting the known values:
2.1 x 10^13 = (x * (2x)^2) / (0.69 - x) * (0.69)^4
Simplifying the equation, we get:
2.1 x 10^13 = 4x^3 / (0.69 - x) * (0.69)^4
Now, we can solve this equation to find the value of 'x', which represents the solubility of CuBr in 0.69 M NH3.
To calculate the solubility of CuBr in 0.69 M NH3, we need to consider the equilibrium between the CuBr solid and its dissolved ions in the ammonia solution. The equation for this equilibrium is:
CuBr(s) ⇌ Cu+(aq) + Br-(aq)
First, we need to write the equilibrium constant expression for the dissociation of CuBr:
Ksp = [Cu+][Br-]
Now, we consider the reactant NH3, which is a weak base and can also react with Cu+ to form a complex ion. The equation for this reaction is:
Cu+(aq) + 4 NH3(aq) ⇌ Cu(NH3)4^2+(aq)
The formation constant for this complex ion is denoted as Kf.
Now, the total concentration of Cu+ can be calculated by considering the original amount of CuBr that dissolves and the amount that comes from the complex ion:
[Cu+] = [Cu(NH3)4^2+] + [CuBr]
Next, we need to write an expression for [Cu(NH3)4^2+] in terms of [Cu+] and [NH3]. Since [Cu(NH3)4^2+] will depend on [Cu+], which is what we want to calculate, we initially assume that all Cu+ ions react with NH3 to form the complex ion:
[Cu(NH3)4^2+] = 4[Cu+]
Substituting this expression into the [Cu+] equation, we get:
[Cu+] = 4[Cu+] + [CuBr]
Solving for [Cu+], we find:
[Cu+] = [CuBr] / 3
We now have an expression for [Cu+] in terms of [CuBr], which allows us to rewrite the equilibrium constant expression:
Ksp = ([CuBr] / 3)[Br-]
Next, we need to express the concentration of Br- in terms of the initial concentration of CuBr and the concentration of NH3. Since CuBr is a strong electrolyte, it completely dissociates into Cu+ and Br-. Hence, the concentration of Br- is equal to the initial concentration of CuBr:
[Br-] = [CuBr]
We can now rewrite the equilibrium constant expression:
Ksp = ([CuBr] / 3)[CuBr]
Since Ksp is a constant, we can substitute the given values into the equation, rearrange, and solve for [CuBr]:
Ksp = ([CuBr] / 3)[CuBr]
[CuBr] = (3Ksp / [CuBr])^(1/2)
Now, substitute the given value of Ksp and solve for [CuBr]:
[CuBr] = (3 * Ksp / 0.69)^(1/2)
By plugging in the value for Ksp and calculating, you can find the solubility of CuBr in 0.69 M NH3. Keep in mind the units while performing the calculation.