Chapter 6, Problem 6.49P

Interpretation Introduction

Interpretation:

${\text{K}}_b$ reaction for Cyanide ion $\left({\text{CN}}^{\text{-}}\right)$ has to be written and the value of ${\text{K}}_b$ for HCN has to be calculated.

Concept introduction:

Consider a weak-acid equilibrium reaction,

$\text{HA}\stackrel{{\text{K}}_{\text{a}}}{\rightleftharpoons }{\text{H}}^{\text{+}}{\text{+A}}^{\text{-}}$

The acid dissociation constant ${\text{K}}_{\text{a}}$ can be given as,

${\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{A}}^{\text{-}}\right]}{\left[\text{HA}\right]}$

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,

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