Chapter 13, Problem 70E

Interpretation Introduction

(a)

Interpretation:

The volume of $\text{Hg}{\left({\text{NO}}_3\right)}_2$ required for complete precipitation if $25.0\text{mL}$ of $0.170M$ potassium iodide reacts completely with $0.209M$ mercury (II) nitrate solution is to be calculated.

Concept introduction:

Molar concentration also termed as molarity, the amount concentration or substance concentration is the concentration of a substance, in terms of a solute in a solution. It is the amount of substance per unit volume of solution. Stoichiometry is the relationship between the quantities in a balanced chemical equation.

Answer to Problem 70E

The volume of $\text{Hg}{\left({\text{NO}}_3\right)}_2$ required for complete precipitation is $\text{10}\text{.15}\text{mL}$.

Explanation of Solution

The balanced equation for the reaction of $\text{Hg}{\left({\text{NO}}_3\right)}_2$ and potassium iodide is shown below.

$\text{Hg}{\left({\text{NO}}_3\right)}_2\left(aq\right)+2\text{KI}\left(aq\right)\to {\text{HgI}}_2\left(s\right)+2{\text{KNO}}_3\left(aq\right)$

The weight of $\text{Hg}{\left({\text{NO}}_3\right)}_2$ is calculated as shown below.

$\text{W}={\text{V}}_{\text{KI}}\times \text{Molar}\text{concentration}\times \frac{{\text{n}}_{\text{Hg}{\left({\text{NO}}_3\right)}_2}}{{\text{n}}_{\text{KI}}}\times \text{MM}$ …(1)

Where,

• ${\text{V}}_{\text{KI}}$ is the volume of potassium iodide.

• $\frac{{\text{n}}_{\text{Hg}{\left({\text{NO}}_3\right)}_2}}{{\text{n}}_{\text{KI}}}$ is the number of moles of $\text{Hg}{\left({\text{NO}}_3\right)}_2$ and $\text{KI}$.

• $\text{MM}$ is the molar mass of $\text{Hg}{\left({\text{NO}}_3\right)}_2$.

Substitute the values ${\text{V}}_{\text{KI}}$ as $25.0\text{mL}$, $\frac{{\text{n}}_{\text{Hg}{\left({\text{NO}}_3\right)}_2}}{{\text{n}}_{\text{KI}}}$ as $\frac{1\text{mol Hg}{\left({\text{NO}}_3\right)}_2}{2\text{mol}\text{KI}}$, $\text{Molar}\text{concentration}$ as $\frac{\text{0}\text{.170 mol KI}}{1000\text{mL solution}}$ and $\text{MM}$ as $324.59\text{g Hg}{\left({\text{NO}}_3\right)}_2$ in equation (1) as shown below.

$\begin{array}{rll}\text{W}& =& 25.0\text{mL}\times \frac{\text{0}\text{.170 mol KI}}{1000\text{mL solution}}\times \frac{1\text{mol Hg}{\left({\text{NO}}_3\right)}_2}{2\text{mol}\text{KI}}\times \frac{324.59\text{g Hg}{\left({\text{NO}}_3\right)}_2}{1\text{mol}\text{Hg}{\left({\text{NO}}_3\right)}_2}\\ & =& \frac{25\times 0.170\times 324.59}{1000\times 2}\text{g}\\ & =& \text{0}\text{.689}\text{g}\end{array}$

Molarity is defined as the number of moles present in one liter of solution. General expression is shown below.

$\text{Molarity}=\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}}$ … (2)

The equation (2) is rearranged to find the volume of $\text{Hg}{\left({\text{NO}}_3\right)}_2$ as shown below.

$\text{Volume}\text{of}\text{solution}=\frac{\text{Moles}\text{of}\text{solute}}{\text{Molarity}}$ … (3)

Substitute the value of $0.209M$ $\text{Hg}{\left({\text{NO}}_3\right)}_2$, $\text{MM}$ is $324.59\text{g Hg}{\left({\text{NO}}_3\right)}_2$, weight is $\text{0}\text{.689}\text{g}$ in rearranged equation as shown below.

$\begin{array}{rll}\text{Volume of Hg}{\left({\text{NO}}_3\right)}_2& =& \frac{0.689\text{g Hg}{\left({\text{NO}}_3\right)}_2}{324.5\text{g Hg}{\left({\text{NO}}_3\right)}_2}\times \frac{1}{0.209\text{mol}/\text{L}}\\ & =& \frac{0.01015\text{L}\times \text{1000}\text{mL}}{\text{1}\text{L}}\\ \text{= 10}\text{.15}\text{mL}\end{array}$

Conclusion

The volume of $\text{Hg}{\left({\text{NO}}_3\right)}_2$ required for complete precipitation is found to be $\text{10}\text{.15}\text{mL}$$9.76\text{mL}$.

Interpretation Introduction

(b)

Interpretation:

The mass of ${\text{HgI}}_2$ precipitate is to be calculated.

Concept introduction:

Molar concentration also termed as molarity, the amount concentration or substance concentration is the concentration of a substance, in terms of a solute in a solution. It is the amount of substance per unit volume of solution. Stoichiometry is the relationship between the quantities in a balanced chemical equation.

Answer to Problem 70E

The mass of ${\text{HgI}}_2$ precipitate is $0.926\text{g}$.

Explanation of Solution

The weight of $\text{Hg}{\left({\text{NO}}_3\right)}_2$ is calculated as shown below.

$\text{W}={\text{V}}_{\text{KI}}\times \frac{\text{Molar}}{\text{Concentration}}\times \frac{{\text{n}}_{{\text{HgI}}_2}}{{\text{n}}_{\text{KI}}}\times \text{MM}$ …(1)

Where,

• ${\text{V}}_{\text{KI}}$ is the volume of potassium iodide.

• $\frac{\text{Molar}}{\text{Concentration}}$ is the molar concentration of potassium iodide.

• $\frac{{\text{n}}_{{\text{HgI}}_2}}{{\text{n}}_{\text{KI}}}$ is the number of moles of ${\text{HgI}}_2$ and $\text{KI}$.

• $\text{MM}$ is the molar mass of ${\text{HgI}}_2$.

Substitute the values of ${\text{V}}_{\text{KI}}$ as $25.0\text{mL}$, $\frac{\text{Molar}}{\text{Concentration}}$ as $\frac{\text{0}\text{.170 mol KI}}{1000\text{mL solution}}$, $\frac{{\text{n}}_{{\text{HgI}}_2}}{{\text{n}}_{\text{KI}}}$ as $\frac{1{\text{mol HgI}}_2}{2\text{mol}\text{KI}}$ and $\text{MM}$ as $324.59{\text{g HgI}}_2$ in equation (1) as shown in equation below.

$\begin{array}{rll}\text{W}& =& 25.0\text{mL}\times \frac{\text{0}\text{.170 mol KI}}{1000\text{mL solution}}\times \frac{1{\text{mol HgI}}_2}{2\text{mol}\text{KI}}\times \frac{324.59{\text{g HgI}}_2}{1\text{mol}\text{Hg}{\left({\text{NO}}_3\right)}_2}\\ & =& \frac{25\times 0.170\times 454.39}{1000\times 2}\text{g}\\ \text{= 0}\text{.926}\text{g}\end{array}$

Conclusion

The mass of ${\text{HgI}}_2$ precipitate is found to be $0.926\text{g}$.