Skip to main content
Chemistry LibreTexts

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, 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 \(\ce{C12H22O11}\) 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.


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 \(\mathrm{1.10\times250 = 275\: g}\).

\(\mathrm{Amount\: of\: sugar = \dfrac{0.131\: mol/L}{0.250\: L} = 0.0327\: mol}\)

\(\mathrm{Amount\: of\: water = \dfrac{(275 - 11.2)\,g}{18\: g/mol} = 14.7\: mol}\)


\(\mathrm{Mole\: percentage = \dfrac{0.0327}{14.7 + 0.0327}\times 100 = 0.22 \%}\)

At this level, the concentration can be expressed as 2200 ppm by mole. However, the concentration is 4.1 % by weight or 41000 ppm.


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 -

\(\mathrm{Concentration =\dfrac{1\: mol}{22.4\: L}= 0.0446\: M}\)


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 -

\(\mathrm{Amount = 0.025\: L \times 0.1234\: mol/L = 3.085\times10^{-3}\: mol= 0.1234\: g}\)


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 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.

\(Amount_1 = C_1 V_1\)
\(Amount_2 = C_2 V_2\)

If Amount1 is equivalent to Amount2, then we have

\(C_1 V_1 = C_2 V_2\)

If Amount1 is equivalent to n Amount2, 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 -

\(\mathrm{[NaOH] = \dfrac{0.02500\: L \times 0.1234\: mol/L}{0.02345\:L} = 0.1316\: M}\)


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 \(\ce{NaOH}\) .


\(\mathrm{[OH] = 0.1316\: M}\) 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

  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.

    Hint: \(\mathrm{100 \times \dfrac{10.0}{250 + 10} =\: ?\: \%}\)

    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 \(\ce{C12H22O11=\, 342}\)).

    Hint: 0.12 M

    \(\mathrm{\dfrac{10\: g}{0.250\: L}\times\dfrac{1\: mol}{342\: g}=\: ?\: mol/L\: or\: M}\)

    Exercise -

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

  3. You dissolved 13.0 g of \(\ce{NaCl}\) to make 2.00 L of solution. Calculate the molarity. Atomic wt: \(\mathrm{Na = 23.0}\)\(\mathrm{Cl = 35.5}\).

    Hint: 0.111 M

    \(\mathrm{\dfrac{13.0\: g}{2.00\: L}\times\dfrac{1\: mol}{(23.0+35.5)\: g}=\:?\: mol/L}\)

    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 \(\ce{H2SO4}\) to prepare a 250.0 mL solution. Calculate the concentration of this solution. Molar mass: \(\mathrm{H_2SO_4 = 98.0}\).

    Hint: 0.24 M

    Caution -
    Never pour water into concentrated sulfuric acid.

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

    Hint: 1.01 g

    Skill -
    Apply the equation: \(\mathrm{Amount = C \times V}\).

    \(\ce{Ca(HCO3)2}\) is present in hard-water.

  6. A stock sulfuric acid solution contains 98.0 % of \(\ce{H2SO4}\), 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: \(\mathrm{H_2SO_4 = 98.1}\).

    Hint: 544 mL

    \(\mathrm{5.0\times2 = 10\: mol\: H_2SO_4\times\dfrac{98.1\: g}{1\: mol}\times\dfrac{100\: g\: stock}{98\: g\: H_2SO_4}\times\dfrac{1\: mL}{1.84\: g}=\: ?\: mL}\)

    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 % \(\ce{HNO3}\) by weight. What is the molarity? Molar mass: \(\mathrm{HNO_3 = 63.01}\).

    Hint: 16 M

    Assume you have 1.00 L solution.

    \(\mathrm{\dfrac{1000\: mL\: solution}{1\: L}\times\dfrac{1.423\: g}{1\: mL}\times\dfrac{70.9\: g\: HNO_3}{100\: g\: solution}\times\dfrac{1\: mol}{63.01\: g\: HNO_3}=\: ?\: mol/L}\)

    Skill -

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

  8. If you want to make 250.0 mL of 0.100 M calcium chloride solution, how many moles of \(\ce{CaCl2}\) will you need?

    Hint: 0.025 mol

    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 \(\ce{CaCl2}\) are needed? Molar mass: \(\mathrm{CaCl_2 = 111.1\: g/mol}\) .

    Hint: 2.78 g

    \(\mathrm{0.250\: L\times\dfrac{0.1\: mol}{1\: L}\times\dfrac{111.1\: g}{1\: mol}=\: ?\: g\: (CaCl_2)}\)

    Skill -

    Prepare a solution of definite concentration.

  10. What is the molarity of \(\ce{Na+}\) in a 0.123 M solution of \(\ce{Na2SO4}\)?

    Hint: 0.246 M

    Skill -
    Explain the dissolution as ionization:

    \(\mathrm{Na_2SO_4 \rightarrow 2 Na^+ + SO_4^{2-}}\)

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

    Hint: 10.0 mL

    How many mL will contain 6 millimole \(\ce{Al^3+}\) ions?

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

    Hint: 50.0 mL

    Prepare solution of certain concentration by dilution.

  13. How many mL of 0.321 M \(\ce{HCl}\) solution is required to neutralize 5.0 mL of 1.284 M \(\ce{KOH}\) solution?

    Hint: 20.0 mL

    Skill -
    Performing calculations of titration results.

  14. If 10.0 mL \(\ce{HNO3}\) completely neutralize 25.0 mL of 0.351 M \(\ce{KOH}\), calculate the molarity of \(\ce{HNO3}\).

    Hint: 0.878 M

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

  15. Given a solution containing 0.242 g of barium chloride, \(\ce{BaCl2}\), how many mL of 0.0581 M \(\ce{H2SO4}\) will completely precipitate the barium ions, \(\ce{Ba^2+}\)? Molar mass: \(\mathrm{BaCl_2 = 208.3}\).

    Hint: 20.0 mL

    Skill -
    Performing volumetric analysis.

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

    Hint: 21.4%

    Skill -

    Convert mol \(\ce{H2SO4}\) to mol \(\ce{NaOH}\) to weight \(\ce{NaOH}\) to % \(\ce{NaOH}\).