The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10–4, 1.8 × 10–4 and 4.8 × 10–9 respectively. Calculate the ionization constants of the corresponding conjugate base. ?

To find the ionization constants of the conjugate bases of HF, HCOOH, and HCN, we can use the relationship between the ionization constant of an acid (Ka) and the ionization constant of its conjugate base (Kb). This relationship is given by the equation:

Understanding the Relationship

The product of the ionization constants of a conjugate acid-base pair is equal to the ion product of water (Kw) at a given temperature. At 298 K, Kw is approximately 1.0 × 10^-14. The equation can be expressed as:

Ka × Kb = Kw

From this equation, we can rearrange it to find Kb:

Kb = Kw / Ka

Calculating Kb for Each Acid

Now, let's calculate the Kb for each of the acids provided:

• For HF:

Ka = 6.8 × 10^-4

Kb = 1.0 × 10^-14 / 6.8 × 10^-4

Kb ≈ 1.47 × 10^-11

• For HCOOH:

Ka = 1.8 × 10^-4

Kb = 1.0 × 10^-14 / 1.8 × 10^-4

Kb ≈ 5.56 × 10^-11

• For HCN:

Ka = 4.8 × 10^-9

Kb = 1.0 × 10^-14 / 4.8 × 10^-9

Kb ≈ 2.08 × 10^-6

Summary of Results

To summarize, the ionization constants of the conjugate bases are as follows:

• For F^- (from HF): Kb ≈ 1.47 × 10^-11
• For HCOO^- (from HCOOH): Kb ≈ 5.56 × 10^-11
• For CN^- (from HCN): Kb ≈ 2.08 × 10^-6

This method allows us to determine the strength of the conjugate bases based on the strength of their corresponding acids. The lower the Kb value, the weaker the conjugate base, which is consistent with the strength of the original acid. If you have any further questions or need clarification on any part of this process, feel free to ask!