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

Short Answer

Expert verified
The equation relating the pOH of a buffer to the pKb of its base component is: \[pOH = pK_b -\log_{10} \frac{[B]}{[HB^+]}\] This equation is analogous to the Henderson-Hasselbalch equation for bases and pOH instead of acids and pH.

Step by step solution

01

Recall the definition of pOH and pKb

pOH is the opposite of the base 10 logarithm of the hydroxide ion concentration, i.e. \[pOH = -\log_{10} [\mathrm{OH}^-]\] pKb is the opposite of the base 10 logarithm of the base dissociation constant (\(K_b\)), which is defined as: \[pK_{b} = -\log_{10}K_{b}\]
02

Write the Kb expression for the base component of the buffer

Let's consider a generic base B and its conjugate acid HB+. The ionization of B in water is given as: \[B + H_2O \longleftrightarrow HB^+ + OH^-\] The equilibrium constant for this reaction, Kb, can be expressed as: \[K_b = \frac{[HB^+][OH^-]}{[B]}\]
03

Manipulating the Kb expression to involve pOH and pKb

From the definition of pOH and pKb, we can rewrite the expression for Kb: \[K_b = 10^{-pK_b}\] and \([OH^-] = 10^{-pOH}\) Now, we substitute these expressions into the Kb expression: \[10^{-pK_b} = \frac{[HB^+][10^{-pOH}]}{[B]}\]
04

Solve for pOH

We need to rearrange the equation to find the expression for pOH: \[pOH = -\log_{10} \frac{[HB^+]}{[B]} + pK_b\] Now, we have the equation relating the pOH of a buffer to the pKb of its base component: \[pOH = pK_b -\log_{10} \frac{[B]}{[HB^+]}\] This equation is the analogous to the Henderson-Hasselbalch equation, but for bases and pOH instead of acids and pH.

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