C1V1 = C2V2 is a simple formula, but it has a huge importance for chemistry students and for anyone working in the chemistry laboratory. C1V1 = C2V2 is known as the dilution formula or the dilution equation. It is used for preparing different dilutions of a stock solution.
A stock solution is a concentrated solution prepared by weighing an exact amount of a reagent. Several different dilutions can be prepared from the stock solution by adding the desired solvent in a specific volume.
The dilution formula helps determine the volume of solvent that we need to add in order to prepare a new concentration C2 from the original concentration C1 of the stock solution.
So, in this article, you will learn how to use C1V1 = C2V2 with examples.
To understand this equation, it is helpful to know the following terms:
C1: This represents the concentration of the original, undiluted solution. It is usually measured in units of concentration such as moles per liter (M) or grams per liter (g/L).
V1: This represents the volume of the original, undiluted solution. It is usually measured in units of volume such as liters (L) or milliliters (mL).
C2: This represents the concentration of the diluted solution after it has been mixed with a solvent. It is also usually measured in units of concentration such as moles per liter (M) or grams per liter (g/L).
V2: This represents the volume of the diluted solution after it has been mixed with a solvent. It is also usually measured in units of volume such as liters (L) or milliliters (mL).
What are the units used in the C1V1 = C2V2 equation?
As we already identified, the symbols C1 and C2 represent concentrations, while V1 and V2 represent solution volumes. Therefore, the units of concentration, such as grams per liter (g/L), moles per liter (mol/L) or moles per decimeter cube (mol/dm3), where 1 liter = 1 dm3, are used for C1 and C2. The concentration unit is also sometimes written as M, where M stands for molarity, i.e., moles of solute per litre of solution.
On the other hand, the units for V1 and V2 are those used for representing the volume of a solution, i.e., liter (L), milliliters (mL) or decimeter cubic (cm3) where I dm3 = 1 L = 1000 mL.
What is important to note here is that the concentration and volume units used in the dilution equation (C1V1 = C2V2) must be consistent with each other. If the initial concentration (C1) is expressed in mol/L, then you will get the final concentration (C2) in mol/L as well.
Similarly, the units used for volume must be the same on both sides of the equation. If the final volume V2 of solution required is given in L, then you will determine V1 in L using the dilution equation.
Maintaining consistency in units is a fundamental stoichiometric requirement. The dilution equation obeys the law of conservation, i.e., the total number of moles of solute (the substance being dissolved) is conserved during the dilution process.
Inconsistency in units may lead to an unbalanced equation and, thus, inaccurate results, which must be avoided at all costs.
That being said, let us move ahead and show you how to use C1V1 = C2V2 through the examples given below.
The dilution formula can be rearranged in one of the above forms based on which quantity is unknown.
For example, let’s say you have a solution with a concentration of C1 = 5 M and a volume of V1 = 100 mL. You want to dilute this solution by adding 200 mL of solvent, so the volume of the diluted solution will be V2 = 300 mL (100 mL of the original solution + 200 mL of solvent). Now calculate the concentration of the diluted solution.
To calculate the concentration of the diluted solution (C2), you can plug these values into the dilution equation:
⇒ C1V1 = C2V2
⇒ 5 M * 100 mL = C2 * 300 mL
∴ C2 = 1.67 M
Result: This means that the concentration of the diluted solution is 1.67 M.
Another example is –
You have a solution with a concentration of C1 = 10 g/L and a volume of V1 = 50 mL. You want to dilute this solution by adding 250 mL of solvent, so the volume of the diluted solution will be V2 = 300 mL (50 mL of the original solution + 250 mL of solvent). Now calculate the concentration of the diluted solution.
To calculate the concentration of the diluted solution (C2), you can use the dilution equation:
∴ C1V1 = C2V2
⇒ 10 g/L * 50 mL = C2 * 300 mL
∴ C2 = 1.67 g/L
Result: This means that the concentration of the diluted solution is 1.67 g/L.
