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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

Expert verified
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.

Step by step solution

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.

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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

Calculate the molarity of each of the following solutions: (a) \(29.0 \mathrm{~g}\) of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) in \(545 \mathrm{~mL}\) of solution, (b) \(15.4 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) in \(74.0 \mathrm{~mL}\) of solution, (c) \(9.00 \mathrm{~g}\) of sodium chloride \((\mathrm{NaCl})\) in \(86.4 \mathrm{~mL}\) of solution.

(a) Determine the chloride ion concentration in each of the following solutions: \(0.150 \mathrm{M} \mathrm{BaCl}_{2}, 0.566 \mathrm{M} \mathrm{NaCl}\), \(1.202 \mathrm{M} \mathrm{AlCl}_{3}\) (b) What is the concentration of a \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\) solution that is \(2.55 \mathrm{M}\) in nitrate ion?

Sulfites (compounds containing the \(\mathrm{SO}_{3}^{2-}\) ions) are used as preservatives in dried fruits and vegetables and in wine making. In an experiment to test for the presence of sulfite in fruit, a student first soaked several dried apricots in water overnight and then filtered the solution to remove all solid particles. She then treated the solution with hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) to oxidize the sulfite ions to sulfate ions. Finally, the sulfate ions were precipitated by treating the solution with a few drops of a barium chloride \(\left(\mathrm{BaCl}_{2}\right)\) solution. Write a balanced equation for each of the preceding steps.

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) \(\mathrm{H}_{2}+\mathrm{Br}_{2} \longrightarrow 2 \mathrm{HBr}\)

The following "cycle of copper" experiment is performed in some general chemistry laboratories. The series of reactions starts with copper and ends with metallic copper. The steps are as follows: (1) A piece of copper wire of known mass is allowed to react with concentrated nitric acid [the products are copper(II) nitrate, nitrogen dioxide, and water]. (2) The copper(II) nitrate is treated with a sodium hydroxide solution to form copper(II) hydroxide precipitate. (3) On heating, copper(II) hydroxide decomposes to yield copper(II) oxide. (4) The copper(II) oxide is combined with concentrated sulfuric acid to yield copper(II) sulfate. (5) Copper(II) sulfate is treated with an excess of zinc metal to form metallic copper. (6) The remaining zinc metal is removed by treatment with hydrochloric acid, and metallic copper is filtered, dried, and weighed. (a) Write a balanced equation for each step and classify the reactions. (b) Assuming that a student started with \(65.6 \mathrm{~g}\) of copper, calculate the theoretical yield at each step. (c) Considering the nature of the steps, comment on why it is possible to recover most of the copper used at the start.

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