Calculate the solubility of silver chloride in a 0.010 mol/L solution of sodium chloride at 25 degrees Celsius. At SATP, Ksp AgCl(s) = 1.8 x 10^-10
AgCl <==> Ag^+ + Cl^-
NaCl ==> Na^+ + Cl^-
Ksp = (Ag^+)(Cl^-) = 1.8 x 10^-10
Let S = solubility of AgCl, then
(Ag^+) = S
(Cl^-) = S+0.01
Solve for S.
Note: A similar problem to this post (0.1 M NaCl instead of 0.01 M NaCl) was on a couple of days ago; the one who posted said that the answer came back incorrect. Check my work. Check myu thinking.
To calculate the solubility of silver chloride (AgCl) in a 0.010 mol/L solution of sodium chloride (NaCl) at 25 degrees Celsius, we need to consider the effect of the common ion (chloride ion, Cl-) on the solubility.
The solubility product constant (Ksp) for AgCl at SATP (Standard Ambient Temperature and Pressure) is given as 1.8 x 10^-10. The equation for the dissolution of AgCl in water is:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Using the concept of equilibrium, we can write an expression for the solubility of AgCl, represented as "s":
Ksp = [Ag+][Cl-] = (s)(s) = s^2
Now, we need to consider the effect of the common ion, Cl-, from the NaCl solution. The concentration of Cl- in the 0.010 mol/L NaCl solution is also 0.010 mol/L.
Therefore, the new expression for the solubility of AgCl, considering the common ion effect, is:
Ksp = [Ag+][Cl-] = (s + 0.010)(0.010)
Substituting the value of Ksp = 1.8 x 10^-10, we can solve for the solubility of AgCl (s):
1.8 x 10^-10 = (s + 0.010)(0.010)
Dividing both sides by 0.010 gives:
1.8 x 10^-10 / 0.010 = s + 0.010
Simplifying further, we have:
1.8 x 10^-8 = s + 0.010
Subtracting 0.010 from both sides of the equation gives:
1.8 x 10^-8 - 0.010 = s
Simplifying:
s = -0.010 + 1.8 x 10^-8
Therefore, the solubility of silver chloride (AgCl) in a 0.010 mol/L solution of sodium chloride (NaCl) at 25 degrees Celsius is approximately equal to -0.010 + 1.8 x 10^-8.
To calculate the solubility of silver chloride (AgCl) in a 0.010 mol/L solution of sodium chloride (NaCl), we need to make use of the solubility product constant (Ksp) for AgCl.
The solubility product constant (Ksp) is an equilibrium expression that relates to the solubility of a sparingly soluble salt. For AgCl(s), the Ksp is given as 1.8 x 10^-10 at SATP (Standard Ambient Temperature and Pressure), which means that the product of the concentrations of silver ions (Ag+) and chloride ions (Cl-) in a saturated solution of AgCl is equal to this value.
Let's assume that x mol/L of AgCl will dissolve in the solution. Since AgCl dissociates into Ag+ and Cl- ions, we can write the following expression:
Ksp = [Ag+][Cl-]
Now, let's consider the concentration of chloride ions ([Cl-]) in a 0.010 mol/L solution of sodium chloride. Since NaCl is a strong electrolyte and completely dissociates into Na+ and Cl-, the concentration of chloride ions is also 0.010 mol/L.
Substituting the given values into the expression:
1.8 x 10^-10 = (x)(0.010)
Solving for x:
x = (1.8 x 10^-10) / (0.010)
x = 1.8 x 10^-8 mol/L
Therefore, the solubility of silver chloride (AgCl) in a 0.010 mol/L solution of sodium chloride at 25 degrees Celsius is approximately 1.8 x 10^-8 mol/L.