A stock solution is a concentrated solution that can be diluted for various laboratory applications. Dilution involves adding more solvent, typically water, to a solution to decrease its concentration. For instance, when a dark purple solution is gradually mixed with water, the color lightens to a fuchsia hue, indicating a reduction in concentration. This visual change exemplifies the dilution process, where the original solution becomes less concentrated as more solvent is introduced. Understanding this concept is crucial in laboratory settings, as it allows for the preparation of solutions with desired concentrations for experiments and analyses.
13. Solutions
Dilutions
13. Solutions
Dilutions: Videos & Practice Problems
Topic summary
Dilutions describe making a solution less concentrated by adding more solvent, usually water. A concentrated stock solution or standard solution is the starting solution, and after dilution the concentration decreases from a larger molarity to a smaller one. In dilution problems, the same compound is present before and after dilution, but its concentration and volume change.
The central relationship is \(M_1V_1=M_2V_2\) . Here, M1 and V1 are the molarity and volume before dilution, while M2 and V2 are after dilution. Because dilution lowers concentration, \(M_1\) is always greater than \(M_2\). The final volume is found from \(V_2=V_1+\text{volume of solvent added}\) .
Understanding dilutions also connects to concentration itself, since molarity is based on moles of solute per liter of solution. For multiple or series dilutions, the same dilution equation is applied step by step until the final concentration is found.
In Dilutions, a solvent (usually water) is added to a concentrated solution.
Concentrated & Diluted Solutions
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Concept
Dilutions
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1mDilutions Video Summary
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Example
Dilutions Example 1
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1mDilutions Example 1 Video Summary
To determine the concentration of solutions based on the number of solute moles represented by spheres, we can calculate the molarity for each solution. Molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters:
Molarity (M) = \(\frac{\text{moles of solute}}{\text{liters of solution}}\)
In this example, we have three solutions:
Solution A: Contains 5 spheres (moles of solute) in 1 liter of solution. Thus, the molarity is:
M_A = \(\frac{5 \text{ moles}}{1 \text{ L}} = 5 \text{ M}\)
Solution B: Contains 3 spheres in 2 liters of solution. Therefore, the molarity is:
M_B = \(\frac{3 \text{ moles}}{2 \text{ L}} = 1.5 \text{ M}\)
Solution C: Contains 6 spheres in 3 liters of solution, giving us:
M_C = \(\frac{6 \text{ moles}}{3 \text{ L}} = 2 \text{ M}\)
Now, to arrange the solutions from least concentrated to most concentrated based on their molarity values, we find:
1. Solution B: 1.5 M
2. Solution C: 2 M
3. Solution A: 5 M
Thus, the order from least concentrated to most concentrated is B, C, and A.
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Concept
Dilutions
Video duration:
58sDilutions Video Summary
Understanding dilution is essential in chemistry, as it allows us to create solutions with lower concentrations from more concentrated ones. The process of dilution can be quantitatively described using the equation:
\( M_1 V_1 = M_2 V_2 \)
In this equation, \( M_1 \) and \( V_1 \) represent the molarity and volume of the solution before dilution, while \( M_2 \) and \( V_2 \) represent the molarity and volume after dilution. It is important to note that \( M_1 \), the molarity of the concentrated solution, is always greater than \( M_2 \), the molarity of the diluted solution.
The final volume after dilution, \( V_2 \), is determined by the initial volume \( V_1 \) plus the volume of solvent added. This relationship can be expressed as:
\( V_2 = V_1 + V_{\text{solvent}} \)
By applying these principles, one can effectively prepare solutions with desired concentrations, which is a fundamental skill in various scientific applications.
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Example
Dilutions Example 2
Video duration:
2mDilutions Example 2 Video Summary
To determine the volume of a concentrated solution needed to prepare a diluted solution, we can apply the dilution equation, which is expressed as:
\( M_1 V_1 = M_2 V_2 \)
In this equation, \( M_1 \) represents the molarity of the concentrated solution, \( V_1 \) is the volume of the concentrated solution we need to find, \( M_2 \) is the molarity of the diluted solution, and \( V_2 \) is the volume of the diluted solution.
