10.5: Quantifying Solutions- Concentration Units

Concentration

To define a solution accurately, it is necessary to specify its concentration, which refers to the amount of solute dissolved in a given quantity of solvent. Terms such as dilute and concentrated are commonly used to describe solutions with relatively low or high amounts of solute, respectively; however, these are relative descriptors and their interpretation can vary depending on the context.

Percent Composition

There are several ways of expressing the concentration of a solution by using a percentage. The mass/mass percent concentration, (m/m)%, is defined as the mass of a solute divided by the mass of a solution times 100 (Equation \(\PageIndex{1}\) ):

\[\mathrm{ (m/m)\% = \dfrac{mass\: of\: solute \:(g)}{mass\: of\: solution \:(g)}\times100\%} \]

If you can measure the masses of the solute and the solution, determining the mass/mass percent is easy. Each mass must be expressed in the same units to determine the proper concentration.

Example \(\PageIndex{1}\)

A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. What is the mass/mass percent concentration of the solution?

Solution

We can substitute the quantities given in the equation for mass/mass percent:

\(\mathrm{(m/m)\%=\dfrac{36.5\: g}{355\: g}\times100\%=10.3\%}\)

For gases and liquids, volumes are relatively easy to measure, so the concentration of a liquid or a gas solution can be expressed as a volume/volume percent concentration, (v/v)%, the volume of a solute divided by the volume of a solution times 100 (Equation \(\PageIndex{2}\) ):

\[\mathrm{(v/v)\% = \dfrac{volume\: of\: solute \:(mL)}{volume\: of\: solution \:(mL)}\times100\%} \]

Again, the units of the solute and the solution must be the same. A hybrid concentration unit, mass/volume percent concentration, (m/v)%, is commonly used for intravenous (IV) fluids (Figure \(\PageIndex{1}\)). It is defined as the mass in grams of a solute, divided by volume in milliliters of solution times 100 (Equation \(\PageIndex{3}\)):

\[\mathrm{(m/v)\% = \dfrac{mass\: of\: solute\: (g)}{volume\: of\: solution\: (mL)}\times100\%} \]

Molarity

Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Such concentration units are useful for discussing chemical reactions in which a solute is a product or a reactant. Molar mass can then be used as a conversion factor to convert amounts in moles to amounts in grams.

Molarity is defined as the number of moles of a solute dissolved per liter of solution:

\[\mathrm{molarity=\dfrac{number\: of\: moles\: of\: solute}{number\: of\: liters\: of\: solution}} \nonumber \]

Molarity is abbreviated M (often referred to as “molar”), and the units are often abbreviated as mol/L. It is important to remember that “mol” in this expression refers to moles of solute and that “L” refers to liters of solution. For example, if you have 1.5 mol of NaCl dissolved in 0.500 L of solution, its molarity is therefore

\[\mathrm{\dfrac{1.5\: mol\: NaCl}{0.500\: L\: solution}=3.0\: M\: NaCl} \nonumber \]

which is read as “three point oh molar sodium chloride.” Sometimes (aq) is added when the solvent is water, as in “3.0 M NaCl(aq).”

Before a molarity concentration can be calculated, the amount of the solute must be expressed in moles, and the volume of the solution must be expressed in liters.

If the amount of solute is given in mass units, it must first be converted to moles before using the definition of molarity to calculate concentration. For example, to determine the molar concentration of a solution containing 22.4 g of HCl dissolved in 1.56 L of solution, we use a flowchart showing mass of solute leading to moles of solute, which then leads to molarity of solution with the addition of volume in liters (Figure \(\PageIndex{2}\)).

Step 1: Convert the mass of solute to moles using the molar mass of HCl (36.46 g/mol):

\[22.4\cancel{gHCl}\times \dfrac{1\: mol\: HCl}{36.46\cancel{gHCl}}=0.614\, mol\; HCl \nonumber \]

Step 2: Use the definition of molarity to determine the concentration:

\[M \: =\: \dfrac{0.614\: mol\: HCl}{1.56L\: solution}=0.394\, M HCl \nonumber \]

Using Molarity in Calculations

The definition of molarity can also be used to calculate a needed volume of solution, given its concentration and the number of moles desired, or the number of moles of solute (and subsequently, the mass of the solute), given its concentration and volume. As in the percent concentration, molarity can also be expressed as a conversion factor.

