Chapter 16.5, Problem 16.9WE

(a)

Interpretation Introduction

Interpretation:

The pOH of the given aqueous solutions 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 pOH as follows,

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

(b)

Interpretation Introduction

Interpretation:

The pOH of the given aqueous solutions 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 pOH as follows,

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

To Calculate: The pOH of the aqueous solution with 0.013 M ${\text{Ba(OH)}}_{\text{2}}$

(c)

Interpretation Introduction

Interpretation:

The pOH of the given aqueous solutions 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 pOH as follows,

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

To Calculate: The pOH of the aqueous solution with $9.2\times {10}^{-5}M$ $\text{KOH}$