Chapter 16.4, Problem 6PPA

(a)

Interpretation Introduction

Interpretation:

The concentration of hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ of the given solutions has to be calculated.

Concept Information:

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 concentration of hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ of the given solutions at ${\text{25}}^{\circ }\text{C}$

Answer to Problem 6PPA

Answer

• The concentration of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ of the given solution (a) is $9.5\times {10}^{-14}M$
• The concentration of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ of the given solution (b) is $7.2\times {10}^{-6}M$
• The concentration of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ of the given solution (c) is $1.0\times {10}^{-7}M$

Explanation of Solution

Given data:

The pOH of the given solution is 13.02

Calculation of hydroxide ion $[O{H}^{-}]$:

The hydroxide ion concentration is calculated as follows,

$\begin{array}{l}{\text{[OH}}^{\text{-}}\text{]}\text{=}{\text{10}}^{\text{-pOH}}\\ ={\text{10}}^{\text{-13}\text{.02}}\\ =9.5\times {10}^{-14}M\end{array}$

Therefore, the concentration of hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$= $9.5\times {10}^{-14}M$

(b)

Interpretation Introduction

Interpretation:

The concentration of hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ of the given solutions has to be calculated.

Concept Information:

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 concentration of hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ of the given solutions at ${\text{25}}^{\circ }\text{C}$

Answer to Problem 6PPA

Answer

• The concentration of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ of the given solution (a) is $9.5\times {10}^{-14}M$
• The concentration of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ of the given solution (b) is $7.2\times {10}^{-6}M$
• The concentration of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ of the given solution (c) is $1.0\times {10}^{-7}M$

Explanation of Solution

Given data:

The pOH of the given solution is 5.14

Calculation of hydroxide ion $[O{H}^{-}]$:

The hydroxide ion concentration is calculated as follows,

$\begin{array}{l}{\text{[OH}}^{\text{-}}\text{]}\text{=}{\text{10}}^{\text{-pOH}}\\ ={\text{10}}^{\text{-5}\text{.14}}\\ =7.2\times {10}^{-6}M\end{array}$

Therefore, the concentration of hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$= $7.2\times {10}^{-6}M$

(c)

Interpretation Introduction

Interpretation:

The concentration of hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ of the given solutions has to be calculated.

Concept Information:

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 concentration of hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ of the given solutions at ${\text{25}}^{\circ }\text{C}$

Answer to Problem 6PPA

Answer

• The concentration of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ of the given solution (a) is $9.5\times {10}^{-14}M$
• The concentration of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ of the given solution (b) is $7.2\times {10}^{-6}M$
• The concentration of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ of the given solution (c) is $1.0\times {10}^{-7}M$

Explanation of Solution

Given data:

The pOH of the given solution is 6.98

Calculation of hydroxide ion $[O{H}^{-}]$:

The hydroxide ion concentration is calculated as follows,

$\begin{array}{l}{\text{[OH}}^{\text{-}}\text{]}\text{=}{\text{10}}^{\text{-pOH}}\\ ={\text{10}}^{\text{-6}\text{.98}}\\ =1.0\times {10}^{-7}M\end{array}$

Therefore, the concentration of hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$= $1.0\times {10}^{-7}M$