Molarity is defined as moles solute per liter solution. There is an understood 1 in the denominator of the conversion factor. For example, a 3.0 M solution of sucrose means that there are three moles of sucrose dissolved in every liter of solution. Mathematically, this is stated as follows:

3.0 moles sucrose in 1 L solution

Dividing both sides of this expression by either side, we generate two possible conversion factors:

\(\mathrm{\dfrac{3.0\: mol\: sucrose}{1\: L\: solution}}\) or \(\mathrm{\dfrac{1\: L\: solution}{3.0\: mol\: sucrose}}\)

The first conversion factor can be used to convert from volume (L) of solution to moles solute, and the second converts from moles of solute to volume (L) of solution (Figure \(\PageIndex{3}\)).

For example, suppose we are asked how many moles of sucrose are present in 0.108 L of a 3.0 M sucrose solution. The given volume (0.108 L) is multiplied by the first conversion factor to cancel the L units, and find that 0.32 moles of sucrose are present.

\[0.108\cancel{L\, solution}\times \dfrac{3.0\, mol\, sucrose}{\cancel{1L\, solution}}=0.32\, mol\, sucrose \nonumber \]

How many liters of 3.0 M sucrose solution are needed to obtain 4.88 mol of sucrose? In such a conversion, we multiply the given (4.88 moles sucrose) with the second conversion factor. This cancels the moles units and converts it to liters of solution.

\[4.88\cancel{mol\, sucrose}\times \dfrac{1\, L\, solution}{\cancel{3.0\, mol\, sucrose}}=1.63\, L\, solution \nonumber \]

Example \(\PageIndex{9}\)

Ethylene glycol (C₂H₆O₂) is mixed with water to make auto engine coolants. How many grams of C₂H₆O₂ are in 5.00 L of a 6.00 M aqueous solution?

Solution

1. To solve for the volume, multiply the "given" (0.450 mol of dimethylamine) with the molarity conversion factor (0.0753 M). Use the proper conversion factor to cancel the unit "mol" and get the unit volume (L) of solution:

\(\mathrm{0.450\: mol\: dimethylamine\times\dfrac{1\: L\: solution}{0.0753\: mol\: dimethylamine}=5.98\: L\: solution}\)

2. The strategy in solving this problem is to convert the given volume (5.00 L) using the 6.00 M (conversion factor) to solve for moles of ethylene glycol, which can then be converted to grams.

Step 1: Convert the given volume (5.00 L) to moles ethylene glycol.

\(\mathrm{5.00\: L\: solution\times\dfrac{6.00\: mol\: C_2H_6O_2}{1\: L\: solution}=30.0\: mol\: C_2H_6O_2}\)

Step 2: Convert 30.0 mols C₂H₆O₂ to grams C₂H₆O₂. Molar mass of C₂H₆O₂= 62.08 g/mol

\(\mathrm{30.0\: mol\: C_2H_6O_2\times\dfrac{62.08\: g\: C_2H_6O_2}{1\: mol\: C_2H_6O_2}=1,860\: g\: C_2H_6O_2}\)

The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows (Figure \(\PageIndex{4}\)):

\(\mathrm{5.00\: L\: solution\times\dfrac{6.00\: mol\: C_2H_6O_2}{1\: L\: solution}\times\dfrac{62.08\: g\: C_2H_6O_2}{1\: mol\: C_2H_6O_2}=1,860\: g\: C_2H_6O_2}\)

The final answer is rounded off to 3 significant figures. Thus, there are 1,860 g of C₂H₆O₂ in the specified amount of engine coolant.

Note: Dimethylamine has a “fishy” odor. In fact, organic compounds called amines cause the odor of decaying fish.

Using Molarity in Stoichiometry Problems

Of all the ways of expressing concentration, molarity is the one most commonly used in stoichiometry problems because it is directly related to the mole unit. Consider the following chemical equation:

\[\ce{HCl(aq) + NaOH(aq) → H2O(ℓ) + NaCl(aq)}\label{Eq1}  \]

Suppose we want to know how many liters of aqueous HCl solution will react with a given mass of NaOH. A typical approach to answering this question is as follows (Figure \(\PageIndex{5}\)):

Example \(\PageIndex{12}\)

How many milliliters of a 2.75 M HCl solution are needed to react with 185 g of NaOH? The balanced chemical equation for this reaction is as follows:

\[\ce{HCl(aq) + NaOH(aq) → H2O(ℓ) + NaCl(aq)} \nonumber \]

Solution

We will follow the flowchart to answer this question. First, we convert the mass of NaOH to moles of NaOH using its molar mass, 40.00 g/mol:

\(\mathrm{185\: g\: NaOH\times\dfrac{1\: mol\: NaOH}{40.00\: g\: NaOH}=4.63\: mol\: NaOH}\)

Using the balanced chemical equation, we see that there is a one-to-one ratio of moles of HCl to moles of NaOH. We use this to determine the number of moles of HCl needed to react with the given amount of NaOH:

\(\mathrm{4.63\: mol\: NaOH\times\dfrac{1\: mol\: HCl}{1\: mol\: NaOH}=4.63\: mol\: HCl}\)

Finally, we use the definition of molarity to determine the volume of 2.75 M HCl needed:

\(\mathrm{2.75\: M\: HCl=\dfrac{4.63\: mol\: HCl}{volume\: of\: HCl\: solution}}\)

\(\mathrm{volume\: of\: HCl=\dfrac{4.63\: mol\: HCl}{2.75\: M\: HCl}=1.68\: L\times\dfrac{1,000\: mL}{1\: L}=1,680\: mL}\)

We need 1,680 mL of 2.75 M HCl to react with the NaOH.

