Question

Calculate the concentration of each of the following diluted solutions: a. \(1.0 \mathrm{~L}\) of a \(4.0 \mathrm{M} \mathrm{HNO}_{3}\) solution is added to water so that the final volume is \(8.0 \mathrm{~L}\). b. Water is added to \(0.25 \mathrm{~L}\) of a \(6.0 \mathrm{M}\) NaF solution to make 2.0 L of a diluted NaF solution. c. A \(50.0-\mathrm{mL}\) sample of an \(8.0 \%(\mathrm{~m} / \mathrm{v}) \mathrm{KBr}\) solution is diluted with water so that the final volume is \(200.0 \mathrm{~mL}\). d. A \(5.0\) -mL sample of a \(50.0 \%\) (m/v) acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) solution is added to water to give a final volume of \(25 \mathrm{~mL}\).

Short answer

a. 0.50 M, b. 0.75 M, c. 2.0% (m/v), d. 10.0% (m/v)

Step-by-step solution

Title - Determine the formula

The dilution formula to use is: \( C_1 V_1 = C_2 V_2 \), where \(C_1\) is the initial concentration, \(V_1\) is the initial volume, \(C_2\) is the final concentration, and \(V_2\) is the final volume.

Title - Calculate the concentration for part a

Using \( C_1 = 4.0 \,M \), \( V_1 = 1.0 \,L \), and \( V_2 = 8.0 \,L \):\( C_2 = \frac{C_1 V_1}{V_2} = \frac{4.0 \,M \times 1.0 \,L}{8.0 \,L} = 0.50 \,M \).

Title - Calculate the concentration for part b

Using \( C_1 = 6.0 \,M \), \( V_1 = 0.25 \,L \), and \( V_2 = 2.0 \,L \):\( C_2 = \frac{C_1 V_1}{V_2} = \frac{6.0 \,M \times 0.25 \,L}{2.0 \,L} = 0.75 \,M \).

Title - Calculate the concentration for part c

Using \( C_1 = 8.0\text{\text{\textbackslash(\textbackslash)% (m/v)} } \), \( V_1 = 50.0 \,mL \), and \( V_2 = 200.0 \,mL \):\( C_2 = \frac{C_1 V_1}{V_2} = \frac{8.0\text{\text{\textbackslash(\textbackslash)% (m/v)} } \times 50.0 \,mL}{200.0 \,mL} = 2.0\text{\text{\textbackslash(\textbackslash)% (m/v)} } \).

Title - Calculate the concentration for part d

Using \( C_1 = 50.0\text{\text{\textbackslash(\textbackslash)% (m/v)} } \), \( V_1 = 5.0 \,mL \), and \( V_2 = 25 \,mL \):\( C_2 = \frac{C_1 V_1}{V_2} = \frac{50.0\text{\text{\textbackslash(\textbackslash)% (m/v)} } \times 5.0 \,mL}{25 \,mL} = 10.0\text{\text{\textbackslash(\textbackslash)% (m/v)} } \).

Key concepts

Concentration

Concentration tells us how much solute is in a given amount of solvent or solution. It's like counting candies in a jar - a higher concentration means more candies! Depending on what you're measuring, concentration can be expressed in different ways:

• Molarity (M) - This is moles of solute per liter of solution. For example, a 2M NaCl solution has 2 moles of salt in one liter of water.
• Mass/Volume Percentage (% m/v) - This shows how much mass of solute is in a certain volume of solution. For instance, an 8% (m/v) solution has 8 grams of solute in 100 milliliters of solution.

Understanding concentration is essential because it helps in preparing solutions and performing chemical reactions correctly. For example, if you're working with acids or bases in a lab, knowing the concentration ensures safety and accuracy.

Solution Dilution

Diluting a solution means you’re adding more solvent to lower the concentration of the solute. Think of it like adding more water to your lemonade to make it less sweet.
The key formula for dilution is:

\[ C_1 V_1 = C_2 V_2 \]
Where:

• \( C_1 \) is the initial concentration
• \( V_1 \) is the initial volume
• \( C_2 \) is the final concentration
• \( V_2 \) is the final volume

Let's break it down using an example. If you have 1 liter of a 4M HNO₃ solution and dilute it to 8 liters, the concentration changes. Using the formula:
\( C_2 = \frac{C_1 \times V_1}{V_2} = \frac{4.0 \ M \times 1.0 \ L}{8.0 \ L} = 0.50 \ M \)
So, the new concentration is 0.50M. This formula helps in adjusting the solution’s strength precisely.

Molarity

Molarity is a common way to express the concentration of a solution. It is represented by the symbol 'M' and defined as the number of moles of solute per liter of solution.

Formula for calculating molarity is:

\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
Molarity makes it easy to relate concentration to the number of particles. For example, in part b of the exercise, if you dilute 0.25 liters of a 6M NaF solution to 2 liters, here’s the calculation:

Using the dilution formula:

\[ M_2 = \frac{M_1 \times V_1}{V_2} = \frac{6.0 \ M \times 0.25 \ L}{2.0 \ L} = 0.75 \ M \]
The new concentration is 0.75M. Molarity is used widely in chemistry because it directly relates to the molecular scale of chemical reactions.

Mass/Volume Percentage

Mass/Volume Percentage (m/v%) is another way to describe solution concentration. It shows the mass of the solute in grams per 100 milliliters of the solution.
Here’s the formula:

\[ \text{(m/v)%} = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100 \]
For instance, in part c of the exercise, a 50mL sample of an 8% (m/v) KBr solution is diluted to 200mL. To find the new concentration after dilution, you apply the dilution formula:

\[ C_2 = \frac{C_1 V_1}{V_2} = \frac{8.0 \%(m/v) \times 50.0 \ mL}{200.0 \ mL} = 2.0 \%(m/v) \]
The new concentration of potassium bromide is 2.0% (m/v). This method is particularly useful in medicine and biology where small concentrations are crucial.

Seeing these calculations side-by-side helps in understanding how different concentration units operate and why they are essential in various scientific fields.