Concentration of Solutions

In chemistry, we rarely work with pure substances. Most often, we work with mixtures, and the most common type of mixture is a solution, where one substance (the solute) is dissolved evenly throughout another (the solvent).

The concentration of a solution is a measure of how much solute is present in a given amount of solvent or solution. Qualitatively, we can describe a solution with a low concentration as being dilute (like weak tea) and one with a high concentration as being concentrated (like strong tea).

While concentration can be expressed in many ways (such as mass percentage), the most useful for chemical reactions is molar concentration.

Molar Concentration

The molar concentration of a solution is a measure of the moles of solute per unit volume of solution. The formal IUPAC term for this quantity is amount concentration (symbol c), and its defining equation is:

\[ c = \dfrac{n(\mathrm{solute})}{V(\mathrm{solution})} \] where n is the amount of solute in moles (mol) and V is the total volume of the solution in liters (L).

The standard unit for this quantity is moles per liter (mol L−1). When expressed in these units, molar concentration is most commonly known as molarity, and the unit is abbreviated with the symbol M.

\[ \class{mjx-molar}{\mathrm{M}} \equiv \mathrm{mol~L^{-1}} \]

NoteA Note on Symbols: The Molar M

While the full capital ‘M’ is very common for molarity, this textbook adheres to the formal IUPAC and NIST standard, which recommends a small capital letter (M) to avoid confusion with the SI prefix “Mega-”.

NoteA Note on Notation: Square Brackets […]

It is common to see the concentration of a species represented by placing its chemical formula in square brackets, such as [NaCl]. While many resources use this as a direct shorthand for molarity, this textbook will adhere to the formal IUPAC convention where the symbol for amount concentration is a lowercase italic c, such as c(NaCl). We will reserve the use of square brackets for the specific context of equilibrium constants in a later chapter.

The most fundamental use of molarity is to calculate the concentration of a solution prepared in the lab.

Practice

A solution is prepared by dissolving 83.5 g magnesium chloride (M = 95.21 g mol−1) in enough water to make a 500.0 mL solution. Find the molar concentrations (in mol L−1) of the following aqueous solutes:

  1. magnesium chloride
  2. magnesium(2+)
  3. chlorine(1−)
Solution

\[ \begin{align*} \mathrm{MgCl_2(s)} &\longrightarrow \mathrm{Mg^{2+}(aq)} + 2~\mathrm{Cl^-(aq)} \end{align*} \]

1. c(MgCl2)

Find the molar concentration (c) by dividing the amount of MgCl2 (in mol) by the total volume (in L) of the mixture.

\[ \begin{align*} c(\mathrm{MgCl_2}) &= n(\mathrm{MgCl_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= m(\mathrm{MgCl_2}) ~ M(\mathrm{MgCl_2})^{-1} ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= 83.\bar{5}~\mathrm{g} ~ \left ( \dfrac{\mathrm{mol}}{95.2\bar{1}~\mathrm{g}} \right ) \left ( \dfrac{1}{500.\bar{0}~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 1.7\bar{5}4~\mathrm{mol~L^{-1}} = 1.7\bar{5}4~\class{mjx-molar}{\mathrm{M}} \end{align*} \]

2. c(Mg2+)

This can be solved in two ways.

One way is to find the molar concentration (c) of Mg2+ by dividing the amount of Mg2+ by the total volume (in L) of the mixture. Determine the amount of Mg2+ by using the mole ratio (r) with the moles of MgCl2.

\[ \begin{align*} c(\mathrm{Mg^{2+}}) &= n(\mathrm{Mg^{2+}}) ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= n(\mathrm{MgCl_2}) ~ r(\mathrm{Mg^{2+}, MgCl_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= m(\mathrm{MgCl_2}) ~ M(\mathrm{MgCl_2})^{-1} ~ r(\mathrm{Mg^{2+}, MgCl_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= 83.\bar{5}~\mathrm{g} ~ \left ( \dfrac{\mathrm{mol}}{95.2\bar{1}~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{1~mol~Mg^{2+}}}{\mathrm{1~mol~MgCl_2}} \right ) \left ( \dfrac{1}{500.\bar{0}~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 1.7\bar{5}4~\mathrm{mol~L^{-1}} = 1.7\bar{5}4~\class{mjx-molar}{\mathrm{M}} \end{align*} \]

