Chapter 4, Problem 122CP

In the spectroscopic analysis of many substances, a series of standard solutions of known concentration are measured to generate a calibration curve. How would you prepare standard solutions containing 10.0, 25.0, 50.0, 75.0, and 100. ppm of copper from a commercially produced 1000.0-ppm solution? Assume each solution has a final volume of 100.0 mL. (See Exercise 123 for definitions.)

Interpretation Introduction

Interpretation: The volume of stock solution of each sub-division has to be calculated.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

$\text{Volume}\text{of}\text{solution(in}\text{litres)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Molarity(in}\text{M)}}$

Answer to Problem 122CP

Volume of stock solution is $\text{10mL}$

Explanation of Solution

Record the given data

Mass of solute= $\text{100ppm standard}$

Volume of stock solution= $\text{1000}\text{ppm}$

The mass of the solute and volume of stock solution are recorded as shown above.

To calculate the volume of stock solution to make $\text{100}\text{mL}\text{of}\text{100ppm}\text{solution}$ $\text{Cu}\text{in}\text{water}$

$\begin{array}{l}\frac{\text{100}\text{.}\text{µg}\text{Cu}}{\text{mL}}{\text{×100 mL=1×10}}^{\text{4}}\text{µg}\text{Cu}\text{needed}\\ {\text{1×10}}^{\text{4}}\text{µg}\text{Cu×}\frac{\text{1}\text{mL}\text{stock}}{\text{1000}\text{µg}\text{Cu}}\text{=10mL }\text{of}\text{stock}\text{solution}\end{array}$

Transfer $\text{10mL}$ of the $\text{1000ppm}$ stock solution to a $\text{100mL}$ volumetric flask, and dilute to the mark.

$\text{Required}\text{volume}\text{stock}\text{solution}\text{is}\text{10mL}$

The volume of stock solution is calculated by plugging in the values of concentration of stock solution and mass of copper along with required volume of the solution. The volume of stock solution copper is $\text{10mL}$.

Interpretation Introduction

Interpretation: The volume of stock solution of each sub-division has to be calculated.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

$\text{Volume}\text{of}\text{solution(in}\text{litres)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Molarity(in}\text{M)}}$

Answer to Problem 122CP

Volume of stock solution is $\text{7}\text{.5mL}$

Explanation of Solution

Record the given data

Mass of solute= $\text{75ppm standard}$

Volume of stock solution= $\text{1000}\text{ppm}$

The mass of the solute and volume of stock solution are recorded as shown above.

To calculate the volume of stock solution to make $\text{100}\text{mL}\text{of}\text{75ppm}\text{solution}$ $\text{Cu}\text{in}\text{water}$

$\begin{array}{l}\frac{\text{75}\text{.}\text{µg}\text{Cu}}{\text{mL}}\text{×100 mL=7}{\text{.5×10}}^{\text{3}}\text{µg}\text{Cu}\text{needed}\\ \text{7}{\text{.5×10}}^{\text{3}}\text{µg}\text{Cu×}\frac{\text{1}\text{mL}\text{stock}}{\text{1000}\text{µg}\text{Cu}}\text{=7}\text{.5mL }\text{of}\text{stock}\text{solution}\end{array}$

Transfer $\text{5mL}$ of the $\text{1000ppm}$ stock solution to a $\text{100mL}$ volumetric flask, and dilute to the mark.

$\text{Required}\text{volume}\text{stock}\text{solution}\text{is}\text{7}\text{.5mL}$

The volume of stock solution is calculated by plugging in the values of concentration of stock solution and mass of copper along with required volume of the solution. The volume of stock solution copper is $\text{7}\text{.5mL}$.

Interpretation Introduction

Interpretation: The volume of stock solution of each sub-division has to be calculated.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

$\text{Volume}\text{of}\text{solution(in}\text{litres)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Molarity(in}\text{M)}}$

Answer to Problem 122CP

Volume of stock solution is $\text{5mL}$

Explanation of Solution

Record the given data

Mass of solute= $\text{50ppm standard}$

Volume of stock solution= $\text{1000}\text{ppm}$

The mass of the solute and volume of stock solution are recorded as shown above.

