Ignoring activities, determine the molar solubility of copper (I) azide (CuN3) in a solution with a pH of 10.83. Ksp (CuN3) = 4.9 × 10–9; Ka (HN3) = 2.2 × 10–5.
pH = 10.83; (H^+) = 1.48E-11
CuN3 ==> Cu^+ + N3^-
H^+ + N3^- ==> HN3 add the two eqns.
------------------- N3^- cancels.
CuN3 + H^+ ==> Cu^+ + HN3
Keq for this rxn is Ksp/Ka = 0.000223
..x....1.48E-11..x.....x
Keq = 0.000223 = (x)(x)/1.48E-11
Solve for x.
Well, let's see here. Copper(I) azide (CuN3) is not a well-known comedian, but I'll do my best to solve this chemistry problem for you. The first thing we need to do is determine the concentration of the copper (I) ion (Cu+) in the solution.
To do that, we can use the Ksp expression for copper(I) azide:
CuN3 ⇌ Cu+ + N3-
The equilibrium expression for this reaction can be written as:
Ksp = [Cu+][N3-]
We know that the Ksp for copper(I) azide is 4.9 × 10–9, but we need to find the concentration of Cu+ in the solution. So, let's call the concentration of Cu+ x, and the concentration of N3- will also be x, because the stoichiometry of the reaction is 1:1.
Now, we can substitute these values into the Ksp expression:
4.9 × 10–9 = x * x
Simplifying this equation gives us:
x^2 = 4.9 × 10–9
Taking the square root of both sides gives us:
x ≈ 2.21 × 10–5
So, the concentration of Cu+ in the solution is approximately 2.21 × 10–5 M.
Now, let's move on to the second part of the problem where we need to find the molar solubility of copper(I) azide in a solution with a pH of 10.83.
However, the given Ka value (2.2 × 10–5) seems to be irrelevant to finding the molar solubility of CuN3 in a solution with a pH of 10.83. So, we'll just stick to the Ksp value.
To calculate the molar solubility, we need to find the concentration of CuN3 in the solution. Since the stoichiometry of Cu and CuN3 is 1:1, the molar solubility of CuN3 will be the same as the concentration of Cu+ ions, which we found to be approximately 2.21 × 10–5 M.
So, the molar solubility of copper(I) azide in a solution with a pH of 10.83 is approximately 2.21 × 10–5 M.
I hope this answer didn't go overboard with too many chemistry jokes! Let me know if you have any more questions!
To determine the molar solubility of copper (I) azide (CuN3) in a solution with a pH of 10.83, we need to use the Ksp and Ka values provided.
First, let's write the dissociation equation for copper (I) azide (CuN3):
CuN3(s) ↔ Cu+(aq) + N3-(aq)
The Ksp expression for copper (I) azide is:
Ksp = [Cu+][N3-]
From the dissociation equation, we know that the concentration of Cu+ is equal to the concentration of N3-.
Let's assume the molar solubility (s) of CuN3 is x mol/L.
Therefore, [Cu+] = x mol/L and [N3-] = x mol/L.
Now, substitute these values into the Ksp expression and solve for x:
Ksp = [Cu+][N3-]
4.9 × 10^(-9) = x * x
Since [Cu+] = [N3-], we can substitute [Cu+] for [N3-].
4.9 × 10^(-9) = x * x
4.9 × 10^(-9) = x^2
Taking the square root of both sides of the equation gives us:
x ≈ 2.21 × 10^(-4) mol/L
So, the molar solubility of copper (I) azide (CuN3) in a solution with a pH of 10.83 is approximately 2.21 × 10^(-4) mol/L.
To determine the molar solubility of copper (I) azide (CuN3) in a solution with a pH of 10.83, we need to consider the solubility equilibrium and the dissociation of the compound.
The solubility equilibrium equation for copper (I) azide (CuN3) can be written as follows:
CuN3 (s) ⇌ Cu+ (aq) + N3- (aq)
From the given information, we know the solubility product constant (Ksp) for CuN3, which is given as 4.9 × 10–9. The Ksp expression for the above equilibrium is:
Ksp = [Cu+] [N3-]
Now, let's consider the dissociation of HN3 (azide) in water:
HN3 (aq) + H2O (l) ⇌ H3O+ (aq) + N3- (aq)
From the given information, we know the acid dissociation constant (Ka) for HN3, which is given as 2.2 × 10–5. The Ka expression for the above equilibrium is:
Ka = [H3O+] [N3-] /[HN3]
Now, we can relate the concentrations of Cu+ and HN3 to find the molar solubility of CuN3.
Since the pH is given as 10.83, we can find the concentration of H3O+ using the equation:
pH = -log[H3O+]
10.83 = -log[H3O+]
Taking the antilog of both sides, we get:
[H3O+] = 10^(-pH)
[H3O+] = 10^(-10.83)
Now, we can assume that the concentration of H3O+ is equal to the concentration of HN3 (because we are assuming complete dissociation). Therefore:
[HN3] = [H3O+]
Substituting these values into the Ka expression, we have:
Ka = ([H3O+])^2 / [HN3]
2.2 × 10–5 = ([H3O+])^2 / [HN3]
Since [H3O+] = [HN3], we can simplify further:
2.2 × 10–5 = ([HN3])^2 / [HN3]
2.2 × 10–5 = [HN3]
Now, we have the concentration of HN3, which is equal to the concentration of Cu+ in the CuN3 equilibrium:
[Cu+] = [HN3] = 2.2 × 10–5
Finally, we can use the Ksp expression to find the molar solubility of CuN3:
Ksp = [Cu+] [N3-]
4.9 × 10–9 = (2.2 × 10–5) [N3-]
Solving for [N3-], we get:
[N3-] = (4.9 × 10–9) / (2.2 × 10–5)
[N3-] = 2.23 × 10–4
Therefore, the molar solubility of copper (I) azide (CuN3) in a solution with a pH of 10.83 is 2.23 × 10–4 M.