When two or more substances are mixed together in a manner that is homoge-neous and uniform at the molecular level, the mixture is called a The component (usually a liquid) that is present in much larger quantity than the others is called the solvent; the other components are the The concentra-tion of a soluconcentra-tion describes the amount of solute present in a given amount of solution. When a solution is involved in a reaction, the stoichiometric calcula-tions must take into account two quantities not previously discussed: the con-centration of the solution, and its volume.
PREPARATION OF SOLUTIONS
The ways in which chemists most frequently express concentrations involve the mole as the concentration unit (rather than the gram), because reactions are between molecules as the basic entities.
Molarity
A solution that contains one mole of solute per liter of solution is known as a one molar solution; it is abbreviated 1.00 M. In general,
Preparation of
moles of solute molarity of solution = M =
liter of solution
It is simple to prepare solutions of known molarity from solids and non-volatile liquids that can be weighed on an analytical balance, and then dissolved and diluted to a known volume in a volumetric flask. When a reagent-grade sample is accurately weighed and diluted with care, as in the following prob-lem, the resulting solution is said to be standard
PROBLEM:
Prepare 250.0 ml of a M solution, using solid SOLUTION:
From the given volume and concentration you can calculate how many grams of to weigh out:
wt. of needed = (0.2500 (l69.9
liter A mole AgNO
= 5.309 g
Transfer the 5.309 g to a 250.0 ml volumetric flask, dissolve it in some distilled water, then dilute to the mark (see p 86). Shake vigorously to get a uniform solution. Don't add 250.0 ml of water to the weighed sample, because the resulting solution may actually be larger or smaller than 250.0 ml due to interaction of solute and solvent.
Many crystals contain "water of crystallization," which must be included in the weight of the material weighed out for preparing a solution. Allowance is made for this by using the molecular weight of the hydrate in your calculations, not the mole weight of the anhydrous form.
PROBLEM:
Prepare 100.0 ml of M starting with solid • SOLUTION:
From the given volume and concentration of you can calculate the moles of required. Furthermore, the formula shows that 1 mole of
is required per mole of Thus the weight (W) of needed is = (0.1000 ( 0.2000 ( ,
V liter V mole V mole
= 4.992 g needed
Transfer the 4.992 g to a 100 ml volumetric flask, dissolve it in some distilled water, then dilute to the mark. The fact that some of the water in the solution
Stoichlometry III: Calculations Based on Concentrations of Solutions
comes from the weighed sample is irrelevant. The source of the water is never a matter of concern.
Many times it is convenient to obtain a substance from its solution. What volume of solution should you use to get the quantity of solute you want?
PROBLEM:
What volume of 0.250 M will be needed in order to obtain 8.10 g of
SOLUTION:
If V = liters of solution needed, it must supply the number of moles contained in g of
Moles of needed =
mole
Moles of in V liters = (V liters)
= 0.250V moles 0.250V = 0.0500 V = 0.200 liter = 200 ml needed
This problem illustrates the two most common ways of calculating moles of a compound: (a) weight divided by mole weight, and (b) molarity times volume in liters.
Molality
When discussing the colligative properties of a solution (Chapter 21), it is more important to relate the moles of solute to a constant amount of solvent rather than to the volume of the solution, as in the case of molarity. In practice this is accomplished by using a kilogram of solvent instead of a liter of solution as the reference. A solution that contains one mole of solute per kilogram of solvent is known as a one solution; it is abbreviated 1.00 m. In general,
of solution of solute kg or solvent
Because the density of water is approximately 1 the molarities and molalities of water solutions will have about the same value. This will not be true for most other solvents.
Preparation of Solutions
It is simple to prepare solutions of known molality. The volume of the final solution doesn't enter into it at all.
PROBLEM:
Prepare a 2.00 naphthalene solution using 50.0 g as the solvent.
SOLUTION:
You are given the molality of the solution and the weight of the solvent, from which you can find the number of moles of needed.
... moles
kg
x = moles needed
Weight of needed = (0.100 moles 128
mole /
= 12.8 g
To prepare the dissolve 12.8 g in 50.0 g If you knew that the density of is 1.59 you could measure out
1.59 ml
Mole Fraction
Another way of expressing concentrations that is used commonly with gases (see p 162) and colligative properties (see p 328) is mole fraction, which is defined (for a given component) as being the moles of component in question divided by the total moles of all components in solution. For a solution that has three components (A, B, and C), the mole fraction of A is given by
c A moles of A
mole fraction of A = =
moles A + moles B + moles C If you think about it, it's also easy to make a solution of a given mole fraction.
PROBLEM:
Prepare a 0.0348 mole fraction solution of sucrose mole weight = 342 using 100 g (that is, 100 ml) of water.
SOLUTION:
You are given the mole fraction of sucrose and the moles of water
192 Stolchlometry III Calculations Based on Concentrations of Solutions
g = 5 55 moles 18 mole
What you don't know x the moles of sucrose needed By definition, x moles sucrose
mole fraction = 0 0348 =
(5 55 moles + mole sucrose)
(5 55)(0 0348)
x = mole sucrose
Weight of sucrose needed = (0 200 mole sucrose) (342 mole
= 68 4 g sucrose
Prepare the solution by dissolving 68 4 g sucrose 100 g water
Commercial Concentrated Solutions
Many solutions can't be made accurately, or at all, by weighing out the solute and dissolving the proper amount of solvent For pure substances such as HC1 and are gases, and are fuming hygroscopic corrosive liquids, NaOH and KOH avidly absorb water and from the In such circumstances, the customary procedure to purchase the chemicals
the form of extremely concentrated solutions, then dilute them to the desired strength The following problem typical
PROBLEM.
Commercial concentrated labeled as having a density of 1 84 and being 96 0% by weight Calculate the of solution SOLUTION:
a typical conversion problem which we want to go from grams of solution per liter to moles of per liter
960
liter A g 98 1 g liter
- 0
PROBLEM:
What are (a) the and (b) the mole fraction of the commercial solution the previous
Preparation of Solutions
SOLUTION:
(a) To find the molality we need to know, for a given amount of solution, the moles of the kg of If we take a liter of solution, we shall have
1840 g of solution, of which 4.0% is water, so (0.040)( 1840 g) = 74 g Because there are moles of in this liter, we have
(b) To find the mole fraction, we need to know (for a given amount of solution) the moles of and the moles of If we take a liter of solution, we shall have 18.0 moles of and
= moles of mole
Therefore the mole fraction is
_ __
(18.0 moles + (4.1 moles
Note that it is not possible to convert from molarity to molality or mole fraction unless some information about the density or weight composition of the solution is given.
Dilution
One of the most common ways to prepare a solution is to dilute a concentrated solution that has already been prepared. There is a fundamental principle that underlies all dilutions: the number of moles of solute is the same after dilution as before. It is only the moles of solvent that have been changed (increased).
This principle makes dilution calculations simple. If and are the molarities before and after dilution, and and are the initial and final volumes of solution, then
moles of solute before dilution = moles of solute after dilution ,.
The following problem illustrates the use of this equation.
PROBLEM:
What volume of 18.0 is needed for the preparation of 2.00 liters of 3.00