To calculate the volume of stock solution to make $\text{100}\text{mL}\text{of}5\text{0ppm}\text{solution}$ $\text{Cu}\text{in}\text{water}$

$\begin{array}{l}\frac{\text{50}\text{.}\text{µg}\text{Cu}}{\text{mL}}{\text{×100 mL=5×10}}^3\text{µg}\text{Cu}\text{needed}\\ {\text{5×10}}^3\text{µg}\text{Cu×}\frac{\text{1}\text{mL}\text{stock}}{\text{1000}\text{µg}\text{Cu}}\text{=5mL }\text{of}\text{stock}\text{solution}\end{array}$

Transfer $\text{5mL}$ of the $\text{1000ppm}$ stock solution to a $\text{100mL}$ volumetric flask, and dilute to the mark.

$\text{Required}\text{volume}\text{stock}\text{solution}\text{is}5\text{mL}$

The volume of stock solution is calculated by plugging in the values of concentration of stock solution and mass of copper along with required volume of the solution. The volume of stock solution copper is $\text{5mL}$.

Interpretation Introduction

Interpretation: The volume of stock solution of each sub-division has to be calculated.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

$\text{Volume}\text{of}\text{solution(in}\text{litres)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Molarity(in}\text{M)}}$

Answer to Problem 122CP

Volume of stock solution is $\text{2}\text{.5mL}$

Explanation of Solution

Record the given data

Mass of solute= $\text{25ppm standard}$

Volume of stock solution= $\text{1000}\text{ppm}$

The mass of the solute and volume of stock solution are recorded as shown above.

To calculate the volume of stock solution to make $\text{100}\text{mL}\text{of}25\text{ppm}\text{solution}$ $\text{Cu}\text{in}\text{water}$

$\begin{array}{l}\frac{\text{25}\text{.}\text{µg}\text{Cu}}{\text{mL}}\text{×100 mL=2}{\text{.5×10}}^3\text{µg}\text{Cu}\text{needed}\\ \text{2}{\text{.5×10}}^3\text{µg}\text{Cu×}\frac{\text{1}\text{mL}\text{stock}}{\text{1000}\text{µg}\text{Cu}}\text{=2}\text{.5mL }\text{of}\text{stock}\text{solution}\end{array}$

Transfer $\text{2}\text{.5mL}$ of the $\text{1000ppm}$ stock solution to a $\text{100mL}$ volumetric flask, and dilute to the mark.

$\text{Required}\text{volume}\text{stock}\text{solution}\text{is}\text{2}\text{.5mL}$

The volume of stock solution is calculated by plugging in the values of concentration of stock solution and mass of copper along with required volume of the solution. The volume of stock solution copper is $\text{2}\text{.5mL}$.

Interpretation Introduction

Interpretation: The volume of stock solution of each sub-division has to be calculated.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

$\text{Volume}\text{of}\text{solution(in}\text{litres)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Molarity(in}\text{M)}}$

Answer to Problem 122CP

Volume of stock solution is $\text{1}\text{mL}$

Explanation of Solution

Record the given data

Mass of solute= $\text{10}\text{ppm standard}$

Volume of stock solution= $\text{1000}\text{ppm}$

The mass of the solute and volume of stock solution are recorded as shown above.

To calculate the volume of stock solution to make $\text{100}\text{mL}\text{of}\text{10}\text{ppm}\text{solution}$ $\text{Cu}\text{in}\text{water}$

$\begin{array}{l}\frac{\text{10}\text{.}\text{µg}\text{Cu}}{\text{mL}}{\text{×100 mL=1×10}}^{\text{3}}\text{µg}\text{Cu}\text{needed}\\ {\text{1×10}}^{\text{3}}\text{µg}\text{Cu×}\frac{\text{1}\text{mL}\text{stock}}{\text{1000}\text{µg}\text{Cu}}\text{= 1mL }\text{of}\text{stock}\text{solution}\end{array}$

Transfer $\text{1}\text{mL}$ of the $\text{1000}\text{ppm}$ stock solution to a $\text{100mL}$ volumetric flask, and dilute to the mark.

$\text{Required}\text{volume}\text{stock}\text{solution}\text{is}\text{1mL}$

The volume of stock solution is calculated by plugging in the values of concentration of stock solution and mass of copper along with required volume of the solution. The volume of stock solution copper is $\text{1}\text{mL}$.