Which is incorrect: Ka”Kb-Kw Ka + Kb = 14 Ka Kb = 10^-14 at room temperature PKa + PKb = 14
The Correct Answer and Explanation is:
Correct Answer:
Ka + Kb = 14 is incorrect.
Explanation
In acid-base chemistry, several relationships link the acid dissociation constant (Ka) and the base dissociation constant (Kb) of a conjugate acid-base pair, especially in aqueous solutions at room temperature (typically 25°C or 298 K). The following are correct and well-established:
- Ka × Kb = Kw
This equation shows that the product of the acid and base dissociation constants of a conjugate acid-base pair is equal to the ionization constant of water (Kw). At room temperature, Kw = 1.0 × 10⁻¹⁴. This is a foundational concept in equilibrium chemistry. - pKa + pKb = 14
This is the logarithmic version of the first relationship. Since pKa = -log(Ka) and pKb = -log(Kb), and since log(Ka × Kb) = log(Kw), it follows that:
pKa + pKb = -log(Ka) – log(Kb) = -log(Ka × Kb) = -log(10⁻¹⁴) = 14
This is also valid at 25°C. - Ka × Kb = 10⁻¹⁴ at room temperature
This is simply another way of stating the Kw relationship mentioned above. It is true at 25°C, where the ionization of water produces H₃O⁺ and OH⁻ ions in a concentration of 1.0 × 10⁻⁷ M each, leading to Kw = [H₃O⁺][OH⁻] = 10⁻¹⁴.
Now, the incorrect statement:
Ka + Kb = 14 is not valid. You cannot simply add Ka and Kb and expect it to equal 14. Ka and Kb are exponential values, not linear, so their addition has no direct chemical meaning. This statement likely confuses the correct logarithmic form pKa + pKb = 14.
In summary, Ka + Kb = 14 is incorrect because the addition of dissociation constants does not correspond to any known principle in acid-base equilibrium.
