How to use (M1V1=M2V2) formula for dilution in chemistry?

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What does (M1V1 = M2V2) mean and where it is used in chemistry?

The equation (M1V1 = M2V2) is used to solve the problems related to dilution in chemistry where –

  • M1 represents the molarity of an initial concentrated solution.
  •  V1 represents the volume of the initial concentrated solution.
  • M2 represents the molarity of the final diluted solution.
  • V2 represents the volume of the final diluted solution.

m1v1 = m2v2, dilution equation

Dilution: It is a process in which the concentration of a solute in the solution is decreased by adding more solvent to the solution. In the dilution process, the concentration of the solution decreases while the volume of the whole solution increased.

The amount of dilute remains constant in the dilution process. (the number of moles is the same before and after dilution).

Why is the (M1V1 = M2V2) formula used in the dilution problem, explain?

To understand this equation, we have to focus on the definition of Molarity. “Molarity is defined as the moles of a solute per liters of a solution”.

∴ Molarity = Moles of solute/liters of solution

We can also write it as –

⇒ Moles of solute = (Molarity) × (Liters of solution)

∴ Moles of solute = M × V      [∴ M represents molarity and V represents Volume]

In dilution process, the number of moles is the same before and after dilution. That means (M×V) remains same before and after concentration change.

So, according to this, we can write the initial and final conditions as –

M1V1 = M2V2

The above equation represents the dilution equation.

Does M1V1 = M2V2 have to be in liters?

The short answer, it doesn’t matter whether should be M1V1 or M2V2 in liters or not.

What’s really matter is –

  • The initial and final concentrations must have the same units.
  • If your initial concentration is in moles per liter and final concentration should also have the same unit(moles per liter).
  • Also, if the concentration is in moles per liter, then, it doesn’t necessary that the volume should also be in liters. But the thing is, the initial volume and final volume must have the same unit.
  • Let’s say if V1 is in milliliter then V2 should also be in milliliter, and vice versa, if the given volume is in liter then another volume should also be in liter.

How to solve dilution problems using (M1V1 = M2V2)?

It’s really easy, all you have to do is rearrange the formula(M1V1=M2V2) according to the question needs and find the value of the unknown one.

In most of the dilution problems, you have to ask to find either the concentration or volume in either the initial side or final side of the equation.

For example, If in a given problem, the value of M1 is x, for M2, it is y, and for V2, it is z. Now find the value of V1 =?

∴ We have, M1V1 = M2V2

Put given values in the above equation-

⇒ xV1 = yz

∴ V1 = yz/x       Answer

That’s how simple it is to solve the problem of dilution using the equation M1V1 = M2V2.

Examples of dilution problems using (M1V1 = M2V2)?

Let’s take some examples to understand the process of solving dilution problems using the equation (M1V1 = M2V2).

Q1. A solution having a molarity of 6 M and a volume of 4 liters, after the dilution, the volume of the solution becomes 8 liters, then, calculate the molarity of the dilute solution?

Answer – In the given problem, the initial molarity is 6 M, and the initial volume is 4 liters.

So, we have, M1 = 6 and V1 = 4

Also, it is given, that after the dilution, the volume of the solution becomes 8 liters.

So, V2 = 8

Now we have to find the molarity of the dilute solution which means M2 =?

Use the equation, M1V1 = M2V2

∴ M2 = M1V1/V2

= (6 × 4)/8

= 3

∴ So, the molarity of dilute solution (M2) is 3.

Q2. A solution having a molarity of 8 M and a volume of 4 liters. If 2 liters of solvent are added to the solution, calculate the molarity of the final solution?

Answer = Given, M1 = 8, V1 = 4

“If 2 liters of solvent are added to solution”

∴ V2 = V1 + 2

= 4 + 2

V2 = 6 liters

Now we have to find M2.

Use the equation, M1V1 = M2V2

∴ M2 = M1V1/V2

= (8 × 4)/6

= 16/3

∴ M2 = 5.33 

Q3. How many mL of a 2.50 M NaOH solution are required to make 525 mL of a 0.150 M NaOH solution?

Answer –

In given problem, we have given, M1 = 2.50, M2 = 0.150, V2 = 525

We have to find, V1 =?

Use the equation, M1V1 = M2V2

∴ V1 = M2V2/M1

= (0.150 × 525)/2.50

= 31.5

∴ V1 = 31.5 mL

Q4. If 45.0 mL of a 6.00 M HCl solution is diluted to a final volume of 0.250 L, what is the final concentration? 

Answer –

Given, M1 = 6, V1 = 45 mL

Also, the final volume (V2) is 0.250 L, we have to find, M2 =?

We have one problem here, in the given question, V1 is in milliliter and V2 is in Liter, so, both have different units.

According to the dilution equation concept –

The initial and final volume must have same units.

For example – If V1 is in liter then V2 must also be in liters.

So, what we have to do is convert the milliliter unit to liter.

V1 = 45 mL = 0.045 Liter

Now put these values in the dilution equation-

⇒ M1V1 = M2V2

∴ M2 = M1V1/V2

= (6 × 0.045)/0.250

= 1.08

∴ M2 = 1.08 M

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