What is the concentration of Ag+ in 0.0543M AgNO3 and 1.1657CN-?

Ag+ +2CN- <-> [Ag(CN)sub2]-
kf = 5.3x10^18

To find the concentration of Ag+ in the given solution, we need to consider the equilibrium expression for the reaction between Ag+ and CN- to form [Ag(CN)sub2]-:

Ag+ + 2CN- ↔ [Ag(CN)sub2]-

The equilibrium constant for this reaction, kf, is given as 5.3x10^18.

Now, let's assume that 'x' represents the change in concentration of Ag+ in the reaction. At equilibrium, the concentration of Ag+ will be 0.0543 - 'x' since it reacts with CN-.

Considering the stoichiometry of the reaction, the change in concentration of [Ag(CN)sub2]- will be '2x' because 1 mole of Ag+ reacts with 2 moles of CN- to form 1 mole of [Ag(CN)sub2]-.

Using the equilibrium constant equation for the reaction, we can write:

kf = [Ag(CN)sub2]- / (Ag+ * [CN-]²)

Plugging in the given values:

5.3x10^18 = (2x) / ((0.0543 - x) * (1.1657)^2)

Simplifying the equation, we get:

5.3x10^18 * (0.0543 - x) * (1.1657)^2 = 2x

Now, we can solve this equation to find the value of 'x' and then calculate the concentration of Ag+ at equilibrium. However, note that this is a quadratic equation, and solving it can be a bit complex.

If you have access to software like MATLAB or a graphing calculator, you can plot the quadratic equation and find the value of x. Alternatively, you can use an online quadratic equation solver to determine the value of 'x'.

Once you have the value of 'x', you can substitute it back into the expression [Ag+ concentration] = 0.0543 - 'x' to calculate the concentration of Ag+.