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How would you prepare \(60.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HNO}_{3}\) from a stock solution of \(4.00 \mathrm{M} \mathrm{HNO}_{3}\) ?

Short Answer

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To prepare 60.0 mL of 0.200 M \( \mathrm{HNO}_3 \), dilute 3.00 mL of 4.00 M stock solution with water to 60.0 mL.

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01

Determine the Required Moles of Solute

First, calculate the moles of solute needed for the desired solution. Use the formula: \[\text{Moles of solute} = M_{desired} \times V_{desired}\]where \(M_{desired} = 0.200 \space \text{M}\) and \(V_{desired} = 60.0 \space \text{mL}\) or \(0.0600 \space \text{L}\). Thus, \[\text{Moles of solute} = 0.200 \times 0.0600 = 0.0120 \space \text{moles}\]
02

Calculate the Volume of Stock Solution Needed

Using the moles of solute from Step 1, determine the volume of the stock solution necessary. Rearrange the formula for molarity to solve for volume:\[V_{stock} = \frac{\text{Moles of solute}}{M_{stock}}\]where \(M_{stock} = 4.00 \space \text{M}\). Then,\[V_{stock} = \frac{0.0120}{4.00} = 0.0030 \space \text{L} = 3.00 \space \text{mL}\]
03

Mix the Solutions

Transfer 3.00 mL of the 4.00 M \( \mathrm{HNO}_3 \) stock solution into a container. Add distilled water to dilute it to a total volume of 60.0 mL. Stir the solution to ensure it is mixed uniformly.

Key Concepts

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

Molarity
Molarity is a key concept in chemistry that measures the concentration of a solution. It is defined as the number of moles of solute present in one liter of solution. The formula for molarity is \[M = \frac{\text{moles of solute}}{\text{liters of solution}} \] where:
  • M is the molarity.
  • The moles of solute is the amount of substance dissolved.
  • The liters of solution is the total volume of the mixture.
Understanding molarity helps with preparing solutions of a specified strength and is widely used in lab settings. For example, if you have a 0.200 M nitric acid solution, it means there are 0.200 moles of HNO3 per liter of solution. For practical purposes, especially in a lab, it's essential to calculate molarity accurately to ensure that the solution behaves as expected during experiments.
Dilution
Dilution refers to the process of reducing the concentration of a solute in a solution, generally by adding more solvent to it. This process is useful when you need a lower concentration than what's available from a stock solution. The calculation follows the principle of conservation of moles, meaning the total amount of solute remains constant before and after dilution. The formula for dilution is:\[M_1 V_1 = M_2 V_2 \]where:
  • \(M_1\) and \(M_2\) are the molarity of the stock and diluted solutions respectively.
  • \(V_1\) and \(V_2\) are their volumes.
In the given exercise, we used dilution to convert a 4.00 M stock solution into a 0.200 M solution by adding a calculated amount of water. Mastery of dilution techniques is critical for working with different concentrations based on experimental requirements.
Stock Solution
A stock solution is a highly concentrated form of a reagent, stored in a lab for ease of use. It allows researchers to quickly prepare solutions of varying concentration by diluting the stock solution to the desired level. Stock solutions save time and resources, as they eliminate the constant need for measuring out solids and dissolving them. When preparing a desired solution concentration, you often calculate how much of the concentrated stock solution is needed. Then, you dilute it to reach your required concentration. For instance, the given problem involves a stock solution with a molarity of 4.00 M, which we diluted to create a 0.200 M solution. Properly labeled and maintained stock solutions are crucial for efficiency and accuracy in any laboratory setting. They allow for fast, accurate dilution and minimize the risk of errors in solution preparation.
Nitric Acid (HNO3)
Nitric Acid (HNO3) is a strong, highly corrosive mineral acid used in a wide array of industries and laboratory settings. Due to its reactivity, care must be taken in its handling and use in experiments. It is widely employed in the preparation of fertilizers, explosives, and for various industrial processes. In laboratory practices, HNO3 is often kept as a stock solution, which can be diluted as needed for specific reactions or tests. The handling of nitric acid requires consideration of safety due to its potential hazards:
  • It's corrosive to skin and metals.
  • It can cause severe burns.
  • Inhalation can cause respiratory irritation.
Proper storage and protective measures such as gloves and safety goggles are necessary when working with HNO3. Despite its dangers, it is an essential reagent that must be well managed to safely benefit from its chemical properties.

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Most popular questions from this chapter

Classify the following redox reactions as combination. decomposition, or displacement: (a) \(2 \mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\) (b) \(\mathrm{Mg}+2 \mathrm{AgNO}_{3} \longrightarrow \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{Ag}\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{2} \longrightarrow \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{H}_{2}+\mathrm{Br}_{2} \longrightarrow 2 \mathrm{HBr}\)

If \(30.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{CaCl}_{2}\) is added to \(15.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{AgNO}_{3},\) what is the mass in grams of \(\mathrm{AgCl}\) precipitate?

Acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right)\) is a monoprotic acid commonly known as "aspirin." A typical aspirin tablet, however, contains only a small amount of the acid. In an experiment to determine its composition, an aspirin tablet was crushed and dissolved in water. It took \(12.25 \mathrm{~mL}\) of \(0.1466 \mathrm{M} \mathrm{NaOH}\) to neutralize the solution. Calculate the number of grains of aspirin in the tablet (one grain \(=0.0648 \mathrm{~g}\) ).

Acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) is an important ingredient of vinegar. A sample of \(50.0 \mathrm{~mL}\) of a commercial vinegar is titrated against a \(1.00 \mathrm{M} \mathrm{NaOH}\) solution. What is the concentration (in \(M\) ) of acetic acid present in the vinegar if \(5.75 \mathrm{~mL}\) of the base is needed for the titration?

Give a chemical explanation for each of the following: (a) When calcium metal is added to a sulfuric acid solution, hydrogen gas is generated. After a few minutes, the reaction slows down and eventually stops even though none of the reactants is used up. Explain. (b) In the activity series, aluminum is above hydrogen, yet the metal appears to be unreactive toward hydrochloric acid. Why? (Hint: Al forms an oxide, \(\mathrm{Al}_{2} \mathrm{O}_{3},\) on the surface.) (c) Sodium and potassium lie above copper in the activity series. Explain why \(\mathrm{Cu}^{2+}\) ions in a \(\mathrm{CuSO}_{4}\) solution are not converted to metallic copper upon the addition of these metals. (d) A metal M reacts slowly with steam. There is no visible change when it is placed in a pale green iron(II) sulfate solution. Where should we place \(\mathrm{M}\) in the activity series? (e) Before aluminum metal was obtained by electrolysis, it was produced by reducing its chloride \(\left(\mathrm{AlCl}_{3}\right)\) with an active metal. What metals would you use to produce aluminum in that way?

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