Chapter 8, Problem 8.68EP

(a)

Interpretation Introduction

Interpretation:

The volume of water should be added to yield the given solution has to be determined.

Concept Introduction:

Dilution equation:

Dilution equation is given by,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{where,}\\ {\text{C}}_{\text{s}}\text{-}\text{concentration}\text{of}\text{stock}\text{solution}\\ {\text{V}}_{\text{s}}\text{-}\text{volume}\text{of}\text{stock}\text{solution}\\ {\text{C}}_{\text{d}}\text{-}\text{concentration}\text{of}\text{diluted}\text{solution}\\ {\text{V}}_{\text{d}}\text{-}\text{volume}\text{of}\text{diluted}\text{solution}\end{array}$

Answer to Problem 8.68EP

The volume of water should be added to yield the given solution is 175 mL.

Explanation of Solution

Given information:

Volume of solution = 25.0 mL

Initial concentration = 1.00 M

Required concentration = 0.125 M

Final volume of solution = ?

Apply dilution equation to calculate the volume of water should be added to yield the given solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{rearrange}\text{the}\text{above}\text{equation,}\\ {\text{V}}_{\text{d}}\text{=}\text{diluted}\text{volume}\text{=}{\text{V}}_{\text{s}}\text{×}\frac{{\text{C}}_{\text{s}}}{{\text{C}}_{\text{d}}}\\ \text{=}\text{25}\text{.0}\text{mL}\text{×}\frac{\text{1}\text{.00}\text{M}}{\text{0}\text{.125}\text{M}}\text{=}\text{200}\text{mL}\end{array}$

Now subtract the original volume (V_d) from the final volume (V_s)

$\begin{array}{l}\text{Volume}\text{of}\text{water}\text{added}\text{=}{\text{V}}_{\text{d}}\text{-}{\text{V}}_{\text{s}}\\ \text{=}\text{200}\text{mL}\text{-}\text{25}\text{.0}\text{mL}\text{=}\text{175}\text{mL}\end{array}$

Therefore, the volume of water should be added to yield the given solution is 175 mL.

(b)

Interpretation Introduction

Interpretation:

The volume of water should be added to yield the given solution has to be determined.

Concept Introduction:

Dilution equation:

Dilution equation is given by,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{where,}\\ {\text{C}}_{\text{s}}\text{-}\text{concentration}\text{of}\text{stock}\text{solution}\\ {\text{V}}_{\text{s}}\text{-}\text{volume}\text{of}\text{stock}\text{solution}\\ {\text{C}}_{\text{d}}\text{-}\text{concentration}\text{of}\text{diluted}\text{solution}\\ {\text{V}}_{\text{d}}\text{-}\text{volume}\text{of}\text{diluted}\text{solution}\end{array}$

Answer to Problem 8.68EP

The volume of water should be added to yield the given solution is 395 mL.

Explanation of Solution

Given information:

Volume of solution = 5.00 mL

Initial concentration = 10.0 M

Required concentration = 0.125 M

Final volume of solution = ?

Apply dilution equation to calculate the volume of water should be added to yield the given solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{rearrange}\text{the}\text{above}\text{equation,}\\ {\text{V}}_{\text{d}}\text{=}\text{diluted}\text{volume}\text{=}{\text{V}}_{\text{s}}\text{×}\frac{{\text{C}}_{\text{s}}}{{\text{C}}_{\text{d}}}\\ \text{=}\text{5}\text{.00}\text{mL}\text{×}\frac{\text{10}\text{.0}\text{M}}{\text{0}\text{.125}\text{M}}\text{=}\text{400}\text{mL}\end{array}$

Now subtract the original volume (V_d) from the final volume (V_s)

$\begin{array}{l}\text{Volume}\text{of}\text{water}\text{added}\text{=}{\text{V}}_{\text{d}}\text{-}{\text{V}}_{\text{s}}\\ \text{=}\text{400}\text{mL}\text{-}\text{5}\text{mL}\text{=}\text{395}\text{mL}\end{array}$

Therefore, the volume of water should be added to yield the given solution is 395 mL.