The other way is to simply use the mole ratio r(Mg2+,MgCl2) with the molar concentration of MgCl2.

\[ \begin{align*} c(\mathrm{Mg^{2+}}) &= c(\mathrm{MgCl_2}) ~ r(\mathrm{Mg^{2+}, MgCl_2}) \\[1.5ex] &= \left ( \dfrac{1.7\bar{5}4~\mathrm{moL~MgCl_2}}{\mathrm{L}} \right ) \left ( \dfrac{\mathrm{mol~Mg^{2+}}}{\mathrm{mol~MgCl_2}} \right ) \\[1.5ex] &= 1.7\bar{5}4~\mathrm{mol~L^{-1}} = 1.7\bar{5}4~\class{mjx-molar}{\mathrm{M}} \end{align*} \]

3. c(Cl−)

This can be solved in two ways.

One way is to find the molar concentration (c) by dividing the amount of Cl− by the total volume (in L) of the mixture. Determine the amount of Cl− by using the mole ratio (r) with the moles of MgCl2. Note: You could also use the mole ratio with the moles of Mg2+ that we solved for above.

\[ \begin{align*} c(\mathrm{Cl^-}) &= n(\mathrm{Cl^-})~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{MgCl_2}) ~ r(\mathrm{Cl^-, MgCl_2}) ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= m(\mathrm{MgCl_2}) ~ M(\mathrm{MgCl_2})^{-1} ~ r(\mathrm{Cl^-, MgCl_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= 83.\bar{5}~\mathrm{g} ~ \left ( \dfrac{\mathrm{mol}}{95.2\bar{1}~\mathrm{g}} \right ) \left ( \dfrac{\displaystyle \mathrm{2~mol~Cl^-}}{\mathrm{mol~MgCl_2}} \right ) \left ( \dfrac{1}{500.\bar{0}~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 3.5\bar{0}8~\mathrm{mol~L^{-1}} = 3.5\bar{0}8~\class{mjx-molar}{\mathrm{M}} \end{align*} \]

or by simply using the mole ratio with the molar concentration of MgCl2 (or Mg2+).

\[ \begin{align*} c(\mathrm{Cl^-}) &= c(\mathrm{MgCl_2}) ~ r(\mathrm{Cl^-, MgCl_2}) \\[1.5ex] &= \left ( \dfrac{1.7\bar{5}4~\mathrm{mol~MgCl_2}}{\mathrm{L}} \right ) \left ( \dfrac{\displaystyle \mathrm{2~mol~Cl^-}}{\mathrm{mol~MgCl_2}} \right ) \\[1.5ex] &= 3.5\bar{0}8~\mathrm{mol~L^{-1}} = 3.5\bar{0}8~\class{mjx-molar}{\mathrm{M}} \end{align*} \]

Molarity as a Conversion Factor

While weighing solids is essential for preparing solutions, it is often more convenient to work with solutions. Chemists prepare stock solutions of a precisely known molarity to be able to quickly dispense a specific number of moles by measuring a volume.

The relationship c = n/V can be rearranged to find the moles of solute in a solution or the volume of solution needed to provide a certain number of moles of solute:

  • Finding Moles from a Known Volume: \[n = c \times V\]
    • Application: A student dispenses 25.0 mL from a 0.150 M NaCl stock solution. By doing so, they have delivered 0.00375 moles of NaCl.

  • Finding Volume for a Desired Number of Moles: \[V = \dfrac{n}{c}\]
    • Application: A procedure requires 0.0750 moles of KMnO4. A chemist can obtain this amount by dispensing 375 mL from a 0.200 M stock solution.

Practice

A chemist is working to separate an aqueous mixture containing dissolved silver and copper ions. The first step is to selectively precipitate the silver as solid silver chloride (AgCl) by adding a sodium chloride solution. In a subsequent step, the copper will be recovered from the remaining solution.

Crucially, any excess chloride ions must be avoided, as they would interfere with the subsequent copper recovery process. Therefore, the chemist must add the exact stoichiometric amount of NaCl to react with the silver, but no more.

