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Solution Stoichiometry

Key words:

solution, solute, and solvent

Skills to develop

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.

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 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, m3 etc).

In an era of environmental concerns and pollution sensitivity brought about by sensitive chemical analysis, the units ppm (part per million 1/106), ppb (part per billion 1/109, and ppt (part per trillion 1/1012) 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 C12H22O11 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 L
The concentration is 0.131 M


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 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

Thus, the mole percentage = --------------- x 100 = 0.22 %
                             14.7 + 0.0327
At this level, the concentration can be expressed as 2200 ppm by mole. However, the concentration is 4.1 % by weight or 41000 ppm.


What is the weight percentage?

Example 3

A well known fact is that at standard temperature and pressure, one mole of ideal gas occupy 22.4 L. What is the concentration in M of this ideal gas?

Hint -

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


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

Solution Stoichiometry

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

Amount = C * 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 NaOH (formula weight = 40) is contained in 25.0 mL of a solution whose concentration is 0.1234 M (same as mol/L).

Hint -

Amount = 0.025 L * 0.1234 mol/L = 3.085x10-3 mol
= 0.1234 g


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:

2 HCl + Ca(OH)2 = CaCl2 + 2 H2O
one mole of HCl reacts with half a mole of Ca(OH)2. Half mole of Ca(OH)2 is the equivalence to one mole of HCl.

Similarly for the reaction involving oxidation reactions,

KMnO4 + 5 FeCl2 + 8 HCl = KCl + MnCl2 + 5 FeCl3 + 4 H2O One mole of KMnO4 is equivalent to 5 moles of FeCl2. This reaction equation shows the requirement of 8 moles of HCl, but in reality, KMnO4 will oxidize Cl- into chlorine, Cl2. We used HCl for simplicity in balancing the equation.

If the concentrations of the two solutions are C1 and C2, and the volumes at the equivalence points are V1 and V2 for solute 1 and 2 respectively, their amounts can be calculated.

Amount1 = C1 V1
Amount2 = C2 V2
If Amount1 is equivalent to Amount2, then we have C1 V1 = C2 V2
If Amount1 is equivalent to n Amount2, then we have C1 V1 = n C2 V2
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 HCl acid with a concentration of 0.1234 M is neutralized by 23.45 mL of NaOH, what is the concentration of the base?

Hint -

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


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

Example 6

If 25.00 mL HCl acid with a concentration of 0.1234 M is neutralized by 23.45 mL of Ca(OH)2, what is the concentration of the base?

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.


[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.

Confidence Building Problems

  1. You dissolved 10.0 g of sugar in 250 mL of water (the volume of a cup of coffee). Calculate the weight percentage of sugar in the solution. Assume the density of solution to be 1.0 g/mL.

    Calculate and estimate concentration.

  2. You dissolved 10.0 g of sugar to make 250 mL of solution. Calculate the concentration in M. (Molar mass of sugar C12H22O11 = 342).
      10 g   1 mol
    -------  -----  =  ? mol/L or M.
    0.250 L  342 g

    Exercise -
    What is the concentration if you dilute the solution to 1.0 L?

  3. You dissolved 13.0 g of NaCl to make 2.00 L of solution. Calculate the molarity. Atomic wt: Na, 23.0; Cl, 35.5.
    13.0 g     1 mol
    ------  ------------- = ? mol/L
    2.00 L  (23.0+35.5) g

    Exercise -
    What is the concentration if the water evaporated, and the volume is reduced to 1.00 L?

  4. You diluted 10.00 mL of 6.0 M H2SO4 to prepare 250.0 mL solution. Calculate the concentration of this solution. Molar mass: H2SO4, 98.0

    Caution -
    Never pour water into concentrated sulfuric acid.

  5. How much calcium carbonate Ca(HCO3)2 is present in 24.0 L of tap water if analysis indicates that the tap water contains 42.0 ppm Ca(HCO3)2? Assume tap water density to be 1.00 g/mL.

    Skill -
    Apply the equation: Amount = C * V.
    Ca(HCO3)2 is present in hard-water.

  6. A stock sulfuric acid solution contains 98.0 % of H2SO4, and its density is 1.840 g/mL. How many mL of this acid is required to prepare 5.0 L of 2.0 M solution? Molar mass: H2SO4, 98.1.
                         98.1 g  100 g stock  1 mL
    5.0*2 = 10 mol H2SO4 ------  -----------  ------  = ? mL.
                         1 mol   98 g H2SO4  1.84 g

    Practical information -
    Stock sulfuric acid is 18.4 M. Calculate it.

  7. A bottle of nitric acid has a density of 1.423 g/mL, and contains 70.9 % HNO3 by weight. What is the molarity? Molar mass: HNO3 = 63.01

    Assume you have 1.00 L solution.

    1000 mL solution 1.423 g   70.9 g HNO3      1 mol
    ---------------- --------- --------------- ------------ = ? mol/L
        1 L          1 mL     100 g solution  63.01 g HNO3

    Skill -
    Many problem solutions require assumptions. Make the appropriate assumption in problem solving.

  8. If you want to make 250.0 mL 0.100 M calcium chloride solution, how many moles of CaCl2 will you need?

    Other units -
    You need 25 mmole or 0.025 mol.

  9. If you want to make 250.0 mL of 0.100 M calcium chloride solution, how many grams of CaCl2 are needed? Molar mass: CaCl2, 111.1 g/mol.
            0.1 mol  111.1 g
    0.250 L -------  -------  =  ? g (CaCl2)
              1 L     1 mol

    Skill -
    Prepare a solution of definite concentration.

  10. What is the molarity of Na+ in a 0.123 M solution of Na2SO4?

    Skill -
    Explain the dissolution as ionization:

    Na2SO4 = 2 Na+ + SO42-

  11. How many mL of 0.200 M aluminum chloride (AlCl3) solution will contain 6.00 millimole of Cl- ions?

    How many mL will contain 6 millimole Al3+ ions?

  12. How many mL of stock HNO3 solution (16.0 M) is required to make 400.0 mL 2.0 M solution?

    Prepare solution of certain cencentration by dilution.

  13. How many mL of 0.321 M HCl solution is required to neutralize 5.0 mL 1.284 M KOH solution?

    Skill -
    Performing calculations of titration results.

  14. If 10.0 mL HNO3 completely neutralize 25.0 mL of 0.351 M KOH, calculate the molarity of HNO3.

    Further consideration -
    What is the answer if sulfuric acid is used?

  15. Given a solution containing 0.242 g of barium chloride, BaCl2, how many mL of 0.0581 M H2SO4 will completely precipitate the barium ions, Ba2+? Molar mass: BaCl2, 208.3

    Skill -
    Performing volumetric analysis.

  16. A 0.5404 g sample containing NaOH and NaCl is dissolved, and titrated with standard 0.1010 M H2SO4. If 14.32 mL of the acid is required, calculate the percentage of NaOH in the sample? Molar mass: NaOH, 40.0.

    Skill -
    Convert mol H2SO4 to mol NaOH to weight NaOH to % NaOH