Chapter 14, Problem 33QRT

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) $pH=1.00$

The relationship between $pHandpOH$ is given below.

  $pH+pOH=14$

Therefore,

  $\begin{array}{l}pH+pOH=14\\ \\ pOH=14-pH=14-1=13.\end{array}$

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

  $\begin{array}{l}pH=-\log \left[H_3{O}^{+}\right]\\ \\ \left[H_3{O}^{+}\right]={10}^{-pH}={10}^{-1}=1.0\times 10^{-1}M.\\ \\ pOH=-\log \left[O{H}^{-}\right]\\ \\ \left[O{H}^{-}\right]={10}^{-pOH}={10}^{-13}=1.0\times 10^{-13}M.\end{array}$

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

(b) $pH=10.5$

The relationship between $pHandpOH$ is given below.

  $pH+pOH=14$

Therefore,

  $\begin{array}{l}pH+pOH=14\\ \\ pOH=14-pH=14-10.5=3.5.\end{array}$

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

  $\begin{array}{l}pH=-\log \left[H_3{O}^{+}\right]\\ \\ \left[H_3{O}^{+}\right]={10}^{-pH}={10}^{-10.5}=3.0\times 10^{-11}M.\\ \\ pOH=-\log \left[O{H}^{-}\right]\\ \\ \left[O{H}^{-}\right]={10}^{-pOH}={10}^{-3.5}=3.0\times 10^{-4}M.\end{array}$

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

(c) $\left[H_3{O}^{+}\right]=\text{1}{\text{.8×10}}^{\text{-4}}\text{M}$

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

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left[1.8\times {10}^{-4}\right]=3.74.$

Then,

  $\begin{array}{l}pH+pOH=14\\ \\ pOH=14-pH=14-3.74=10.26.\\ \\ \left[O{H}^{-}\right]={10}^{-pOH}={10}^{-10.26}=5.5\times 10^{-11}M.\end{array}$

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

(d) $\left[O{H}^{-}\right]=2.{\text{3×10}}^{\text{-5}}\text{M}$

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

  $pOH=-\log \left[O{H}^{-}\right]=-\log \left[2.3\times {10}^{-5}\right]=4.64.$

Then,

  $\begin{array}{l}pH+pOH=14\\ \\ pH=14-pOH=14-4.64=9.36.\\ \\ \left[H_3{O}^{+}\right]={10}^{-pH}={10}^{-9.36}=4.4\times 10^{-10}M.\end{array}$

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

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

$pH$	$\left[H_3{O}^{+}\right]M$	$\left[O{H}^{-}\right]M$	Nature
(a) $1.00$	$1.0\times 10^{-1}$	$1.0\times 10^{-13}$	Acidic
(b) $10.5$	$3.0\times 10^{-11}$	$3.0\times 10^{-4}$	Basic
(c) $3.74$	$\text{1}{\text{.8×10}}^{\text{-4}}$	$5.5\times 10^{-11}$	Acidic
(d) $4.64$	$4.4\times 10^{-10}$	$2.{\text{3×10}}^{\text{-5}}$	Acidic