The sample contains 14.01 g of dissolved silver nitrate (AgNO3). What volume (in mL) of a 0.500 M aqueous NaCl stock solution should the chemist add to precipitate the silver without adding an excess of chloride?

Solution

Concept: Molarity as a Conversion Factor

The balanced chemical equation for the reaction is: \[ \begin{align*} \mathrm{Ag^+(aq)} + \mathrm{NaCl(aq)} \longrightarrow \mathrm{AgCl(s)} + \mathrm{Na^+(aq)} \end{align*} \]

The goal is to find the volume (V) of the NaCl solution that contains a specific number of moles (n) of the solute. We are given the molar concentration (c). The problem’s setup, which requires a slow, controlled addition of the reactant, illustrates a key reason why chemists use solutions: it is far easier to control the addition of a liquid from a buret or pipet than to add a solid powder slowly and evenly.

We begin with the fundamental definition of molar concentration and rearrange it to solve for the volume.

\[ \begin{align*} c(\mathrm{NaCl}) &= n(\mathrm{NaCl}) ~ V(\mathrm{NaCl})^{-1} \\[1.5ex] V(\mathrm{NaCl}) &= n(\mathrm{NaCl}) ~ c(\mathrm{NaCl})^{-1} \\[1.5ex] &= 0.082\bar{5}~\mathrm{mol~NaCl} \left( \dfrac{\mathrm{L}}{0.50\bar{0}~\mathrm{mol~NaCl}} \right) \left( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right) \\[1.5ex] &= 16\bar{5}.0~\mathrm{mL} \\[1.5ex] &= 165~\mathrm{mL} \end{align*} \]

Concentration as an Intensive Property

The previous section demonstrated a crucial laboratory technique: using the molarity of a stock solution as a conversion factor to dispense a precise amount of solute by measuring a volume. This entire practice is built upon a critical, yet intuitive, principle: the concentration of a sample taken from a solution is identical to the concentration of the original solution.

This reliability stems from the fact that concentration is an intensive property, meaning it does not depend on the quantity of the substance. For comparison, the density of a block of copper is the same whether you have a one-gram piece or a one-kilogram brick.

The physical reason for this consistency is that solutions are homogeneous mixtures. At the particle level, this means the solute particles are uniformly distributed throughout the solvent. The interactive figure below explores this principle from both the macroscopic and microscopic viewpoints, showing how the consistent particle distribution allows any sample to be representative of the whole.

This view illustrates the macroscopic result of taking a sample (called an aliquot) from a stock solution. The uniform and identical color in both containers is
evidence that the concentration has not changed.

This tab presents the particle-level model of the homogeneous stock solution. The solute particles (red dots) are distributed evenly across the solvent, which is visualized as a grid of countable squares. This model allows us to define the solution’s concentration as a specific ratio: one particle per grid square.

This final view demonstrates the consequence of uniform distribution. A portion of the stock solution is selected (dashed box) and shown as a separate sample. By counting the particles and grid squares within this sample, we can see that the concentration ratio is identical to that of the original stock solution, confirming that concentration is an intensive property.

A representative sample taken from a stock solution is formally called an aliquot. The critical conclusion from this demonstration is that the concentration of an aliquot is identical to the concentration of the stock solution from which it was taken. Understanding this principle is the essential foundation for the process of dilution.

Diluting and Concentrating Solutions

With the understanding that an aliquot’s concentration is identical to its parent stock solution, we can now explore the processes that change a solution’s concentration.

A solution is diluted when its concentration is decreased. This can be accomplished by adding more solvent or by removing solute.

  • Adding Solvent (+ Solvent): This is the most common laboratory method. An example is preparing a working solution by adding water to an aliquot of a concentrated stock solution.
  • Removing Solute (− Solute): This is often a purification method. An example is using a water filter, where the filter material selectively traps and removes dissolved impurities (solutes) from the water.

A solution is made more concentrated when its concentration is increased. This can be accomplished by adding more solute or by removing solvent.

