Question

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

Short answer

The derived equation relating pOH to the \(pK_b\) of a weak base in a buffer solution is: \[ pOH = \log{\left( \frac{[B]}{[HB^+]} \right)} + pK_b \] This equation can be used to determine the pOH of a buffer system given the concentration of the base and its conjugate acid, and the \(pK_b\) of the base component.

Step-by-step solution

Write down the generic reaction with a weak base and water.

For a buffer solution composed of a weak base, B, and its conjugate acid, HB+, the following equilibrium reaction takes place when the base is mixed with water: \( B + H_2O \leftrightarrows HB^+ + OH^- \) Here, B is the weak base, \(H_2O\) is water, \(HB^+\) is the conjugate acid of B, and \(OH^-\) is hydroxide ion.

Write down the expression for the base ionization constant.

The base ionization constant, \(K_b\), can be defined as follows. \[ K_b = \frac{[HB^+][OH^-]}{[B]} \] This equation represents the equilibrium between the base (B), the conjugate acid (HB+), and the hydroxide (OH-) ions.

Define pOH and pKb.

The pOH is a measure of the acidity of a solution and is calculated as the negative logarithm of the hydroxide ion concentration. \[ pOH = -\log{[OH^-]} \] Similarly, \(pK_b\) is a measure of the basicity of a substance and is calculated as the negative logarithm of the base ionization constant. \[ pK_b = -\log{K_b} \]

Rewrite the expression for the base ionization constant in terms of pOH and pKb.

We can rewrite the expression for the base ionization constant from Step 2 as follows. \[ 10^{pOH} = \frac{[HB^+]}{[B]} \cdot \frac{1}{K_b} \] Take the negative logarithm of both sides of the equation. \[ -\log{10^{pOH}} = -\log{\left( \frac{[HB^+]}{[B]} \cdot \frac{1}{K_b} \right)} \] Using the properties of logarithms, we can simplify the equation to: \[ pOH = \log{\left( \frac{[B]}{[HB^+]} \right)} + \log{K_b} \] But, since \(pK_b = -\log{K_b}\), we can substitute it into the equation.

Derive the final equation relating pOH and pKb.

Replacing the appropriate terms in the equation, we get the final equation relating pOH and \(pK_b\): \[ pOH = \log{\left( \frac{[B]}{[HB^+]} \right)} + pK_b \] This equation can be used to determine the pOH of a buffer system given the concentration of the base and its conjugate acid, and the \(pK_b\) of the base component.