solubility

concentraion

- Expressing concentrations
- Converting concentrations between various units: g/L, percent, weight percentate, mole percent, mol/L (M), ppm, etc.
- Calculating amounts of solute in a given volume of solution
*Amount = Concentration * Volume* - Preparing a solution of prescribed concentration
- Solving any problem involving solution stoichiometry

The topic **solution stoichiometry** deals with quantities
in chemical reactions taking place in solutions. Once you have mastered
this topic, you will be able to prepare solutions of desirable concentrations,
carry out chemical reactions using correct amounts of solutions, predict
amounts produced, and calculate yields. In order build your skills, you are
going to calculate concentrations and amounts of solute present in the
solution.

Solutions are homogeneous mixtures, the major
component being the **solvent**, and minor components being **solute**.

In terms of stoichiometry, we mainly deal with gases and liquid solutions, but properties solid solutions are emphasized in material sciences.

One of the most important properties of a solution is the **concentration**,
which is the amount of solute dissolved in a given volume or weight of
solution. Since there are different ways to represent the quantities,
concentration can be expressed in different units. Weight of solute per unit
volume (of solution), amount in mole per unit volume, weight percentage,
and mole percentage are some of the ways.

For the application in chemical stoichiometry, the unit of moles per litter
(mol/L) is the most convenient, but the real world uses other expressions
for concentration. We need the skill to convert from one type of units to
another, because we are part of the real world. Furthermore, we should
also have the ability to prepare a solution of desirable concentration,
and in this task, we use quantities measured in conventional units
(g, kg, L, m^{3} etc).

In an era of environmental concerns and pollution sensitivity brought
about by sensitive chemical analysis, the units ppm (part per million
1/10^{6}), ppb (part per billion 1/10^{9}, and
ppt (part per trillion 1/10^{12}) are often used.
It is interesting to note that 0.1 percent is the same as 1000 ppm,
but the latter sounds more serious.

**Example 1
**

*
Hint - *

The molecular weight for sugar C_{12}H_{22}O_{11}
is 342.0 g/mol. (Your should have the skill to calculate molecular weight.)
For solving chemical problems, the units mol/L is the most useful.
Thus, the concentration is:

1 mol 1 11.2 g ------- ------- = 0.131 mol/L (M) 342.0 g 0.250 LThe concentration is 0.131 M

**Discussion**

We have to measure the density in order to express the concentration in percentage and other units.

**Example 2
**

*
Hint - *

The total weight of the solution is 1.10*250 = 275 g.

Amount of sugar = (0.131 mol/L)(0.250 L) = 0.0327 mol. (275 - 11.2)g Amount of water = -------------- = 14.7 mol. 18 g/mol 0.0327 Thus, the mole percentage = --------------- x 100 = 0.22 % 14.7 + 0.0327At this level, the concentration can be expressed as 2200 ppm by mole. However, the concentration is 4.1 % by weight or 41000 ppm.

**Exercise**

What is the weight percentage?

**Example 3
**

*
Hint - *

1 mol Concentration = ------ = 0.0446 M. 22.4 L

**Exercise**

For a gas at 273 K, what is the pressure if the concentration is 1.0 M?

The amount of solute in certain volume of solution is equal to the
volume (*V*) multiplied by the concentration (*C*).

Delivering a desirable amount of substance is best done by using volume. For example, when the concentration is 0.1 M, 1.0 mL solution contains 1.0 micromole.

**Example 4
**

*
Hint - *

*Amount* = 0.025 L * 0.1234 mol/L = 3.085x10^{-3} mol

= 0.1234 g

**Exercise**

If 1.000 mL of this solution (0.1234 M) is diluted to 250.0 mL in a volumetric flask, what is the concentration of the final solution? How much NaOH is there in 0.1 mL of the diluted solution.

In problems involving neutralization such as the reaction equation:

Similarly for the reaction involving oxidation reactions,

If the concentrations of the two solutions are
*C*_{1} and *C*_{2}, and the volumes at the
equivalence points are *V*_{1} and *V*_{2}
for solute 1 and 2 respectively, their amounts can be calculated.

**Example 5
**

*
Hint - *

[NaOH] = 0.02500 L * 0.1234 mol/L /0.02345 = 0.1316 M mol

**Skill**

This type of calculation is used for titrations, an important technique in a laboratory.

**Example 6
**

*
Hint - *

[Ca(OH)_{2}] = 0.5 * 0.02500 L * 0.1234 mol/L /0.02345 = 0.06508 M mol

This concentration is half of that of NaOH.

**Discussion**

[OH] = 0.1316 M in both cases of Example 6 and 5.

Solutions are used in the food we eat, the air we breathe, and the medicine we use. Solutions are also everywhere around us in the environment. Thus, there is a need to keep track of concentrations and amounts of key substances in these solutions so that we are at least aware of how much we eat, drink, or take.

© CChieh@UWaterloo.ca