Question

Provide an expression relating \(K_{\mathrm{w}}\) to \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) of a conjugate acid-base pair.

Short answer

The relation between the ion product of water, acid dissociation constant, and base dissociation constant for a conjugate acid-base pair is \(K_w = K_a \times K_b\).

Step-by-step solution

Write out the Dissociation Equations

Start by writing out the dissociation equations for the acid (\(HA\)) and its conjugate base (\(A^-\)):\n\nFor the acid \(HA\):\n\n\(HA \leftrightarrow H^+ + A^-\)\n\nThe equilibrium constant for this reaction is \(K_a\), so:\n\n\(K_a = [H^+][A^-] / [HA]\)\n\nFor the base \(A^-\):\n\n\(A^- + H_2O \leftrightarrow HA + OH^-\)\n\nThe equilibrium constant for this reaction is \(K_b\), so:\n\n\(K_b = [HA][OH^-] / [A^-]\)

Write out the Ionization of Water

The ionization of water is also an equilibrium, represented as follows:\n\n\(H_2O \leftrightarrow H^+ + OH^-\)\n\nThe equilibrium constant for this reaction is \(K_w\), so:\n\n\(K_w = [H^+][OH^-]\)

Combine the Equations

Now, multiply the expressions for \(K_a\) and \(K_b\) together. Because [H\(^+\)] is in both the \(K_a\) and \(K_b\) expressions, it can be cancelled out. What's left is:\n\n\(K_a \times K_b = [HA][OH^-] / [HA] = [OH^-]\)\n\nSubstitute \(K_w = [H^+][OH^-]\) to get this relation:\n\n\(K_w = [H^+][OH^-] = [H^+] \times K_a \times K_b = K_a \times K_b\)

Conclusion

Therefore, for a conjugate acid-base pair, it can be concluded that the product of the acid dissociation constant (\(K_a\)) and the base dissociation constant (\(K_b\)) is equal to the ion product of water (\(K_w\)).