Chemistry 30

2.4 Dilutions of Standard Solutions

Imagine we have a salt water solution with a certain concentration. That means we have a certain amount of salt (a certain mass or a certain number of moles) dissolved in a certain volume of solution. Next we'll dilute this solution - we do that by adding more water, not more salt:

Before Dilution		After Dilution
Solution1		Solution2
M1 = moles1 litre1		M2 = moles2 litre2
rearrange the equations to find moles:
moles1 = M1× litre1		moles2= M2 × litre2

What stayed the same and what changed between the two solutions?

By adding more water, we changed the volume of the solution. Doing so also changed it's concentration. But the number of moles of solute did not change. So,

moles₁ = moles₂

Therefore

M₁ × litre₁ = M₂ × litre₂

Which we will usually express as:

M1V1= M2V2 where M1 and M2 are the concentrations of the original and diluted solutions and V1 and V2 are the volumes of the two solutions

Preparing dilutions is a common activity in the chemistry lab and elsewhere. Once you understand the above relationship, the calculations are easy to do.

Example:

What volume of concentrated sulfuric acid, 18.0 M, is required to prepare 5.00 L of 0.150 M solution by dilution with water?

Solution:

In a dilution question there are 4 variables - M₁, V₁, M₂ and V₂. You will know three of these values and have to calculate the fourth. Organizing your information is the key to correctly answering dilution questions. The most common mistake students make is in incorrectly pairing up concentrations and volumes. Take time to make sure you do not make this mistake.

M1 = 18.0 M		M2= 0.150 M
V1 = ?		V2 = 5.00 L

Set up the formula, and rearrange to solve for the unknown, V₁:

M1V1	=	M2V2
18.0 × V1	=	0.150 × 5.00
V1	=	0.150 × 5.00 18.0
V1	=	0.0417 L
	=	41.7 mL

Standard Solutions

It is often necessary to have a solution whose concentration is very precisely know. Solutions containing a precise mass of solute in a precise volume of solution are called standard solutions.

To prepare a standard solution a piece of lab equipment called a volumetric flask should be used. These flasks range in size from 10 mL to 2000 mL are are carefully calibrated to a single volume. On the narrow stem is a calibration mark. The precise mass of solute is dissolved in a bit of the solvent and this is added to the flask. Then enough solvent is added to the flask until the level reaches the calibration mark.

Scavenger Hunt Find a picture of a volumetric flask.

Movie Link How to prepare a standard solution

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Example:

Describe how you would prepare 500.0 mL of a 0.100 M standard solution of KNO₃

Solution:

We must first determine the mass of KNO₃ that will be needed to prepare 500.0 mL (0.500 L) of a 0.100 M solution. The molar mass of KNO₃ is 101.1 g·mol^-1:

g

=

101.1 g mol

×

0.100mol L

×

0.500 L

=

5.05 g

To prepare the standard solution we will carefully measure out 5.05 g of KNO₃ and dissolve it in some water (perhaps 200 mL of water). We pour this solution into a 500 mL volumetric flask and add just enough water reach the calibration mark on the flask.

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Complete the practice questions before moving on to the next page.