Q. 10.42 - An Introduction to General, Organic, and Biological Chemistry

Question

Complete the following table :

Step-by-Step Solution

Answer

i) The pH of the component is determined as basic

ii) The pH of the component is determined as neutral

iii) The pH of the component is determined as basic

Step 1 : Introduction

The given is the table with components

The objective is to complete the table

Step 2 : Explanation

The pH scale is a logarithmic scale that measures the acidity of water. $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$of aqueous solutions.

Mathematically expression of $\mathrm{pH}$ is the negative logarithmic (base 10 ) of the $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

$$\begin{gathered} \left[H_3{O}^{+}\right]\left[O{H}^{-}\right]=1.0 \times {10}^{-14} \\ \left\{KW\right\}=[H_3 O^{+}]\left[O{H}^{-}\right] \end{gathered}$$

$\mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

$\mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]$

$\mathrm{pH}+\mathrm{pOH}=14$

$\mathrm{pH}=14-\mathrm{pOH}$

Acidic solution$=\mathrm{pH}<7.0\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]>1.0\times {10}^{-7}\mathrm{M}$

Neutral solution $=\mathrm{pH}=7.0\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.0\times {10}^{-7}\mathrm{M}$

Basic solution$=\mathrm{pH}>7.0\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]<1.0\times {10}^{-7}\mathrm{M}$

Step 3 : Explanation

(i)

$$\begin{gathered} pH=10.00>7.0 \\ so, Basic \\ [H_3 O^{+}]={10}^{-pH} \\ \left[H_3{O}^{+}\right]={10}^{-10} \\ \left[O{H}^{-}\right]=1.0 \times {10}^{-10}M \\ \left[O{H}^{-}\right]==1.0 \times {10}^{-4}M \end{gathered}$$

Step 4 : Explanation

(ii)

$\mathrm{pH}=7.00=7.0$ at, Neutral

$$\begin{gathered} \left[H_3{O}^{+}\right]={10}^{-pH} \\ \left[H_3{O}^{+}\right]={10}^{-7} \end{gathered}$$

$\left[{\mathrm{OH}}^{-}\right]=1.0 \times {10}^{-7}M$

$\left[{\mathrm{OH}}^{-}\right]==1.0\times {10}^{-7}\mathrm{M}$

Step 5 : Explanation

(iii)

$$\begin{gathered} \left[H_3{O}^{+}\right]=6.4 \times {10}^{-12}M \\ \left[O{H}^{-}\right]= \\ \left[O{H}^{-}\right]= \\ [OH-]=1.5625 \times {10}^{-3} M \\ pH=-\log \left[H_3{O}^{+}\right] \\ pH=-\log [6.4 \times 10{}^{-12}]=11.19 \\ pH>7.0 \\ so, Basic \end{gathered}$$