More examples on C1V1 = C2V2 equation
You have a solution with a concentration of C1 = 20 g/L and you want to dilute it to a concentration of C2 = 10 g/L. You have a volume of V2 = 200 mL of the diluted solution. Find the volume of the undiluted solution?
To find the volume of the original, undiluted solution (V1), you can rearrange the dilution equation to solve for V1:
∴ C1V1 = C2V2
⇒ V1 = (C2V2)/C1
⇒ V1 = (10 g/L * 200 mL)/20 g/L
∴ V1 = 100 mL
This means that the volume of the original, undiluted solution is 100 mL.
Joe has a solution with a concentration of C1 = 2 g/L and a volume of V1 = 20 L. He adds 30 L of solvent to this solution. What is the final concentration of the solution?
Joe adds 30 L of solvent to the solution. So, the final volume of the solution is V2 = 50 L (20 L of the original solution + 30 L of solvent).
To find the final concentration of the solution, you can use the dilution equation to solve for C2:
∴ C1V1 = C2V2
⇒ C2 = (C1V1)/V2
⇒ C2 = (2 g/L * 20 L)/50 L
∴ C2 = 0.8 g/L
This means that the final concentration of the solution is 0.8 g/L.
What is the concentration of the diluted solution obtained by adding 1000 mL of distilled water to 500 mL of a 3 mol/L KOH solution?
To solve this problem, we can use the dilution formula: C1V1 = C2V2.
We are given that the initial concentration is 3 mol/L and the initial volume is 500 mL. We are also given that we are adding 1000 mL of distilled water to the initial solution, so the final volume is 500 mL + 1000 mL = 1500 mL.
We want to find the concentration of the final, diluted solution, which is represented by C2 in the dilution formula. We can plug in the known values for C1, V1, and V2 into the dilution formula and solve for C2:
∴ C1V1 = C2V2
⇒ (3 mol/L)(500 mL) = C2(1500 mL)
⇒ C2 = (3 mol/L)(500 mL) / 1500 mL
∴ C2 = 1 mol/L
Therefore, the concentration of the diluted solution is 1 mol/L.
Determine the concentration of a solution produced by diluting 8.00 mL of a 0.20 mol/L solution with 12.0 mL of water.
We are given that the initial concentration is 0.20 mol/L and the initial volume is 8.00 mL. We are also given that we are adding 12.0 mL of water to the initial solution, so the final volume is 8.00 mL + 12.0 mL = 20.0 mL.
We want to find the concentration of the final, diluted solution, which is represented by C2 in the dilution formula. We can plug in the known values for C1, V1, and V2 into the dilution formula and solve for C2:
∴ C1V1 = C2V2
⇒ (0.20 mol/L)(8.00 mL) = C2(20.0 mL)
⇒ C2 = (0.20 mol/L)(8.00 mL) / 20.0 mL
∴ C2 = 0.080 mol/L
Therefore, the concentration of the diluted solution is 0.080 mol/L.
FAQ
What is equation C1V1 = C2V2, and where is it used?
The equation C1V1 = C2V2 is known as the dilution formula or the dilution equation. It is used in chemistry and other scientific fields for preparing dilutions by determining the concentration of a solution after it has been mixed with a known volume of a solvent.
What do the symbols C1, V1, C2, and V2 represent in the dilution formula (C1V1 = C2V2)?
C stands for concentration, while V denotes volume. So in the dilution formula;
C1 = initial concentration of the solution
C2 = final concentration of the solution
V1 = volume of C1 required to prepare C2
V2 = final volume of the solution prepared to have concentration C2.
Vishal Goyal is the founder of Topblogtenz, a comprehensive resource for students seeking guidance and support in their chemistry studies. He holds a degree in B.Tech (Chemical Engineering) and has four years of experience as a chemistry tutor. The team at Topblogtenz includes experts like experienced researchers, professors, and educators, with the goal of making complex subjects like chemistry accessible and understandable for all. A passion for sharing knowledge and a love for chemistry and science drives the team behind the website. Let's connect through LinkedIn: https://www.linkedin.com/in/vishal-goyal-2926a122b/
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