Chapter 14, Problem 34QRT

Interpretation Introduction

Interpretation:

The interconversions given below have to be done and in each case whether the solution is acidic or basic has to be estimated.

Concept Introduction:

The relationship between $pHandpOH$ is given below.

  $pH+pOH=14$

Other relations are given below.

  $\begin{array}{l}pH=-\log \left[H_3{O}^{+}\right]\\ \\ pOH=-\log \left[O{H}^{-}\right]\\ \\ K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]={10}^{-14}\end{array}$

If the $pH$ of a solution is less than $7$, then the solution is acidic.  If the $pH$ of a solution is greater than $7$, then the solution is basic.

Explanation of Solution

(a) $\left[H_3{O}^{+}\right]=6.1{\text{×10}}^{\text{-7}}\text{M}$

The $pH$ of the solution can be calculated as given below.

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left[6.1\times {10}^{-7}\right]=6.21.$

Then,

  $\begin{array}{l}pH+pOH=14\\ \\ pOH=14-pH=14-6.21=\mathrm{7.79.}\\ \\ \left[O{H}^{-}\right]={10}^{-pOH}={10}^{-7.79}=1.6\times 10^{-8}M.\end{array}$

As the $pH$ of the solution is less than $7$, the solution is acidic.

(b) $\left[O{H}^{-}\right]=2.{\text{2×10}}^{\text{-9}}\text{M}$

The $pOH$ of the solution can be calculated as given below.

  $pOH=-\log \left[O{H}^{-}\right]=-\log \left[2.2\times {10}^{-9}\right]=8.66.$

Then,

  $\begin{array}{l}pH+pOH=14\\ \\ pH=14-pOH=14-8.66=5.34.\\ \\ \left[H_3{O}^{+}\right]={10}^{-pH}={10}^{-5.34}=4.5\times 10^{-6}M.\end{array}$

As the $pH$ of the solution is greater than $7$, the solution is basic.

(c) $pH=4.67$

The relationship between $pHandpOH$ is given below.

  $pH+pOH=14$

Therefore,

  $\begin{array}{l}pH+pOH=14\\ \\ pOH=14-pH=14-4.67=9.33.\end{array}$

The concentration of $H_3{O}^{+}$ and $O{H}^{-}$ ion can be calculated as given below.

  $\begin{array}{l}pH=-\log \left[H_3{O}^{+}\right]\\ \\ \left[H_3{O}^{+}\right]={10}^{-pH}={10}^{-4.67}=2.1\times 10^{-5}M.\\ \\ pOH=-\log \left[O{H}^{-}\right]\\ \\ \left[O{H}^{-}\right]={10}^{-pOH}={10}^{-9.33}=4.7\times 10^{-10}M.\end{array}$

As the $pH$ of the solution is less than $7$, the solution is acidic.

(d) $\left[H_3{O}^{+}\right]=2.5{\text{×10}}^{\text{-2}}\text{M}$

The $pH$ of the solution can be calculated as given below.

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left[2.5\times {10}^{-2}\right]=1.60.$

Then,

  $\begin{array}{l}pH+pOH=14\\ \\ pOH=14-pH=14-1.60=\mathrm{12.40.}\\ \\ \left[O{H}^{-}\right]={10}^{-pOH}={10}^{-12.40}=4.0\times 10^{-13}M.\end{array}$

As the $pH$ of the solution is less than $7$, the solution is acidic.

(e) $pH=9.12$

The relationship between $pHandpOH$ is given below.

  $pH+pOH=14$

Therefore,

  $\begin{array}{l}pH+pOH=14\\ \\ pOH=14-pH=14-9.12=4.88.\end{array}$

The concentration of $H_3{O}^{+}$ and $O{H}^{-}$ ion can be calculated as given below.

  $\begin{array}{l}pH=-\log \left[H_3{O}^{+}\right]\\ \\ \left[H_3{O}^{+}\right]={10}^{-pH}={10}^{-9.12}=7.6\times 10^{-10}M.\\ \\ pOH=-\log \left[O{H}^{-}\right]\\ \\ \left[O{H}^{-}\right]={10}^{-pOH}={10}^{-4.88}=1.3\times 10^{-5}M.\end{array}$

As the $pH$ of the solution is greater than $7$, the solution is basic.

Now, the table can be filled as given below.

Now, the complete table can be given as shown below.

$pH$	$\left[H_3{O}^{+}\right]M$	$\left[O{H}^{-}\right]M$	Nature
(a) $6.21$	$6.1{\text{×10}}^{\text{-7}}$	$1.6\times 10^{-8}$	Acidic
(b) $5.34$	$4.5\times 10^{-6}$	$2.{\text{2×10}}^{\text{-9}}$	Acidic
(c) $4.67$	$2.1\times 10^{-5}$	$4.7\times 10^{-10}$	Acidic
(d) $1.60$	$2.5{\text{×10}}^{\text{-2}}$	$4.0\times 10^{-13}$	Acidic
(e) $9.12$	$7.6\times 10^{-10}$	$1.3\times 10^{-5}$	Basic