Problem 11 - Chemistry

Question

Acids and bases can be thought of as chemical opposites (acids are proton donors, and bases are proton acceptors). Therefore, one might think that \(K_{\mathrm{a}}=1 / K_{\mathrm{b}}\). Why isn't this the case? What is the relationship between \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) ? Prove it with a derivation.

Step-by-Step Solution

Answer

In conclusion, the relationship between the acid dissociation constant (Ka) and the base dissociation constant (Kb) is not Ka = \( \frac{1}{K_b} \); rather, the correct relation is Kb = \( \frac{K_{w}}{K_{a}} \), where Kw is the ionic product of water. This relationship is derived by connecting the equilibrium reactions of a generic weak acid and its conjugate base, a generic weak base and its conjugate acid, and the ionic product of water.

Step 1: Define the Ka and Kb constants

Ka and Kb are equilibrium constants used to quantify the strength of acids and bases, respectively. Ka represents how well an acid donates protons (H+ ions) while Kb denotes how well a base accepts protons.

Step 2: Write the general reaction of an acid and its conjugate base

Let 'HA' be a generic weak acid and 'A-' be its conjugate base. The general reaction of an acid and its conjugate base in equilibrium is: HA + H2O ⇌ H3O+ + A- (1)

Step 3: Write the expression for Ka

According to the equilibrium reaction in Step 2, the expression for the Ka, the acid dissociation constant, can be written as: Ka = \(\frac{[H_{3}O^{+}][A^{-}]}{[HA]}\) (2)

Step 4: Write the general reaction of a base and its conjugate acid

For a generic weak base, 'B', and its conjugate acid, 'HB+', the general reaction in equilibrium is: B + H2O ⇌ OH- + HB+ (3)

Step 5: Write the expression for Kb

Using the reaction in Step 4, the expression for the base dissociation constant, Kb, can be written as: Kb = \(\frac{[OH^{-}][HB^{+}]}{[B]}\) (4)

Step 6: Relate Ka and Kb using the ionic product of water, Kw

The ionic product of water, Kw, is the product of concentrations of H3O+ and OH- ions at a given temperature: Kw = [H3O+][OH-] (5) Now, we can relate the Ka and Kb expressions using the Kw equation. Divide both sides of the equilibrium constant Kb by Ka: \(\frac{K_{b}}{K_{a}} = \frac{[OH^{-}][HB^{+}]}{[H_{3}O^{+}][A^{-}]}\) From equation (5), we can replace [OH-] with \(\frac{K_{w}}{[H_{3}O^{+}]}\) in the equation above: \(\frac{K_{b}}{K_{a}} = \frac{K_{w}[HB^{+}]}{[H_{3}O^{+}][A^{-}]}\) Therefore, the relationship between Ka and Kb is: Kb = \(\frac{K_{w}}{K_{a}}\) (6) In conclusion, the relationship between the acid dissociation constant (Ka) and the base dissociation constant (Kb) is not Ka = \(\frac{1}{K_b}\); rather, the correct relation is Kb = \(\frac{K_{w}}{K_{a}}\).