How would you prepare a \(6.00 M \mathrm{HNO}_{3}\) solution if only \(3.00 \mathrm{M}\) and \(12.0 \mathrm{M}\) solutions of the acid are available for mixing?

Short Answer

Expert verified
Mix the two solutions in a 2:1 volume ratio of the 3.00 M to the 12.0 M solution to achieve the final concentration of 6.00 M HNO3. That is, for every 2 volumes of the 3.00 M solution, use 1 volume of the 12.0 M solution.

Step by step solution

01

Identify the target concentration desired

We want to prepare a solution with a concentration of 6.00 M of \(\mathrm{HNO}_{3}\). This is the target concentration for the final solution.
02

Apply the principle of mixtures

We use the equation \(C_1V_1 + C_2V_2 = C_fV_f\), where \(C_1\) and \(C_2\) are the concentrations of the original solutions, \(V_1\) and \(V_2\) are the volumes of the original solutions that will be mixed, and \(C_f\) and \(V_f\) are the concentration and volume of the final solution respectively. In this case, \(C_1 = 3.00\ M\), \(C_2 = 12.0\ M\), and \(C_f = 6.00\ M\). We want to find values for \(V_1\) and \(V_2\).
03

Set up the equation for the final volume

Assuming we want to create \(V_f\) liters of the final solution, the equation becomes \(3.00\mathrm{M} \times V_1 + 12.0\mathrm{M} \times V_2 = 6.00\mathrm{M} \times V_f\).
04

Derive a relationship between \(V_1\) and \(V_2\)

To simplify the calculation, we can express one variable in terms of the other. Because the volumes are additive, we can state \(V_1 + V_2 = V_f\). For every volume part of the 12.0 M solution used, we need to use an equal total volume part minus the volume of the 12.0 M solution part of the 3.00 M to keep the volume consistent. We can then express \(V_2\) as \(V_f - V_1\).
05

Substitute \(V_2\) in the initial equation and solve for \(V_1\)

Substituting the expression for \(V_2\) into the equation from Step 3, we get \(3.00\mathrm{M} \times V_1 + 12.0\mathrm{M} \times (V_f - V_1) = 6.00\mathrm{M} \times V_f\). Rearrange and solve this equation to find the relationship between \(V_1\) and \(V_f\).
06

Calculate the final volumes of each solution

After simplifying the equation, we can find a specific ratio in which to mix the volumes of the two original solutions to achieve the desired concentration.

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Key Concepts

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

Concentration of Solutions
Understanding the concentration of solutions is a cornerstone in chemistry, especially when preparing a specific solution from different concentrations. The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent or solution. Imagine if you have a glass of lemonade: the sugar and lemon juice mixed with the water determine the concentration - more sugar means a sweeter (more concentrated) drink.

When a student encounters a problem related to preparing solutions, knowing the concentration concept is crucial. Concentration can be expressed in various ways, such as molarity, molality, or parts per million (ppm), with molarity being the most commonly used in academic settings. By adjusting the concentration, students can learn to control reactions and understand their mechanisms, which is a valuable skill in laboratory and industrial settings.
Molarity
Molarity is a term that comes up frequently in chemistry and is symbolized by 'M'. It's defined as the number of moles of solute per liter of solution. The moles of a substance are a measurement of particles, often molecules or atoms. If you have a 1.00 M sugar solution, this means you have 1 mole of sugar dissolved in 1 liter of water. It's like saying you have a certain number of sugar packets per jug of water — a way of quantifying the 'strength' of your solution.

To calculate molarity, you use the formula:
\( M = \frac{{moles\text{{ of solute}}}}{{volume\text{{ of solution in liters}}}} \). This formula embodies the heart of preparing solutions in chemistry, allowing students to calculate the exact amount to mix to achieve a desired concentration. Remember, accurate measurements are key to getting reliable results in any experiment!
Principle of Mixtures
The principle of mixtures plays a crucial role when combining two solutions to achieve a target molarity. It's based on the concept that when you combine solutions, their volumes are additive, and the amount of substance in the combined volume is the sum from each original solution.

Understanding Volume Addition

When you pour 50 mL of water into a 100 mL of alcohol, you don't get 150 mL of solution due to volume contraction, but when solving chemistry problems, we assume volumes are additive for simplicity. In the exercise given, we assume that mixing 3.00 M and 12.0 M \( \text{{HNO}}_3 \) solutions will give us a simple sum of their volumes.

Applying the Equation

By understanding this principle and applying the equation \( C_1V_1 + C_2V_2 = C_fV_f \), students can calculate the correct proportions of each solution to mix to get the desired concentration. It's like following a recipe to get the perfect flavor - you need to know the right amount of each ingredient to add. Just as a chef uses the principle of mixtures to combine flavors, chemists use it to create solutions of needed strength and volume for their experiments.

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Most popular questions from this chapter

For each pairing below, predict which will dissolve faster and explain why. (a) a teaspoon of sugar or a large crystal of sugar dissolving in hot coffee (b) \(15.0\) grams of copper(II) sulfate in \(100 \mathrm{~mL}\) of water or in \(100 \mathrm{~mL}\) of a \(15 \%\) copper(II) sulfate solution (c) a packet of artificial sweetener dissolving in a cup of iced tea or a cup of hot tea (d) 20 grams of silver nitrate in \(1.5 \mathrm{~L}\) of water sitting on a table or in \(1.5 \mathrm{~L}\) of water sloshing around in a speeding car

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When \(80.5 \mathrm{~mL}\) of \(0.642 M \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) are mixed with \(44.5 \mathrm{~mL}\) of \(0.743 \mathrm{M} \mathrm{KOH}\), a precipitate of \(\mathrm{Ba}(\mathrm{OH})_{2}\) forms. How many grams of \(\mathrm{Ba}(\mathrm{OH})_{2}\) do you expect?

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