In the given problem, we have:
- Concentrated solution: 5.2 M (this is \( M_1 \))
- Diluted solution: 2.7 M (this is \( M_2 \))
- Volume of diluted solution: 3.5 L (this is \( V_2 \))
Since we are dealing with one compound, hydrobromic acid, and two different molarities, this indicates a dilution scenario. To find \( V_1 \), we rearrange the equation:
\( V_1 = \frac{M_2 V_2}{M_1} \)
Substituting the known values into the equation gives:
\( V_1 = \frac{(2.7 \, \text{M})(3.5 \, \text{L})}{5.2 \, \text{M}} \)
Calculating this yields:
\( V_1 = \frac{9.45 \, \text{mol}}{5.2 \, \text{M}} = 1.8173 \, \text{L} \)
To convert liters to milliliters, we use the conversion factor where 1 L = 1000 mL:
\( 1.8173 \, \text{L} \times 1000 \, \text{mL/L} = 1817.3 \, \text{mL} \)
Considering significant figures, since the values 5.2, 3.5, and 2.7 all have two significant figures, we round 1817.3 mL to 1800 mL. Thus, the final answer is:
1800 mL
In summary, when faced with a dilution problem involving a single compound and two molarities, the dilution equation is the key to finding the unknown volume of the concentrated solution needed for preparation.
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Problem
To what final volume would 100 mL of 5.0 M KCl have to be diluted in order to make a solution that is 0.54 M KCl?
A
72 mL
B
289 mL
C
330 mL
D
930 mL
E
1400 mL
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Problem
If 880 mL of water is added to 125.0 mL of a 0.770 M HBrO4 solution what is the resulting molarity?
A
0.096 M
B
0.136 M
C
0.257 M
D
0.892 M
E
1.76 M
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Problem
A student prepared a stock solution by dissolving 25.00 g of NaOH in enough water to make 150.0 mL solution. The student took 20.0 mL of the stock solution and diluted it with enough water to make 250.0 mL solution. Finally taking 75.0 mL of that solution and dissolving it in water to make 500 mL solution. What is the concentration of NaOH for this final solution? (MW of NaOH:40.00 g/mol).
A
0.0500 M
B
0.025 M
C
0.005 M
D
0.500 M
E
0.0100 M
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Dilution in chemistry refers to the process of reducing the concentration of a solute in a solution by adding more solvent, usually water. This results in a less concentrated solution compared to the original. The starting solution, often called a stock or standard solution, is typically concentrated. When solvent is added, the volume increases, but the amount of solute remains the same, leading to a decrease in molarity. This process is important in laboratories to prepare solutions of desired concentrations for experiments.
To calculate the concentration after dilution, you use the dilution equation: . Here, and are the molarity and volume before dilution, and and are after dilution. Since dilution lowers concentration, . The final volume equals the initial volume plus the volume of solvent added.
A stock solution, also called a standard solution, is a concentrated solution prepared in advance and used as the starting point for dilutions. It contains a known concentration of solute. Stock solutions are used because they allow chemists to prepare solutions of various lower concentrations by simply adding solvent. This saves time and ensures accuracy since the concentration of the stock is well-defined. By diluting the stock solution, you can obtain the desired concentration for experiments without having to weigh and dissolve solute each time.
To find the volume of solvent to add during dilution, you first decide the desired final concentration and volume. Using the dilution equation , you can solve for the final volume . Then, subtract the initial volume from to find the volume of solvent to add: . This ensures the final solution has the correct concentration.
The relationship between molarity and volume in dilution is described by the equation . This means the product of the initial molarity and volume equals the product of the final molarity and volume. When you dilute a solution, the volume increases because you add solvent, but the amount of solute stays the same. Therefore, the molarity decreases proportionally. This inverse relationship allows you to calculate any one of the four variables if the other three are known, making it a fundamental concept in preparing solutions of desired concentrations.
