Equilibrium expression: AgCl(s) ⇌ Ag+(aq) + Cl-(aq) KSP = 1.8 x 10-10 The solubility of AgCl in: i) pure water: solubility = 1.3 x 10-5 M
ii) a 0.50 M NaCl solution solubility = 3.6 x 10-10 M as KSP = [Ag+][Cl-]Total where [Cl-]Total = [Cl-]AgCl + [Cl-]NaCl
Equilibrium expression: AgCl(s) ⇌ Ag+(aq) + Cl-(aq) KSP = 1.8 x 10-10 The solubility of AgCl in: i) pure water: solubility = 1.3 x 10-5 M
ii) a 0.50 M NaCl solution solubility = 3.6 x 10-10 M as KSP = [Ag+][Cl-]Total where [Cl-]Total = [Cl-]AgCl + [Cl-]NaCl
I agree with 1.3E-5M for the solubility of AgCl in pure water.
Also I agree with 3.6E-10 M for the solubility of AgCl in 0.5M NaCl but nothing you posted after 3.6E-10 makes sense to me. You don't have a question; I assume you just want an answer check.
To find the solubility of AgCl in different solutions, we can use the equilibrium expression and the concept of the solubility product constant (Ksp).
i) Solubility of AgCl in pure water (H2O):
The equilibrium expression for the dissolution of AgCl in water is:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
The Ksp expression is: Ksp = [Ag+][Cl-]
Given Ksp = 1.8 x 10^(-10) and we assume the solubility of AgCl as x in moles per liter, we can set up the equation:
Ksp = x * x = x^2
Since x is the molar solubility of AgCl, we solve the equation for x:
x^2 = 1.8 x 10^(-10)
Taking the square root of both sides:
x = √(1.8 x 10^(-10)) ≈ 1.34 x 10^(-5) M
Therefore, the solubility of AgCl in pure water is approximately 1.34 x 10^(-5) M.
ii) Solubility of AgCl in a 0.50 M NaCl solution:
To determine the solubility of AgCl in a NaCl solution, we need to consider the additional Cl- ions from the NaCl.
The equilibrium expression for the dissolution of AgCl in this case becomes:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
With the presence of NaCl, the concentration of Cl- ions from NaCl can be represented as [Cl-]NaCl = 0.50 M. Let [Cl-]AgCl represent the concentration of Cl- ions coming from the dissolution of AgCl (which is the solubility of AgCl), we can write:
[Cl-]Total = [Cl-]AgCl + [Cl-]NaCl
Here, [Cl-]Total refers to the total concentration of Cl- ions in the solution.
The Ksp expression is still the same: Ksp = [Ag+][Cl-]Total
Given Ksp = 1.8 x 10^(-10), we can substitute the values:
1.8 x 10^(-10) = [Ag+][Cl-]Total
Since [Cl-]Total = [Cl-]AgCl + [Cl-]NaCl and [Cl-]NaCl = 0.50 M:
1.8 x 10^(-10) = [Ag+]( [Cl-]AgCl + 0.50 M)
The solubility of AgCl is represented by [Cl-]AgCl, so we can isolate it:
[Cl-]AgCl = (1.8 x 10^(-10) )/[Ag+]- 0.5 M
Given the solubility of AgCl in pure water is 1.3 x 10^(-5) M, which is the concentration of [Ag+] in the AgCl + NaCl solution, we can substitute it into the equation:
[Cl-]AgCl = (1.8 x 10^(-10) )/(1.3 x 10^(-5))- 0.5 M
Simplifying the equation:
[Cl-]AgCl ≈ 3.6 x 10^(-10) M
Therefore, the solubility of AgCl in a 0.50 M NaCl solution is approximately 3.6 x 10^(-10) M.