Question

Calculate the concentration of Ag+ when 0.0050 M of KSCN saturates with AgSCN using activity coefficients. Ksp = 1.1x10^12 (Show work)

Answer 1

To calculate the concentration of Ag+ when 0.0050 M of KSCN saturates with AgSCN using activity coefficients, we need to consider the equilibrium reaction:

AgSCN ⇌ Ag+ + SCN-

The solubility product constant (Ksp) expression for this reaction is:

Ksp = [Ag+][SCN-]

Given that the concentration of KSCN is 0.0050 M, we can assume that the concentration of SCN- is also 0.0050 M since KSCN is a 1:1 electrolyte.

Let's assume the concentration of Ag+ is x M when AgSCN reaches saturation. Therefore, the concentration of SCN- will also be x M.

Using the Ksp expression, we have:

Ksp = [Ag+][SCN-] 1.1x10^12 = x * x

Simplifying the equation, we get:

1.1x10^12 = x^2

Taking the square root of both sides, we have:

√(1.1x10^12) = √(x^2) 3.32x10^6 = x

Therefore, the concentration of Ag+ when 0.0050 M of KSCN saturates with AgSCN using activity coefficients is approximately 3.32x10^6 M.