Chapter 16.7, Problem 16.7.2SR

Interpretation Introduction

Interpretation:

The ${\text{K}}_b$ of the given $\text{0}\text{.12 M}$ solution of base has to be calculated

Concept Information:

Base ionization constant: ${\text{K}}_{\text{b}}$

The ionization of a weak base $\text{B}$ is given by the below equation.

${\text{B}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\rightleftarrows {\text{HB}}^{\text{+}}{}_{(aq)}+{\text{OH}}^{\text{-}}{}_{(aq)}$

The equilibrium expression for the ionization of weak base $\text{B}$ will be,

${\text{K}}_b=\frac{[{\text{HB}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[\text{B}]}$

Where,

$K_b$ is base ionization constant,

$[{\text{OH}}^{-}\text{]}$ is concentration of hydroxide ion

$[{\text{HB}}^{\text{+}}\text{]}$  is concentration of conjugate acid

$[\text{B]}$ is concentration of the base

pOH definition:

The $\text{pOH}$ of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ concentration. $\text{pOH}$ scale is analogous to pH scale.

$\text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}$

Relationship between pH and pOH

$\text{pOH}$ is similar to $\text{pH}$.  The only difference is that in $\text{pOH}$ the concentration of hydroxide ion is used as a scale while in $\text{pH}$, the concentration of hydronium ion is used.

The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation,

$\text{pH}\text{+}\text{pOH}\text{=}\text{14,}\text{at}\text{25}{}^{\text{o}}\text{C}$

As $\text{pOH}$ and $\text{pH}$ are opposite scale, the total of both has to be equal to 14.

To Calculate: The ${\text{K}}_b$ of the given $\text{0}\text{.12 M}$ solution of base