Solution Stoichiometry
Key Words
- solution, solute, and solvent
- solubility
- concentration
Skills to Develop
- Expressing concentrations
- Converting concentrations between various units: g/L, percent, weight percentage, mole percent, mol/L (M), ppm, etc.
- Calculating amounts of solute in a given volume of solution
Amount = Concentration * Volume
Applying this formula to solve titration problems
- Preparing a solution of prescribed concentration
- Solving any problem involving solution stoichiometry
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 of solid solutions are emphasized in material sciences.
Solution Concentration
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 liter (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
In a laboratory, Qem dissolved 11.2 g of sugar in water, and then he poured the solution into a 250-mL volume flask. He added enough water to make the solution exactly 250.00 mL. What is the concentration of the solution he has prepared?
Hint -
The molecular weight for sugar \(\ce{C12H22O11}\) is 342.0 g/mol. (Your should have the skill to calculate molecular weight.) For solving chemical problems, the units mol/L are the most useful. Thus, the concentration is:
\(\mathrm{11.2\: g\times\dfrac{1\: mol}{342.0\: g}\times\dfrac{1}{0.250\: L}= 0.131\: mol/L\: (M)}\)
The 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
If the 0.131-M sugar solution has a density of 1.10 g/mL, what is the concentration in mole percentage?
Hint -
The total weight of the solution is \(\mathrm{1.10\times250 = 275\: g}\).
\(\mathrm{Amount\: of\: sugar = \dfrac{0.131\: mol/L}{0.250\: L} = 0.0327\: mol}\)
\(\mathrm{Amount\: of\: water = \dfrac{(275 - 11.2)\,g}{18\: g/mol} = 14.7\: mol}\)
Thus,
\(\mathrm{Mole\: percentage = \dfrac{0.0327}{14.7 + 0.0327}\times 100 = 0.22 \%}\)
At 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
A well known fact is that at standard temperature and pressure, one mole of ideal gas occupies 22.4 L. What is the concentration in M of this ideal gas?
Hint -
\(\mathrm{Concentration =\dfrac{1\: mol}{22.4\: L}= 0.0446\: M}\)
Exercise
For a gas at 273 K, what is the pressure if the concentration is 1.0 M?
Solution Stoichiometry
The amount of solute in a certain volume of solution is equal to the volume (V) multiplied by the concentration (C).
\(\mathrm{Amount = C \times V}\)
If the units are included as part of your formulation or calculation, you can derive the correct unit to express the amount.
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
How much \(\ce{NaOH}\) (formula weight = 40) is contained in 25.0 mL of a solution whose concentration is 0.1234 M (same as mol/L)?
Hint -
\(\mathrm{Amount = 0.025\: L \times 0.1234\: mol/L = 3.085\times10^{-3}\: mol= 0.1234\: g}\)
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 \(\ce{NaOH}\) is there in 0.1 mL of the diluted solution?
In problems involving neutralization such as the reaction equation
\(\mathrm{2 HCl + Ca(OH)_2 \rightarrow CaCl_2 + 2 H_2O}\),
one mole of \(\ce{HCl}\) reacts with half a mole of \(\ce{Ca(OH)2}\). Half mole of \(\ce{Ca(OH)2}\) is the equivalence to one mole of \(\ce{HCl}\) .
Similarly for the reaction involving oxidation reactions,
\(\mathrm{KMnO_4 + 5 FeCl_2 + 8 HCl \rightarrow KCl + MnCl_2 + 5 FeCl_3 + 4 H_2O}\),
one mole of \(\ce{KMnO4}\) is equivalent to 5 moles of \(\ce{FeCl2}\). This reaction equation shows the requirement of 8 moles of \(\ce{HCl}\), but in reality, \(\ce{KMnO4}\) will oxidize \(\ce{Cl-}\) into chlorine, \(\ce{Cl2}\). We used \(\ce{HCl}\) for simplicity in balancing the equation.
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.
\(Amount_1 = C_1 V_1\)
\(Amount_2 = C_2 V_2\)
If Amount_{1} is equivalent to Amount_{2}, then we have
\(C_1 V_1 = C_2 V_2\)
If Amount_{1} is equivalent to n Amount_{2}, then we have
\(C_1 V_1 = n C_2 V_2\)
These equations are useful for titration calculations. Since their derivation is so trivial, you are expected to derive them whenever required. The two examples below show the application of the above discussion.
Example 5
If 25.00 mL \(\ce{HCl}\) acid with a concentration of 0.1234 M is neutralized by 23.45 mL of \(\ce{NaOH}\), what is the concentration of the base?
Hint -
\(\mathrm{[NaOH] = \dfrac{0.02500\: L \times 0.1234\: mol/L}{0.02345\:L} = 0.1316\: M}\)
Skill
This type of calculation is used for titrations, an important technique in a laboratory.
Example 6
If 25.00 mL \(\ce{HCl}\) acid with a concentration of 0.1234 M is neutralized by 23.45 mL of \(\ce{Ca(OH)2}\), what is the concentration of the base?
Hint -
\(\mathrm{[Ca(OH)_2] = \dfrac{0.5 \times 0.02500\: L \times 0.1234\: mol/L} {0.02345\:L} = 0.06508\: M}\)
This concentration is half of that of \(\ce{NaOH}\) .
Discussion
\(\mathrm{[OH] = 0.1316\: M}\) in both cases of Examples 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.
Confidence Building Problems
- Exercise -
What is the concentration if the water evaporated, and the volume is reduced to 1.00 L?
Contributors
Chung (Peter) Chieh (Chemistry, University of Waterloo)