(c)

Interpretation Introduction

Interpretation:

The volume of water should be added to yield the given solution has to be determined.

Concept Introduction:

Dilution equation:

Dilution equation is given by,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{where,}\\ {\text{C}}_{\text{s}}\text{-}\text{concentration}\text{of}\text{stock}\text{solution}\\ {\text{V}}_{\text{s}}\text{-}\text{volume}\text{of}\text{stock}\text{solution}\\ {\text{C}}_{\text{d}}\text{-}\text{concentration}\text{of}\text{diluted}\text{solution}\\ {\text{V}}_{\text{d}}\text{-}\text{volume}\text{of}\text{diluted}\text{solution}\end{array}$

Answer to Problem 8.68EP

The volume of water should be added to yield the given solution is 47500 mL.

Explanation of Solution

Given information:

Volume of solution = 2500 mL

Initial concentration = 2.50 M

Required concentration = 0.125 M

Final volume of solution = ?

Apply dilution equation to calculate the volume of water should be added to yield the given solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{rearrange}\text{the}\text{above}\text{equation,}\\ {\text{V}}_{\text{d}}\text{=}\text{diluted}\text{volume}\text{=}{\text{V}}_{\text{s}}\text{×}\frac{{\text{C}}_{\text{s}}}{{\text{C}}_{\text{d}}}\\ \text{=}\text{2500}\text{mL}\text{×}\frac{\text{2}\text{.50}\text{M}}{\text{0}\text{.125}\text{M}}\text{=}\text{50000}\text{mL}\end{array}$

Now subtract the original volume (V_d) from the final volume (V_s)

$\begin{array}{l}\text{Volume}\text{of}\text{water}\text{added}\text{=}{\text{V}}_{\text{d}}\text{-}{\text{V}}_{\text{s}}\\ \text{=}\text{50000}\text{mL}\text{-}\text{2500}\text{mL}\text{=}\text{47500}\text{mL}\end{array}$

Therefore, the volume of water should be added to yield the given solution is 47500 mL.

(d)

Interpretation Introduction

Interpretation:

The volume of water should be added to yield the given solution has to be determined.

Concept Introduction:

Dilution equation:

Dilution equation is given by,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{where,}\\ {\text{C}}_{\text{s}}\text{-}\text{concentration}\text{of}\text{stock}\text{solution}\\ {\text{V}}_{\text{s}}\text{-}\text{volume}\text{of}\text{stock}\text{solution}\\ {\text{C}}_{\text{d}}\text{-}\text{concentration}\text{of}\text{diluted}\text{solution}\\ {\text{V}}_{\text{d}}\text{-}\text{volume}\text{of}\text{diluted}\text{solution}\end{array}$

Answer to Problem 8.68EP

The volume of water should be added to yield the given solution is 3.0 mL.

Explanation of Solution

Given information:

Volume of solution = 75.0 mL

Initial concentration = 0.130 M

Required concentration = 0.125 M

Final volume of solution = ?

Apply dilution equation to calculate the volume of water should be added to yield the given solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{rearrange}\text{the}\text{above}\text{equation,}\\ {\text{V}}_{\text{d}}\text{=}\text{diluted}\text{volume}\text{=}{\text{V}}_{\text{s}}\text{×}\frac{{\text{C}}_{\text{s}}}{{\text{C}}_{\text{d}}}\\ \text{=}\text{75}\text{.0}\text{mL}\text{×}\frac{\text{0}\text{.130}\text{M}}{\text{0}\text{.125}\text{M}}\text{=}\text{78}\text{mL}\end{array}$

Now subtract the original volume (V_d) from the final volume (V_s)

$\begin{array}{l}\text{Volume}\text{of}\text{water}\text{added}\text{=}{\text{V}}_{\text{d}}\text{-}{\text{V}}_{\text{s}}\\ \text{=}\text{78}\text{.0}\text{mL}\text{-}\text{75}\text{.0}\text{mL}\text{=}\text{3}\text{mL}\end{array}$

Therefore, the volume of water should be added to yield the given solution is 3.0 mL.