K =[CH3COO^-]/[CN^-]
K =[1.5×10^-5] / [4.5×10^-10]
K=0.3×10^5 =3×10^4

Dissociation Constants and Equilibrium Constant Calculation

Dissociation constants are measures of the extent to which a weak acid or base dissociates in a solution. They are usually denoted by the symbol Ka. In this case, we have two dissociation constants: Ka1 for acetic acid and Ka2 for HCN.

Given:
Ka1 (acetic acid) = 1.5 * 10^-5
Ka2 (HCN) = 4.5 * 10^-10

We are asked to calculate the equilibrium constant for the following reaction:
CN- + CH3COOH ⇌ HCN + CH3COO-

To calculate the equilibrium constant, we need to write the expression for it based on the balanced equation. The equilibrium constant expression is given by:

Kc = [HCN] * [CH3COO-] / [CN-] * [CH3COOH]

Now, let's determine the equilibrium concentrations of the species involved.

Step 1: Let x be the concentration of CN-, CH3COOH, HCN, and CH3COO- at equilibrium.

Step 2: From the balanced equation, we can see that 1 mole of CN- reacts with 1 mole of CH3COOH to give 1 mole of HCN and 1 mole of CH3COO-. Therefore, the concentration of CN- and CH3COOH at equilibrium will be (initial concentration - x).

Step 3: The concentration of HCN and CH3COO- at equilibrium will be x.

Step 4: Substitute these concentrations into the equilibrium constant expression.

Kc = [x] * [x] / [(initial concentration - x)] * [(initial concentration - x)]

Step 5: As the equilibrium constants are very small, we can assume that the change in concentration (x) is negligible compared to the initial concentration. Therefore, we can simplify the expression.

Kc ≈ x^2 / (initial concentration)^2

Step 6: Substitute the given dissociation constants into the equation.

Kc ≈ (4.5 * 10^-10)^2 / (1.5 * 10^-5)^2

Step 7: Calculate the value of Kc.

Kc ≈ 3 * 10^4

Therefore, the equilibrium constant for the given reaction is approximately 3 * 10^4.