The same multi-step problem can also be worked out in a single line, rather than as separate steps, as follows (Figure \(\PageIndex{6}\)):

\(\mathrm{185\: g\: NaOH\times\dfrac{1\: mol\: NaOH}{40.00\: g\: NaOH}\times\dfrac{1\: mol\: HCl}{1\: mol\: NaOH}\times\dfrac{1\: L\: HCl\: solution}{2.75\: mol\: HCl}\times\dfrac{1000\: mL\: HCl\: solution}{1\: L\: HCl\: solution}=1,680\: mL\: HCl\: solution}\)

Our final answer (rounded off to three significant figures) is 1,680 mL HCl solution.

The general steps for performing stoichiometry problems such as this are shown in Figure \(\PageIndex{7}\). You may want to consult this figure when working with solutions in chemical reactions. The double arrows in Figure \(\PageIndex{7}\) indicate that you can start at either end of the chart and, after a series of simple conversions, determine the quantity at the other end.  

Many of the fluids found in our bodies are solutions. The solutes range from simple ionic compounds to complex proteins with typical concentrations (Table \(\PageIndex{1}\)).

Note: Concentrations are approximate and can vary widely (Table \(\PageIndex{1}\)).

Dilutions

When additional water is added to an aqueous solution, the concentration of that solution decreases. This is because the number of moles of the solute does not change, while the volume of the solution increases. We can set up an equality between the moles of the solute before the dilution (1) and the moles of the solute after the dilution (2).

\[\text{mol}_1 = \text{mol}_2 \nonumber \]

Since the moles of solute in a solution is equal to the molarity multiplied by the liters, we can set those equal.

\[M_1 \times L_1 = M_2 \times L_2 \nonumber \]

Finally, because the two sides of the equation are set equal to one another, the volume can be in any units we choose, as long as that unit is the same on both sides. Our equation for calculating the molarity of a diluted solution becomes:

\[M_1 \times V_1 = M_2 \times V_2 \nonumber \]

Suppose that you have \(100. \: \text{mL}\) of a \(2.0 \: \text{M}\) solution of \(\ce{HCl}\). You dilute the solution by adding enough water to make the solution volume \(500. \: \text{mL}\). The new molarity can easily be calculated by using the above equation and solving for \(M_2\).

\[M_2 = \frac{M_1 \times V_1}{V_2} = \frac{2.0 \: \text{M} \times 100. \: \text{mL}}{500. \: \text{mL}} = 0.40 \: \text{M} \: \ce{HCl} \nonumber \]

The solution has been diluted by one-fifth since the new volume is five times as great as the original volume. Consequently, the molarity is one-fifth of its original value.

Dilution Calculations: Molarity and Percent Solutions

Any units of concentration and volume can be used, as long as both concentrations and both volumes have the same unit. For example, if we are using M (molarity), then we can express the equation as follows:

\[\mathrm{Molarity_{initial}\: \times Volume_{initial}\: = Molarity_{final}\: \times Volume_{final}\:}  \]

\[\mathrm{M_1\: \times V_1\: = M_2\: \times M_2\:}  \]

If we are using percent, the dilution equation is as follows:

\[\mathrm{Concentration_{initial}\: \times Volume_{initial}\: = Concentration_{final}\: \times Volume_{final}\:}  \]

\[\mathrm{C_1\: \times V_1\: = C_2\: \times M_2\:}  \]

Example \(\PageIndex{13}\)

A 125 mL sample of 0.900 M NaCl is diluted to 1,125 mL. What is the final concentration of the diluted solution?

Solution

Because the volume units are the same, and we are looking for the molarity of the final solution, we can use (concentration × volume)_initial = (concentration × volume)_final:

(0.900 M × 125 mL) = (concentration × 1,125 mL)

We solve by isolating the unknown concentration by itself on one side of the equation. Dividing by 1,125 mL gives

\(\mathrm{concentration = \dfrac{0.900\: M\times125\: mL}{1,125\: mL}=0.100\: M}\)

as the final concentration.