Lesson 4.7 Working With Solutions
Suggested Reading
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Zumdahl Chapter 4 Section 4.3
Essential Question
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What calculations are needed to make solutions in the lab?
Learning Objectives
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Describe the molarity of solutions.
Perform calculations needed to make solutions.
Solve problems involving concentration, amount of solute and
volume of solution.
Reactions between two solid reactants often proceed very slowly or not at
all. This is because the molecules or ions in a crystal tend to occupy
approximately fixed positions, so that the chance of two molecules or ions
coming together to react is small. For this reason, most reactions are run
in liquid solutions. Reactant molecules are free to move throughout the
liquid, and reaction is much faster. When you run reactions in liquid
solutions, it is convenient to dispense the amounts of reactants by
measuring out volumes of reactant solutions. In this lesson, we will discuss
calculations involved in making solutions. Pay attention because moving
forward you will be preparing your own solutions for use in your lab
work when we are in need of solutions.
Molar Concentration
When we dissolve a substance in a liquid, we call the substance the solute
and the liquid the solvent. For example, ammonia gas dissolves readily in
water, and aqueous ammonia solutions are often used in the laboratory. In
such solutions, ammonia gas is the solute and water is the solvent.
The general term concentration refers to the quantity of solute in a
standard quantity of solution. Qualitatively, we say that a solution is dilute
when the solute concentration is low and concentrated when the solute
concentration is high. Usually these terms are used in a comparative
sense and do not refer to a specific concentration. We say that one
solution is more dilute or less concentrated than another. However, for
commercially available solution, the term concentrated refers to the
maximum concentration available. For example, concentrated aqueous
ammonia contains about 28% NH3 by mass. In this example, the
concentration of NH3 is expressed as the mass percent of solute (the mass
of solute in 100g of solution). In the lab we usually express concentration
in terms of molar concentration, or molarity (M), which is defined as the
moles of solute dissolved in 1 liter (cubic decimeter) of solution. The massmole relationship makes molarity more convenient for working with
laboratory reagents.
An aqueous solution that is 0.15 M NH3 (read this as "0.15 molar NH3")
contains 0.15 mole NH3 per liter of solution. If you want to prepare a
solution that is, for example, 0.200 M CuSO4, you place 0.200 mol of
CuSO4 in a 1.000 L volumetric flask, or a proportional amount in a flask of
a different size. You add a small quantity of water to dissolve the CuSO4.
Then you fill the flask with with additional water to the mark on the neck
and mix the solution. I typically pause and mix the solution when I have
filled the flask half-way. As the solute dissolves it expands and occupies
space in the solution. If you fill the volumetric flask to the mark before this
expansion occurs you can end up with too much solvent and the
concentration may be inaccurate as a result.
Example: Calculating Molarity from Mass and Volume
A sample of NaNO3, weighing 0.38 g is placed in a 50.0 mL volumetric
flask. The flask is then filled with water to the mark of the neck. What is the
molarity of the resulting solution?
Solution:
To calculate the molarity, you need the moles of solute. Therefore, you
first covert grams NaNO3 to moles. The molarity equals the moles of solute
divided by the liters of solution.
The last significant figure is underlined. The molarity is
One of the advantages of molarity as a concentration unit is that, for a
solution of known molarity, the amount of solute is related to the volume of
solution. Rather than having to weigh out a specified mass of substance,
you can instead measure out a definite volume of solution of the
substance, which is usually much easier.
As the next example illustrates, molarity can be used as a factor for
converting from moles of solute to liters of solution, and vice versa.
Example: Using Molarity as a Conversion Factor
An experiment calls for the addition to a reaction vessel of 0.184 g of
sodium hydroxide, NaOH, in aqueous solution. How many millimeters of
0.150 M NaOH should be added?
Solution:
First you need to convert grams NaOH to moles NaOH, because molarity
relates moles of solute to volume of solution. Then, you convert moles of
NaOH to liters of solution, using molarity as a conversion factor. Here,
0.150 M means that 1 L of solution contains 0.150 moles of solute, so the
conversion factor is:
Here is the calculation
You need to add 30.7 mL of a 0.150 M NaOH to the reaction vessel.
You can see from the example that problems like these are solved like all
other stoichiometry problems where dimensional analysis is critical to
achieving the correct answer. Let the units guide you to the solution!
Diluting Solutions
Commercially available aqueous ammonia (28.0% NH3) is 14.8 M NH3.
Suppose that you want as solution that is 1.00 M NH3. You need to dilute
the concentrated solution with a definite quantity of water. For this
purpose, you must know the relationship between the molarity of the
solution before dilution (the initial molarity) and that after dilution (the final
molarity).
To obtain this relationship, first recall the equation defining molarity:
You can rearrange this to give
Moles solute = molarity x liters of solution
The product of molarity and the volume (in liters) gives the moles of solute
in the solution. Writing Mi for the initial molar concentration and Vi for the
initial volume of solution, you get
Moles solute = Mi x Vi
When the solution is diluted by adding water, the concentration and
volume change to Mf and Vf giving
Moles solute = Mf x Vf
Because the moles of solute has not changed during the dilution
Mi x Vi = Mf x Vf
This is a very useful equation! It is among the chemistry equations that I
use most often. The next example illustrates how you can use this formula
to find the volume of a concentrated solution needed to prepare a given
volume of dilute solution.
Example: Diluting Solution
You are given a solution of 14.8 M NH3. How many milliliters of this
solution do you require to give 100.0 mL of 1.00 M NH3 when diluted.
Solution:
You know the final volume (100.0 mL), final concentration (1.00 M) and
initial concentration (14.8 M) Write the dilution formula and rearrange it to
give the initial volume.
Now substitute the known values into the equation
This result mean that you need to add 6.76 mL to a 100 mL volumetric
flask and then fill the the mark with distilled water, pausing half way to mix.
HOMEWORK: book questions page 171 questions 17, 18, 21, 23, 25,
29