Chapter 16, Problem 9PE

Interpretation Introduction

(a)

Interpretation:

The equivalent mass of $\text{HI}$ in the reaction $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}+\text{2HI}\to {\text{CaI}}_{\text{2}}+{\text{2H}}_{\text{2}}\text{O}$ is to be determined.

Concept introduction:

In a chemical reaction, one equivalent of an acid is defined as the quantity that yields one mole of hydrogen ion. One equivalent of base is defined as the quantity that reacts with one mole of hydrogen ion. The number of grams per equivalent is called equivalent mass. The unit of equivalent mass is $\text{g}/\text{eq}$.

Answer to Problem 9PE

The equivalent mass of $\text{HI}$ in the given reaction $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}+\text{2HI}\to {\text{CaI}}_{\text{2}}+{\text{2H}}_{\text{2}}\text{O}$ is $128\text{g}/\text{eq}$.

Explanation of Solution

The given chemical reaction is shown below.

$\text{Ca}{\left(\text{OH}\right)}_{\text{2}}+\text{2HI}\to {\text{CaI}}_{\text{2}}+{\text{2H}}_{\text{2}}\text{O}$

In the above reaction each mole of $\text{HI}$ gives one mole of hydrogen ion and one mole of hydrogen ion is one equivalent of acid. The equivalent mass is calculated by dividing the molar mass by equivalents per mole.

Therefore,

$\text{Equivalent mass of}\text{HI}=\frac{\text{Molar}\text{mass}\text{of}\text{HI}}{\text{equivalents}\text{per}\text{mole}\text{of}\text{HI}}$…(1)

The molar mass of $\text{HI}$ is calculated by using the formula as given below.

$\text{Molar}\text{mass}\text{of}\text{HI}=\left(\begin{array}{c}\text{Molar}\text{mass}\text{of}\text{hydrogen}\text{+}\\ \text{Molar}\text{mass}\text{of}\text{iodine}\end{array}\right)$…(2)

The molar mass of hydrogen is $1\text{g}/\text{mol}$.

The molar mass of iodine is $127\text{g}/\text{mol}$.

Substitute the values of molar masses of hydrogen and iodine in equation (2).

$\begin{array}{rll}\text{Molar}\text{mass}\text{of}\text{HI}& =& \left(\text{1}\text{g}/\text{mol}+127\text{g}/\text{mol}\right)\\ & =& 128\text{g}/\text{mol}\end{array}$

Substitute the values of molar mass and equivalents per mole of $\text{HI}$ in equation (1).

$\begin{array}{rll}\text{Equivalent mass of}\text{HI}& =& \frac{\text{128}\text{g}/\text{mol}}{\text{1}\text{eq}/\text{mol}}\\ & =& 128\text{g}/\text{eq}\end{array}$

Therefore, the equivalent mass of $\text{HI}$ in the given reaction is $128\text{g}/\text{eq}$.

Conclusion

The equivalent mass of $\text{HI}$ in the given reaction is $128\text{g}/\text{eq}$.

Interpretation Introduction

(b)

Interpretation:

The equivalent mass of ${\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}$ in the reaction $\text{NaOH}+{\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}\to {\text{NaH}}_2{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}+{\text{H}}_2\text{O}$ is to be determined.

Concept introduction:

In a chemical reaction one equivalent of an acid is defined as the quantity that yields one mole of hydrogen ion. One equivalent of base is defined as the quantity that reacts with one mole of hydrogen ion. The number of grams per equivalent is called equivalent mass. The unit of equivalent mass is $\text{g}/\text{eq}$.

Answer to Problem 9PE

The equivalent mass of ${\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}$ in the given reaction $\text{NaOH}+{\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}\to {\text{NaH}}_2{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}+{\text{H}}_2\text{O}$ is $181\text{g}/\text{eq}$.

Explanation of Solution

The chemical reaction is given below.

$\text{NaOH}+{\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}\to {\text{NaH}}_2{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}+{\text{H}}_2\text{O}$

In the above reaction one mole of hydroxide ion reacts with one mole of hydrogen ion of the molecule ${\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}$, and results in the formation of one mole of water molecule. Since, one mole of hydrogen ion is one equivalent of acid. Therefore, there is one $\text{eq}/\text{mol}$ of acid ${\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}$. The equivalent mass is calculated by dividing the molar mass by equivalents per mole.

Therefore,

$\text{Equivalent mass of}{\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}=\frac{\text{Molar}\text{mass}\text{of}{\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}}{\text{equivalents}\text{per}\text{mole}\text{of}{\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}}$…(1)

The molar mass of ${\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}$ is calculated by using the formula as given below.

$\text{Molar}\text{mass}\text{of}{\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}=\left(\begin{array}{c}11\times \text{Molar}\text{mass}\text{of}\text{hydrogen}+\\ \text{Molar}\text{mass}\text{of}\text{nitrogen}+\\ 9\times \text{Molar}\text{mass}\text{of}\text{carbon}\\ +3\times \text{Molar}\text{mass}\text{of}\text{oxygen}\end{array}\right)$…(2)

The molar mass of hydrogen is $1\text{g}/\text{mol}$.

The molar mass of carbon is $12\text{g}/\text{mol}$.

The molar mass of nitrogen is $\text{14}\text{g}/\text{mol}$.

The molar mass of oxygen is $\text{16}\text{g}/\text{mol}$.

Substitute the values of molar masses of hydrogen, Carbon, nitrogen and oxygen in equation (2).

$\begin{array}{rll}\text{Molar}\text{mass}\text{of}{\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}& =& \left(\begin{array}{c}11\times 1\text{g}/\text{mol}+\text{14}\text{g}/\text{mol}+\\ 9\times 12\text{g}/\text{mol}+3\times \text{16}\text{g}/\text{mol}\end{array}\right)\\ & =& \left(11+14+108+48\right)\text{g}/\text{mol}\\ & =& 181\text{g}/\text{mol}\end{array}$

Substitute the values of molar mass and equivalents per mole of ${\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}$ in equation (1).

$\begin{array}{rll}\text{Equivalent mass of}{\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}& =& \frac{\text{181}\text{g}/\text{mol}}{\text{1}\text{eq}/\text{mol}}\\ & =& 181\text{g}/\text{eq}\end{array}$

Therefore, the equivalent mass of ${\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}$ in the given reaction is $181\text{g}/\text{eq}$.

Conclusion

The equivalent mass of ${\text{H}}_{\text{3}}{\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{NO}}_{\text{3}}$ in the given reaction is $181\text{g}/\text{eq}$.