Ksp for AgBr is 5x10-13. What is the maximum concentration of silver ion that you can have in a 0.1 M solution of NaBr?
........AgBr ==> Ag^+ + Br^-
I......solid.....0......0
C......solid.....x......x
E......solid.....x......x
........NaBr --> Na^+ +Br^-
I......0.1.......0.....0
C.....-0.1......0.1...0.1
E........0......0.1...0.1
Note that NaBr is completely dissociated while AgCl is not.
Write the Ksp expression for AgBr, and substitute the concentrations from thre above two equations.
(Ag^+) = x
(Br^-) = x from AgBr and 0.1 from NaBr
Solve for x = (Ag^+)
To find the maximum concentration of silver ions (Ag+) in a 0.1 M solution of NaBr, we need to determine if a precipitate will form or not. We can use the concept of the solubility product constant (Ksp) to do this.
The Ksp expression for AgBr is written as:
Ksp = [Ag+][Br-]
Given that the Ksp for AgBr is 5x10^(-13), we can assign this value to Ksp as follows:
5x10^(-13) = [Ag+][Br-]
Since NaBr is a strong electrolyte, it completely dissociates into its ions in solution:
NaBr(s) → Na+(aq) + Br-(aq)
Therefore, the concentration of bromide ions (Br-) in the solution will be equal to the initial concentration of NaBr. In this case, it is 0.1 M.
Now, assuming that x represents the concentration of silver ions (Ag+), we can express the Ksp equation as:
5x10^(-13) = (x)(0.1)
Solving for x:
x = (5x10^(-13))/(0.1)
= 5x10^(-12) M
Therefore, the maximum concentration of silver ions (Ag+) that can exist in the 0.1 M solution of NaBr is 5x10^(-12) M.
To determine the maximum concentration of silver ion (Ag+) in a 0.1 M solution of NaBr, we need to consider the solubility product constant (Ksp) for AgBr.
The Ksp expression for AgBr is given as:
Ksp = [Ag+][Br-]
The balanced equation for the dissociation of AgBr in water is:
AgBr(s) ⇌ Ag+(aq) + Br-(aq)
The Ksp value provided (5x10^-13) represents the product of Ag+ and Br- concentrations when AgBr is at saturation point.
Since NaBr is a soluble ionic compound, it dissociates into Na+ and Br- ions in water. However, AgBr has a much lower solubility, and therefore, we assume it remains in the solid state.
Now, let's assume x represents the concentration of Ag+ in the solution. Since NaBr is a 0.1 M solution, the concentration of Br- ions is also 0.1 M.
Using the Ksp expression, we can write:
5x10^-13 = (x)(0.1)
Rearranging the equation and solving for x, we get:
x = (5x10^-13) / (0.1)
Calculating this expression, we find that the maximum concentration of silver ion (Ag+) in the 0.1 M solution of NaBr is 5x10^-12 M.