Chapter 18.3, Problem 27PP

Interpretation Introduction

(a)

Interpretation:

$pHandpOH$ of the solution is to be found out using given concentration.

Concept introduction:

pH gives the concentration of hydrogen ion in a solution.

It tells about acidity and basicity of a solution.

It is given by:

$\begin{array}{l}pH=-\log \left[H^{+}\right]\\ pOH=-\log \left[O{H}^{-}\right]\\ pH+pOH=14\\ \left[H^{+}\right]isconcentationofhydrogenions\\ \left[O{H}^{-}\right]isconcentationofhydroxylions\end{array}$

Answer to Problem 27PP

$pH=8$ and $pOH=6$.

Explanation of Solution

Given:

$\left[O{H}^{-}\right]=1\times {10}^{-6}M$

$pOH=-\log \left[O{H}^{-}\right]$

Therefore,

$\begin{array}{c}pOH=-\log (1\times {10}^{-6})\\ =6\end{array}$

Therefore,

$\begin{array}{c}pH=14-6\\ =8\end{array}$

Interpretation Introduction

(b)

Interpretation:

$pHandpOH$ of the solution is to be found out using given concentration.

Concept introduction:

pH gives the concentration of hydrogen ion in a solution.

It tells about acidity and basicity of a solution.

It is given by:

$\begin{array}{l}pH=-\log \left[H^{+}\right]\\ pOH=-\log \left[O{H}^{-}\right]\\ pH+pOH=14\\ \left[H^{+}\right]isconcentationofhydrogenions\\ \left[O{H}^{-}\right]isconcentationofhydroxylions\end{array}$

Answer to Problem 27PP

$pH=10.81$

$pOH=3.19$.

Explanation of Solution

Given:

$\left[O{H}^{-}\right]=6.5\times {10}^{-4}M$

$pOH=-\log \left[O{H}^{-}\right]$

Therefore,

$\begin{array}{c}pOH=-\log (6.5\times {10}^{-4})\\ =-\log 6.5-\log {10}^{-4}\\ =-0.81+4\\ =3.19\end{array}$

Therefore,

$\begin{array}{c}pH=14-3.19\\ =10.81\end{array}$

Interpretation Introduction

(c)

Interpretation:

$pHandpOH$ of the solution is to be found out using given concentration.

Concept introduction:

pH gives the concentration of hydrogen ion in a solution.

It tells about acidity and basicity of a solution.

It is given by:

$\begin{array}{l}pH=-\log \left[H^{+}\right]\\ pOH=-\log \left[O{H}^{-}\right]\\ pH+pOH=14\\ \left[H^{+}\right]isconcentationofhydrogenions\\ \left[O{H}^{-}\right]isconcentationofhydroxylions\end{array}$

Answer to Problem 27PP

$pH=8.44$

$pOH=5.56$.

Explanation of Solution

Given:

$\left[H^{+}\right]=3.6\times {10}^{-9}M$

$pH=-\log \left[H^{+}\right]$

Therefore,

$\begin{array}{c}pH=-\log (3.6\times {10}^{-9})\\ =-\log 3.6-\log {10}^{-9}\\ =-0.56+9\\ =8.44\end{array}$

Therefore,

$\begin{array}{c}pOH=14-8.44\\ =5.56\end{array}$

Interpretation Introduction

(d)

Interpretation:

$pHandpOH$ of the solution is to be found out using given concentration.

Concept introduction:

pH gives the concentration of hydrogen ion in a solution.

It tells about acidity and basicity of a solution.

It is given by:

$\begin{array}{l}pH=-\log \left[H^{+}\right]\\ pOH=-\log \left[O{H}^{-}\right]\\ pH+pOH=14\\ \left[H^{+}\right]isconcentationofhydrogenions\\ \left[O{H}^{-}\right]isconcentationofhydroxylions\end{array}$

Answer to Problem 27PP

$pH=1.60$

$pOH=12.40$.

Explanation of Solution

Given:

$\left[H^{+}\right]=2.5\times {10}^{-2}M$

$pH=-\log \left[H^{+}\right]$

Therefore,

$\begin{array}{c}pH=-\log (2.5\times {10}^{-2})\\ =-\log 2.5-\log {10}^{-2}\\ =-0.398+2\\ =1.60\end{array}$

Therefore,

$\begin{array}{c}pOH=14-1.60\\ =12.40\end{array}$