Question

Derive an equation similar to the Henderson-Hasselbalch equation relating the pOH of a buffer to the \(\mathrm{p} K_{b}\) of its base component.

Short answer

The equation relating the pOH of a buffer to the pKb of its base component, similar to the Henderson-Hasselbalch equation, is: \( pOH = pK_b - \log{([B]_0 - x)} \)

Step-by-step solution

Write the formula for base dissociation constant Kb

The base dissociation constant, Kb, is defined as the equilibrium constant for a weak base B and its conjugate acid HB+: \( K_b = \frac{[HB^+][OH^-]}{[B]} \)

Express the concentration of conjugate acid and base species in terms of Kb

To express the concentration of the conjugate acid and base species, let's assume that x moles of OH- ions get dissociated from B at equilibrium. So, \( [B] = [B]_0 - x \) \( [HB^+] = x \) \( [OH^-] = x \) Substitute these expressions into the Kb expression: \( K_b = \frac{x^2}{[B]_0 - x} \)

Calculate the pOH of the solution

The pOH is defined as the negative logarithm of the hydroxide ion concentration: \( pOH = -\log{[OH^-]} \) Since we have expressed [OH-] as x in Step 2, we have: \( pOH = -\log{x} \)

Substitute pOH and pKb into the derived equation

Rearrange the Kb expression from Step 2 and take the negative logarithm of both sides: \( -\log{K_b} = -\log{\frac{x^2}{[B]_0 - x}} \) Substitute the derived expressions of pOH and the relationship between pKb and Kb: \( pK_b = pOH + \log{([B]_0 - x)} \) The equation for pOH in terms of pKb and the concentration of free base and conjugate acid species is: \( pOH = pK_b - \log{([B]_0 - x)} \) Note that this equation is similar to the Henderson-Hasselbalch equation, but it relates the pOH of a solution to the pKb of its base component, rather than the pH and pKa.