Chapter 9, Problem 9.164QP

(a)

Interpretation Introduction

Interpretation:

The required mass of ${\text{KMnO}}_{\text{4}}$ to prepare a final solution should be calculated and reason for why not the very dilute solution are directly prepared should be explained.

Concept introduction:

Volumetric principle:

• In the neutralization reaction, the volume and concentration of initial components are equal to the volume and concentration of the final components.
• In the dilution of the chemical solutions are using volumetric principle.
• In the dilution process, the relationship between initial and final concentrations and volumes of solutions are given in the volumetric equation and it is,

$\begin{array}{l}{\text{M}}_{\text{c}}{\text{×mL}}_{\text{c}}\text{=}{\text{M}}_{\text{d}}{\text{×mL}}_{\text{d}}\\ {\text{M}}_{\text{c}}\text{=}\text{Initial}\text{concentration}\\ {\text{mL}}_{\text{c}}\text{=}\text{Initial}\text{volume}\\ {\text{M}}_{\text{d}}\text{=}\text{diluted}\text{concentration}\\ {\text{mL}}_{\text{d}}\text{=}\text{diluted}\text{volume}\end{array}$

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of the solute is calculated by taken mass of solute divided by molar mass of the solute.

$\text{Mole}\text{=}\frac{\text{Mass}(\text{g)}}{\text{Molar}\text{mass}\text{(g)}}$

Answer to Problem 9.164QP

The required mass of ${\text{KMnO}}_{\text{4}}$ to prepare a final solution is $\text{3}\text{.29}\text{×}{\text{10}}^{\text{-5}}\text{g}{\text{KMnO}}_{\text{4}}$

Explanation of Solution

To record the given data,

Taken mass ${\text{KMnO}}_{\text{4}}$ of in first dilution = $\text{0}\text{.8214 g}$

Final Volume of ${\text{KMnO}}_{\text{4}}$ in first dilution = $\text{500 mL}$

Taken Volume of ${\text{KMnO}}_{\text{4}}$ in second dilution = $\text{2}\text{.000 mL}$

Final Volume of ${\text{KMnO}}_{\text{4}}$ in second dilution = $\text{1000 mL}$

Taken Volume of ${\text{KMnO}}_{\text{4}}$ in third dilution= $\text{10 mL}$

Final Volume of ${\text{KMnO}}_{\text{4}}$ in third dilution = $\text{250 mL}$

The taken mass and volumes of ${\text{KMnO}}_{\text{4}}$ solutions are recorded as shown above.

Calculate the molarity of ${\text{KMnO}}_{\text{4}}$ first diluted solution.

Molar mass of ${\text{KMnO}}_{\text{4}}$ is $\text{158}\text{.04 g}$

$\begin{array}{l}\text{=}\text{0}\text{.8214}\text{g}{\text{KMnO}}_{\text{4}}\text{×}\frac{\text{1}\text{mol}{\text{KMnO}}_{\text{4}}}{\text{158}\text{.04}{\text{KMnO}}_{\text{4}}}\\ \text{=}\text{0}\text{.0051974}\text{mole}\end{array}$

The molarity of ${\text{KMnO}}_{\text{4}}$ is,

$\begin{array}{l}\text{=}\frac{\text{0}\text{.0051974}}{\text{0}\text{.500}\text{L}}\\ \text{=}\text{0}\text{.010395}\text{M}\end{array}$

• The taken mass of ${\text{KMnO}}_{\text{4}}$ is divided by molar mass of ${\text{KMnO}}_{\text{4}}$ to give the mole of taken ${\text{KMnO}}_{\text{4}}.$
• The calculated mole is dividing by volume of solution to give molarity of first diluted ${\text{KMnO}}_{\text{4}}$ solution.
• The molarity of first diluted ${\text{KMnO}}_{\text{4}}$ solution is $\text{0}\text{.010395}\text{M}$.

Calculate the molarity second diluted ${\text{KMnO}}_{\text{4}}$ solution.

$\begin{array}{l}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{=M}}_{\text{2}}{\text{V}}_{\text{2}}\\ \text{1000}\text{mL}\text{=}\text{0}\text{.010395}\text{M×}\text{2}\text{.000}\text{mL}\\ \text{=}\frac{\text{0}\text{.010395}\text{M×}\text{2}\text{.000}\text{mL}}{\text{1000}\text{mL}}\\ \text{=}\text{2}{\text{.079×10}}^{\text{-5}}\text{M}\end{array}$

• The calculated concentration and volume of first diluted ${\text{KMnO}}_{\text{4}}$ solution are plugged in the above equation to give the concentration of second diluted ${\text{KMnO}}_{\text{4}}$ solution.
• The concentration of second diluted ${\text{KMnO}}_{\text{4}}$ solution $\text{2}{\text{.079×10}}^{\text{-5}}\text{M}$

Calculate the molarity second third ${\text{KMnO}}_{\text{4}}$ solution

$\begin{array}{l}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{=M}}_{\text{2}}{\text{V}}_{\text{2}}\\ \text{250}\text{mL}\text{=}\text{0}\text{.00002079}\text{M×}\text{10}\text{.0}\text{mL}\\ \text{=}\frac{\text{0}\text{.00002079}\text{×}\text{10}\text{.0}\text{mL}}{\text{250}\text{mL}}\\ \text{=}\text{8}\text{.32}{\text{×10}}^{\text{-7}}\text{M}\end{array}$

