Chapter 9, Problem 9.FE

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{0}\text{.050}\text{M}$ Sodium butanoate has to be calculated.

Concept Introduction:

Consider a reaction of weak base,

${\text{B+H}}_{\text{2}}\text{O}\stackrel{{\text{K}}_{\text{b}}}{\rightleftharpoons }{\text{BH}}^{\text{+}}{\text{+OH}}^{\text{-}}$

The value of ${\text{K}}_{\text{b}}$ can be calculated as,

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

Where ${\text{K}}_{\text{b}}$ is called as base dissociation constant.

The value of ${\text{K}}_{\text{w}}$ is calculated by the formula,

${\text{K}}_{\text{w}}{\text{=K}}_{\text{a}}{\text{×K}}_{\text{b}}$

The $\text{pH}$ of weak base is calculated using the equation,

$\begin{array}{l}\text{pH=}\left[{\text{H}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{OH}}^{\text{-}}\right]}\\ \\ \text{pH=-log}\left[{\text{H}}^{\text{+}}\right]\end{array}$

To calculate the $\text{pH}$ of $\text{0}\text{.050}\text{M}$ Sodium butanoate