CALCULATE THE EQUILIBRIUM CONSTANT OF THE FOLLOWING EQUILIBRIUM:

CN‾ + CH₃COOH ↔ HCN + CH₃COO‾

GIVEN: pK_b of CN‾ IS 4.69 AND pK_b OF CH₃COO‾ IS 9.25

To calculate the equilibrium constant for the reaction between cyanide ion (CN-) and acetic acid (CH3COOH) to form hydrogen cyanide (HCN) and acetate ion (CH3COO-), we can utilize the provided pKb values for CN- and CH3COO-. The equilibrium constant (K) can be derived from the relationship between the acid dissociation constants (Ka) and the base dissociation constants (Kb).

Understanding the Reaction

The reaction can be represented as follows:

CN- + CH3COOH ⇌ HCN + CH3COO-

In this equilibrium, CN- acts as a base, and CH3COOH acts as an acid. The products are HCN, a weak acid, and CH3COO-, the conjugate base of acetic acid.

Step 1: Calculate the Ka Values

To find the equilibrium constant (K), we first need to convert the given pKb values into Ka values. The relationship between pKa and pKb is given by:

pKa + pKb = 14

From this, we can derive the Ka values:

• For CN-:
• pKb = 4.69 → pKa = 14 - 4.69 = 9.31
• Ka (CN-) = 10^(-pKa) = 10^(-9.31)
• For CH3COO-:
• pKb = 9.25 → pKa = 14 - 9.25 = 4.75
• Ka (CH3COO-) = 10^(-pKa) = 10^(-4.75)

Step 2: Calculate the Equilibrium Constant (K)

The equilibrium constant for the reaction can be expressed in terms of the Ka values of the products and reactants:

K = (Ka of HCN) / (Ka of CH3COOH)

Since we have the Ka values for CN- and CH3COO-, we can express K as:

K = (Ka of HCN) / (Ka of CH3COOH) = (10^(-9.31)) / (10^(-4.75))

Step 3: Perform the Calculation

Now, let's calculate K:

K = 10^(-9.31 + 4.75) = 10^(-4.56)

This gives us:

K ≈ 2.75 x 10^(-5)

Final Thoughts

The equilibrium constant for the reaction CN- + CH3COOH ⇌ HCN + CH3COO- is approximately 2.75 x 10^(-5). This value indicates that at equilibrium, the concentration of products (HCN and CH3COO-) is significantly lower than that of the reactants (CN- and CH3COOH), suggesting that the reaction favors the reactants under standard conditions.