• The calculated concentration and volume of second diluted ${\text{KMnO}}_{\text{4}}$ solution are plugged in the above equation to give the concentration of third diluted ${\text{KMnO}}_{\text{4}}$ solution.
• The concentration of third diluted ${\text{KMnO}}_{\text{4}}$ solution $\text{8}\text{.32}{\text{×10}}^{\text{-7}}\text{M}$

Calculate the mass of ${\text{KMnO}}_{\text{4}}$ in third solution

The mole of ${\text{KMnO}}_{\text{4}}$ in third solution,

$\begin{array}{l}\text{=}\frac{\text{8}\text{.32}{\text{×10}}^{\text{-7}}\text{M}}{\text{1000}\text{mL}\text{soln}}\text{×250}\text{mL}\\ \text{=}\text{2}\text{.08×}{\text{10}}^{\text{-7}}\text{mole}\end{array}$

The mass of ${\text{KMnO}}_{\text{4}}$ in third solution is,

$\begin{array}{l}\text{=}\text{2}{\text{.08×10}}^{\text{-7}}\text{mole×}\frac{\text{158}\text{.04}{\text{KMnO}}_{\text{4}}}{\text{1}\text{mol}{\text{KMnO}}_{\text{4}}}\\ \text{=}\text{3}\text{.29}\text{×}{\text{10}}^{\text{-5}}\text{g}{\text{KMnO}}_{\text{4}}\end{array}$

• The calculated concentration and volume of third diluted ${\text{KMnO}}_{\text{4}}$ solution are plugged in the above equation to give the mole of third diluted ${\text{KMnO}}_{\text{4}}$ solution.
• The mole of third diluted ${\text{KMnO}}_{\text{4}}$ solution $\text{2}\text{.08×}{\text{10}}^{\text{-7}}\text{mole}$
• The calculated mole of third diluted ${\text{KMnO}}_{\text{4}}$ solution is multiplied by molar mass of ${\text{KMnO}}_{\text{4}}$ to give mass of present in third diluted ${\text{KMnO}}_{\text{4}}$ solution.
• The mass of present in third diluted ${\text{KMnO}}_{\text{4}}$ solution is $\text{3}\text{.29}\text{×}{\text{10}}^{\text{-5}}\text{g}{\text{KMnO}}_{\text{4}}$

(b)

Interpretation Introduction

Interpretation:

The required mass of ${\text{KMnO}}_{\text{4}}$ to prepare a final solution should be calculated and reason for why not the very dilute solution are directly prepared should be explained.

Concept introduction:

Volumetric principle:

• In the neutralization reaction, the volume and concentration of initial components are equal to the volume and concentration of the final components.
• In the dilution of the chemical solutions are using volumetric principle.
• In the dilution process, the relationship between initial and final concentrations and volumes of solutions are given in the volumetric equation and it is,

$\begin{array}{l}{\text{M}}_{\text{c}}{\text{×mL}}_{\text{c}}\text{=}{\text{M}}_{\text{d}}{\text{×mL}}_{\text{d}}\\ {\text{M}}_{\text{c}}\text{=}\text{Initial}\text{concentration}\\ {\text{mL}}_{\text{c}}\text{=}\text{Initial}\text{volume}\\ {\text{M}}_{\text{d}}\text{=}\text{diluted}\text{concentration}\\ {\text{mL}}_{\text{d}}\text{=}\text{diluted}\text{volume}\end{array}$

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of the solute is calculated by taken mass of solute divided by molar mass of the solute.

$\text{Mole}\text{=}\frac{\text{Mass}(\text{g)}}{\text{Molar}\text{mass}\text{(g)}}$

Answer to Problem 9.164QP

The weight of solutes to prepare a very dilute solution is too small to directly weigh accurately.  So it results in more errors.  Therefore, the direct preparation of very dilute solution is not a good method.

Explanation of Solution

To record the given data,

Taken mass ${\text{KMnO}}_{\text{4}}$ of in first dilution = $\text{0}\text{.8214 g}$

Final Volume of ${\text{KMnO}}_{\text{4}}$ in first dilution = $\text{500 mL}$

Taken Volume of ${\text{KMnO}}_{\text{4}}$ in second dilution = $\text{2}\text{.000 mL}$

Final Volume of ${\text{KMnO}}_{\text{4}}$ in second dilution = $\text{1000 mL}$

Taken Volume of ${\text{KMnO}}_{\text{4}}$ in third dilution= $\text{10 mL}$

Final Volume of ${\text{KMnO}}_{\text{4}}$ in third dilution = $\text{250 mL}$

The taken mass and volumes of ${\text{KMnO}}_{\text{4}}$ solutions are recorded as shown above.

Explain reason for why not the very dilute solution are directly prepared.

• The weight of solutes to prepare a very dilute solution is too small to directly weigh accurately.  So, it results in more errors.  Therefore, the direct preparation of very dilute solution is not a good method.