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
A stock solution is a concentrated solution that can be diluted by adding more solvent, typically water, to achieve a lower concentration. This process is represented by the equation m1v1=m2v2, where m1 and v1 are the molarity and volume before dilution, and m2 and v2 are after dilution. Dilution decreases molarity, transitioning from a concentrated to a dilute solution.
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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A stock solution is a concentrated solution that serves as a starting point for preparing solutions of lower concentration through dilution. It is important because it allows chemists to store a highly concentrated form of a substance, which can then be diluted to the desired concentration for various laboratory experiments. This approach saves time and resources, as you do not need to prepare fresh solutions from scratch each time. By using a stock solution, you can accurately control the concentration of your working solutions by adding a specific amount of solvent, usually water, to reduce the molarity to the required level.
Dilution decreases the concentration of a solution by adding more solvent, typically water, to the original solution. This increases the total volume while keeping the amount of solute constant, resulting in a lower molarity. For example, if you start with a dark purple concentrated solution and add water, the color becomes lighter, indicating a decrease in concentration. The relationship between the initial and final concentrations and volumes is given by the equation , where and are the molarity and volume before dilution, and and are after dilution.
The formula used to calculate dilutions is . Here, and represent the molarity and volume of the original concentrated solution, while and represent the molarity and volume after dilution. To apply this formula, you need to know three of the four variables and solve for the unknown. For example, if you know the initial concentration and volume, and the desired final concentration, you can calculate the final volume needed. Remember that the final volume is the sum of the initial volume plus the volume of solvent added.
The molarity before dilution () is always greater than the molarity after dilution () because dilution involves adding more solvent to the solution without adding more solute. This increases the total volume of the solution, which spreads the same amount of solute over a larger volume, thereby decreasing the concentration. Since molarity is defined as moles of solute per liter of solution, increasing the volume while keeping moles constant results in a lower molarity. This is why the initial concentrated solution has a higher molarity than the diluted solution.
To determine the volume of solvent to add during dilution, you first use the dilution equation to calculate the final volume needed for the desired concentration. Once you have , subtract the initial volume of the stock solution from it. The difference is the volume of solvent to add: . This ensures the final solution has the desired lower concentration.
