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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 \(100.0 \mathrm{mL}\). (See Exercise 123 for definitions.)

Short Answer

Expert verified
To prepare the standard copper solutions, use the dilution formula \(C_1V_1 = C_2V_2\) where \(C_1\) is the initial concentration (1000 ppm), \(V_1\) is the required initial volume, \(C_2\) is the final concentration, and \(V_2\) is the final volume (100 mL). Calculate the volumes of the 1000 ppm solution needed: 1 mL for 10 ppm, 2.5 mL for 25 ppm, 5 mL for 50 ppm, 7.5 mL for 75 ppm, and 10 mL for 100 ppm. Measure the calculated volumes, add them to appropriate volumetric flasks, add solvent to reach 100 mL, and mix thoroughly to prepare the respective standard solutions.

Step by step solution

01

Calculate the volume of 1000 ppm solution required for the 10 ppm solution

To prepare a 10 ppm copper solution, we can use the dilution formula: \(C_1V_1 = C_2V_2\) \(1000 \text{ ppm} \times V_1 = 10 \text{ ppm} \times 100 \text{ mL}\) Now, to solve for the volume, \(V_1\), required from the 1000 ppm solution: \(V_1 = \frac{10 \text{ ppm} \times 100 \text{ mL}}{1000 \text{ ppm}} = 1 \text{ mL}\) Hence, to prepare a 10 ppm copper solution with a volume of 100 mL, 1 mL of the 1000 ppm copper solution is needed.
02

Calculate the volume of 1000 ppm solution required for the 25 ppm, 50 ppm, 75 ppm, and 100 ppm solutions

Using the same formula used above, we can easily calculate the necessary volumes for the other desired concentrations: For 25 ppm (\(C_2\)): \(V_1 = \frac{25 \text{ ppm} \times 100 \text{ mL}}{1000 \text{ ppm}} = 2.5 \text{ mL}\) For 50 ppm (\(C_2\)): \(V_1 = \frac{50 \text{ ppm} \times 100 \text{ mL}}{1000 \text{ ppm}} = 5 \text{ mL}\) For 75 ppm (\(C_2\)): \(V_1 = \frac{75 \text{ ppm} \times 100 \text{ mL}}{1000 \text{ ppm}} = 7.5 \text{ mL}\) For 100 ppm (\(C_2\)): \(V_1 = \frac{100 \text{ ppm} \times 100 \text{ mL}}{1000 \text{ ppm}} = 10 \text{ mL}\) To prepare the standard solutions, we will need to add the calculated volume of the 1000 ppm solution to the proper amount of solvent (usually water) to achieve a final volume of 100 mL.
03