  • Adding Solute (+ Solute): This is the most direct way to increase concentration, such as dissolving a weighed solid into a solvent to create a stock solution in the first place.
  • Removing Solvent (− Solvent): This is a common technique in organic chemistry. For example, a chemist might use a rotary evaporator to boil off a volatile solvent, leaving behind a more concentrated solution of the desired, non-volatile product.

Of the methods shown, dilution by the addition of solvent is the most common and quantitatively important procedure in a laboratory setting. This process is governed by a simple, yet powerful, underlying principle.

The Principle of Conservation of Solute

The logic of dilution is a direct consequence of the physical process. First, an aliquot of a stock solution is taken with volume, V. This aliquot contains a specific, known amount of solute (n) since the concentration (c) of the stock solution is known (n = c/V). Next, pure solvent is added to increase the total volume of the aliquot.

Because no solute is added or removed during this step, the total amount of solute in the final, diluted solution is identical to the amount that was originally present in the aliquot.

NoteKey Principle of Dilution

The amount of solute, n, is constant before and after dilution.

The Dilution Equation

This core principle provides the basis for all dilution calculations. Let’s define an initial state (before adding solvent) and a final state (after adding solvent). The amount of solute in each state is given by: \[ \begin{align*} n_{\text{initial}} &= c_{\text{initial}} V_{\text{initial}} \quad (\text{solution before dilution}) \\[1.5ex] n_{\text{final}} &= c_{\text{final}} V_{\text{final}} \quad (\text{solution after dilution}) \end{align*} \] Because the amount of solute is conserved during the process (solute is not being added or removed), we can set these two expressions equal to each other: \[ n_{\text{initial}} = n_{\text{final}} \] Substituting the c × V terms for each n gives us the dilution equation, often written more simply with subscript notation: \[ \begin{align*} c_1~V_1 = c_2~V_2 \end{align*} \]

where:

  • c1 is the amount concentration of the aliquot which is equal to the concentration of the stock solution.
  • V1 is the volume of the aliquot taken from the stock solution.
  • c2 is the amount concentration of the final, diluted solution.
  • V2 is the total volume of the final, diluted solution.


Practice

A laboratory procedure requires 250.0 mL of a 1.50 M hydrochloric acid solution. A concentrated stock solution of 12.00 M HCl is available.

  1. What volume (in mL) of the stock solution must be diluted with water to prepare the required solution?
  2. How many moles of HCl(aq) is in the final solution?
Solution

1. Volume of Stock Solution

The problem requires finding the initial volume (V1) needed for a dilution. We can rearrange the dilution equation to solve for V1. Because the equation is a ratio of concentrations, the units for volume do not need to be converted to liters, as long as they are consistent on both sides.

\[ \begin{align*} c_1~V_1 &= c_2~V_2 ~ \longrightarrow \\[1.5ex] V_1 &= c_2 ~ V_2 ~ c_1{^{-1}} \\[1.5ex] &= \left(1.5\bar{0}~\class{mjx-molar}{\mathrm{M}} \right ) \left(250.\bar{0}~\mathrm{mL} \right ) \left( \dfrac{1}{12.0\bar{0}~\class{mjx-molar}{\mathrm{M}}} \right ) \\[1.5ex] &= 31.\bar{2}50~\mathrm{mL} \\[1.5ex] &= 31.2~\mathrm{mL} \end{align*} \]

2. Moles of HCl in Final Solution

The amount of solute (n2) in the final diluted solution can be calculated using the definition of molarity, n = c × V. We use the concentration (c2) and volume (V2) of the final solution. The volume must be converted from milliliters to liters to be consistent with the units of molarity (mol L−1).

\[ \begin{align*} n &= c ~ V \\[1.5ex] &= \left ( \dfrac{1.5\bar{0}~\mathrm{mol}}{\mathrm{L}} \right ) \left ( \dfrac{250.\bar{0}~\mathrm{mL}}{} \right ) \left ( \dfrac{1~\mathrm{L}}{1000~\mathrm{mL}} \right ) \\[1.5ex] &= 0.37\bar{5}00~\mathrm{mol} \\[1.5ex] &= 0.375~\mathrm{mol} \end{align*} \]

This is the amount of solute in the final diluted solution. Based on the principle of dilution, this is also the same amount of solute that was present in the 31.3 mL aliquot of the stock solution taken in Part 1.