Q19.128CP - Chemistry: Molecular Nature Of Matter And Change

Question

The Henderson-Hasselbalch equation gives a relationship for obtaining the pH  of a buffer solution consisting of HA  and  A. Derive an analogous relationship for obtaining the pOH of a buffer solution consisting of B and BH⁺?

Step-by-Step Solution

Answer

The required result is:

 ${\text{pOH = pK}}_{\text{b}}\text{ + }\frac{\left[{\text{BH}}^{\text{ + }}\right]}{\text{[B]}}$

Explanation of the Concept

The Henderson Hasselbalch equation is a rough formula that depicts the relationship between pH or pOH the of a solution as well as pK_a or pK_b the  and the concentration ratio of the dissociated chemical species. When small amounts of strong acid or base are added to the solution, the equilibrium between the weak acid and its conjugate base allows the solution to withstand the strong acid or base.

For a weak base B, the equilibrium established is given in the reaction below.

 $\mathrm{B}+{\mathrm{H}}^{+}\to {\mathrm{BH}}^{+}$ 

The equilibrium constant is for the given reaction is:

${\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{BH}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[\mathrm{B}\right]}$

+ Calculating the Equation

Separate the $\left[{\text{OH}}^{\text{ - }}\right]$  by calculating the equation

 $\left[O{H}^{-}\right]=K_b\frac{\left[B\right]}{\left[B{H}^{+}\right]}$

Using the negative log to solve the equation.

 $-\log \left[O{H}^{-}\right]=-\log K_b-\log \frac{\left[B\right]}{\left[B{H}^{+}\right]}$

Create an analogous relationship to get the pOH of a buffer solution made up of  B and BH⁺.

  ${\text{pOH = pK}}_{\text{b}}\text{ + }\frac{\left[{\text{BH}}^{\text{ + }}\right]}{\text{[B]}}$

Therefore, ${\text{pOH = pK}}_{\text{b}}\text{ + }\frac{\left[{\text{BH}}^{\text{ + }}\right]}{\text{[B]}}$ is final result.