Calculate the solubility of silver chloride in a solution that is 0.140 M in NH3.
1.Write the equations for AgCl, Ag(NH3)2, and NH3. For example, AgCl ==> Ag^+ + Cl^-,
Ag^+ + 2NH3 ==> Ag(NH3)2^+ ,
and the NH3 + H2O ==> NH4^+ + OH^-
2. Write the Ksp for AgCl, Kb for NH3, and Kformation for Ag(NH3)2.
3. Solubility AgCl = (Cl^-) = (Ag^+) + [Ag(NH3)2^+].
4. Write mass balance equations.
For example, [Cl^-]= (Ag^+) + [Ag(NH3)2^+]
0.140 = (NH3) + (NH4^+) + 2[Ag(NH3)2^+]
and (OH^-) = (NH4^+)
5. Make appropriate assumptions to reduce the number of equations (you should reduce the six equations to three) and solve for Ag^+, Ag(NH3)2^+, and Cl^-. Three equations and three unknowns. Post all of your work if you get stuck.
does H2+ exits
To calculate the solubility of silver chloride (AgCl) in a solution that is 0.140 M in ammonia (NH3), we need to consider the formation of the complex ion Ag(NH3)2+ (known as the diamminesilver(I) ion).
The solubility product constant (Ksp) expression for silver chloride is:
Ksp = [Ag+][Cl-]
However, in the presence of NH3, the formation of Ag(NH3)2+ can occur:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
Ag+ (aq) + 2NH3 (aq) ⇌ Ag(NH3)2+ (aq)
The equilibrium constant for the formation of Ag(NH3)2+ is represented as Kf.
We can write the overall equilibrium expression as:
Koverall = Ksp × Kf = [Ag+][Cl-] × [Ag(NH3)2+]
Given that the concentration of NH3 is 0.140 M, we use the balanced equation for the formation reaction to determine the effect of NH3 on the equilibrium concentrations of Ag+ and Cl- ions.
In this equation, the concentration of Ag+ decreases by the stoichiometric coefficient of 1 (as there is only one Ag+ formed) and the concentration of NH3 increases by 2 (as there are two NH3 molecules involved).
To calculate the solubility of AgCl, we will assume that x is the molar solubility of AgCl in the presence of NH3.
Therefore, the concentration of Ag+ is: [Ag+] = x
The concentration of Cl- is: [Cl-] = x
The concentration of NH3 is: [NH3] = 0.140 M + 2x
Substituting these concentrations into the overall equilibrium expression, we get:
Koverall = (x)(x) × [0.140 M + 2x]
Simplifying further and ignoring the contribution of Cl- to changes in concentration due to x being small, we get:
Koverall ≈ x^2 × 0.140 M
Now, we will determine the value of Kf, also known as the overall formation constant for Ag(NH3)2+.
The value of Kf can be looked up in a reference source or determined experimentally. For Ag(NH3)2+, Kf is approximately 1.5 × 10^7.
Finally, we can calculate the solubility of AgCl by setting up an equation using the known values for Koverall and Kf:
Koverall = Ksp × Kf
x^2 × 0.140 M ≈ Ksp × (1.5 × 10^7)
By solving this equation for x, you can determine the solubility of AgCl in the given ammonia solution.