Preparing standard solutions

To prepare each of the standard solutions listed above, follow the steps below: 1. Measure the calculated volume of the 1000 ppm copper solution. 2. Add this volume to an appropriately sized volumetric flask. 3. Add solvent (usually water) to the flask until the final volume of the solution reaches the 100 mL mark. 4. Mix the solution thoroughly. By following these steps for each of the desired concentrations (10 ppm, 25 ppm, 50 ppm, 75 ppm, and 100 ppm), you will have made the necessary calibration standards for spectroscopic analysis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spectroscopy Simplified
Spectroscopy is a fascinating technique that allows us to observe how different substances interact with light. Essentially, it helps scientists and researchers measure the amount of light a substance absorbs or emits. The more light it absorbs or emits, the higher its concentration typically is.
In the context of our exercise, spectroscopy is used to analyze copper concentrations. By measuring how much light copper solutions absorb, we can create a calibration curve. This curve is crucial because it graphically shows the relationship between concentration and absorbance. It allows us to determine unknown concentrations by comparing them to our standard solutions.
A few key benefits of spectroscopy include:
  • High sensitivity, capable of detecting even low concentrations of substances.
  • Speed, allowing for quick analysis of multiple samples.
  • Non-destructive nature, meaning samples are not consumed in the process.
Understanding the basics of spectroscopy enables better analysis and interpretation of results, especially in fields like chemistry and environmental science.
The Art of Solution Dilution
Solution dilution is a handy technique used to lower the concentration of a solution by adding more solvent. This is essential when working with high concentration stock solutions, such as the 1000 ppm copper solution in our exercise.
What you need to know about solution dilution:
  • The dilution formula, \(C_1V_1 = C_2V_2\), where \(C_1\) and \(V_1\) are the concentration and volume of the initial solution, and \(C_2\) and \(V_2\) are the concentration and volume after dilution.
  • This formula helps to calculate how much of the concentrated solution you need to mix with the solvent to achieve a desired lower concentration.
  • It's crucial to measure accurately, as small errors can significantly impact your results.
For example, to dilute the 1000 ppm copper solution to 10 ppm, we use the dilution formula to find that we only need 1 mL of the stock solution and add enough water to reach a final volume of 100 mL.
Mastering Concentration Calculations
Concentration calculations are a vital part of preparing standard solutions in the laboratory. They help you determine the exact amount of solute in a given volume of solution, which is essential for accurate and reliable spectroscopic analysis.
Here’s a breakdown of crucial points regarding concentration calculations:
  • Concentration is generally expressed in parts per million (ppm), which reflects the mass of solute per million parts of solution, often used in very dilute samples.
  • The calculation strategy revolves around manipulating the equation \(C_1V_1 = C_2V_2\).
  • Remember to carefully convert volumes to the same units before performing any calculations to avoid errors. Often, milliliters are used because they are manageable for laboratory work.
By mastering concentration calculations, you are better equipped to prepare mixtures that closely adhere to desired specifications, providing reliable data for your experiments.

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Most popular questions from this chapter

Zinc and magnesium metal each react with hydrochloric acid according to the following equations: $$ \begin{array}{c} \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) \\ \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g) \end{array} $$ A \(10.00-\mathrm{g}\) mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with \(156 \mathrm{mL}\) of \(3.00 \mathrm{M}\) silver nitrate to produce the maximum possible amount of silver chloride. a. Determine the percent magnesium by mass in the original mixture. b. If \(78.0 \mathrm{mL}\) of HCl was added, what was the concentration of the HCl?

Specify which of the following are oxidation-reduction reactions, and identify the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced. a. \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \rightarrow 2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\) b. \(\mathrm{HCl}(g)+\mathrm{NH}_{3}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) c. \(\mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 4 \mathrm{HCl}(a q)+\mathrm{SiO}_{2}(s)\) d. \(\mathrm{SiCl}_{4}(l)+2 \mathrm{Mg}(s) \rightarrow 2 \mathrm{MgCl}_{2}(s)+\mathrm{Si}(s)\) e. \(\mathrm{Al}(\mathrm{OH})_{4}^{-}(a q) \rightarrow \mathrm{AlO}_{2}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(i)\)

A solution is prepared by dissolving 0.6706 g oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) in enough water to make \(100.0 \mathrm{mL}\) of solution. \(\mathrm{A}\) \(10.00-\mathrm{mL}\) aliquot (portion) of this solution is then diluted to a final volume of \(250.0 \mathrm{mL}\). What is the final molarity of the oxalic acid solution?

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CuSO}_{4}(a q)\) b. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CaCl}_{2}(a q)\) c. \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{MgI}_{2}(a q)\) d. \(\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q)+\mathrm{AlBr}_{3}(a q)\)

When organic compounds containing sulfur are burned, sulfur dioxide is produced. The amount of \(\mathrm{SO}_{2}\) formed can be determined by the reaction with hydrogen peroxide: $$ \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) $$ The resulting sulfuric acid is then titrated with a standard NaOH solution. A 1.302 -g sample of coal is burned and the \(\mathrm{SO}_{2}\) is collected in a solution of hydrogen peroxide. It took \(28.44 \mathrm{mL}\) of a \(0.1000-M \mathrm{NaOH}\) solution to titrate the resulting sulfuric acid. Calculate the mass percent of sulfur in the coal sample. Sulfuric acid has two acidic hydrogens.

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