Chapter 16.5, Problem 16.5.5SR

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the given $\text{0}\text{.0095M}$$\text{LiOH}$ solution at ${\text{25}}^{\circ }\text{C}$ has to be calculated

Concept Information:

Strong bases:

Strong bases are formed from alkali metals and alkaline earth metals of Group IA and IIA respectively.

Strong base dissociates into its constituent ions.

For Group IA metal hydroxides, the hydroxide ion concentration is simply the initial concentration of the strong base

For Group IIA metal hydroxides, the hydroxide ion concentration at equilibrium will be twice that of the initial concentration of strong 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.

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

On rearranging, the concentration of hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ can be calculated using 0.0095 M as follows,

${\text{[OH}}^{\text{-}}\text{]}\text{=}{\text{10}}^{\text{-pOH}}$

Relationship between $\text{pH}$ and $\text{pOH}$

$\text{pOH}$ 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 pH of the given 0.0095 M $\text{LiOH}$ solution at ${\text{25}}^{\circ }\text{C}$