Solution Concentrations – SmartStudent Academy

Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for reactions that occur in solution, and are critical for many aspects of our lives, from measuring the correct dose of medicine to detecting chemical pollutants like lead and arsenic.

Chemists use many different ways to define concentrations. In this section, we will cover the most common ways of presenting solution concentration. These include: Molarity and Parts Per Solutions.

Molarity

The most common unit of concentration is molarity, which is also the most useful for calculations involving the stoichiometry of reactions in solution.

The molarity (M) of a solution is the number of moles of solute present in exactly 1 L of solution.

The units of molarity are therefore moles per liter of solution (mol/L), abbreviated as M.

Example Problem:

Calculate the number of moles of sodium hydroxide (NaOH) needed to make 2.50 L of 0.100 M NaOH.

Given: (1) identity of solute = NaOH, (2) volume = 2.50 L, and (3) molarity of solution = 0.100 mol/L (Note: when calculating problems always write out the units of molarity as mol/L, rather than M. This will allow you to cancel out your units when doing the calculation.)

Asked for: amount of solute in moles

Strategy: (1) Rearrange the equation above to solve for the desired unit, in this case for moles. (2) Double check all the units in the equation and make sure they match. Perform any conversions that are needed so that the units match. (3) Fill in values appropriately and do the math.

Solution:

(1) Rearrange the equation above to solve for moles.

(2) Double check all the units in the equation and make sure they match.

The given values for this equation are the volume 2.50 L and the molarity 0.100 mol/L. The volume units for both of these numbers are in Liters (L) and thus, match. Therefore, no conversions need to be made.

(3) Fill in values appropriately and do the math.

Preparation of Solutions

Note that in the example above, we still don’t have enough information to actually make the solution in the laboratory. There is no piece of equipment that can measure out the moles of a substance. For this, we need to convert the number of moles of the sample into the number of grams represented by that number. We can then easily use a balance to weigh the amount of substance needed for the solution. For the example above:

To make a solution, start by addition a portion of the solvent to the flask. Next, weigh out the appropriate amount of solute and slowly add it to the solvent using a volumetric flask. Once it is dissolved in the solvent, the volume of the solution can be brought up to the final solution volume. Volumetric flasks exist in many different sizes to accommodate different solution volumes. Graduated cylinders can also be used to accurately bring a solution to its final volume. Other glassware, including beakers and Erlenmeyer flasks are not accurate enough to make most solutions.

Example Molarity Calculation

The solution contains 10.0 g of cobalt(II) chloride dihydrate, CoCl₂·2H₂O, in enough ethanol to make exactly 500 mL of solution. What is the molar concentration of CoCl₂·2H₂O?

Given: mass of solute and volume of solution

Asked for: concentration (M)

Strategy:

We know that Molarity equals moles/Liter

To calculate Molarity, we need to express:

the mass in the form of moles
the volume in the form of Liters
Plug both into the equation above and calculate

Solution:

Converting the mass into moles. We can use the molar mass to convert the grams of CoCl₂·2H₂O to moles.

The molar mass of CoCl₂2H₂O is 165.87 g/mol (and includes the two water molecules as they are part of the crystal lattice structure of this solid hydrate!)

1. 1. Convert the volume into Liter
2. Plug values into the Molarity equation:
Parts Per Solutions
The most common method of expressing the concentration is based on the quantity of solute in a fixed quantity of solution. The “quantities” referred to here can be expressed in mass, in volume, or both (i.e., the mass of solute in a given volume of solution.) In order to distinguish among these possibilities, the abbreviations (m/m), (v/v) and (m/v) are used.

In most applied fields of Chemistry, (m/m) measure is often used, whereas in clinical chemistry, (m/v) is commonly used, with mass expressed in grams and volume in mL.

One of the more common ways to express such concentrations as “parts per 100“, which we all know as “percent“. Percent solutions define the quantity of a solute that is dissolved in a quantity of solution multiplied by 100. Percent solutions can be expressed in terms of mass solute per mass solution (m/m%), or mass solute per volume of solution (m/v%), or volume of solute per volume of solution (v/v%).

When making a percent solution, it is important to indicate what units are being used, so that others can also make the solution properly. Also, recall that the solution is the sum of both the solvent and the solute when you are performing percent calculations.

Solution = Solute + Solvent

Thus, the following equation can be used when calculating percent solutions:

Example 1:

As an example, a 7.0% v/v solution of ethanol in water, would contain 7 mL of ethanol in a total of 100 mL of solution. How much water is in the solution? In this problem, we know that the:

Solution = Solute + Solvent

Thus, we can fill in the values and then solve for the unknown.

100 mL = 7 mL + X mL of Solvent (in this case water)

shifting the 7 over to the other side, we can see that:

100 mL – 7 mL = 93 mL H₂O

Example

What is the (m/v)% of a solution if 24.0 g of sucrose is dissolved in a total solution of 243 mL?

Example 3

How many grams of NaCl are required to make 625 mL of a 13.5% solution?

For more dilute solutions, parts per million (10⁶ ppm) and parts per billion (10⁹; ppb) are used. These terms are widely employed to express the amounts of trace pollutants in the environment.

Like percentage (“part per hundred”) units, ppm and ppb may be defined in terms of masses, volumes, or mixed mass-volume units. There are also ppm and ppb units defined with respect to numbers of atoms and molecules.

The mass-based definitions of ppm and ppb are given here:

Both ppm and ppb are convenient units for reporting the concentrations of pollutants and other trace contaminants in water. Concentrations of these contaminants are typically very low in treated and natural waters, and their levels cannot exceed relatively low concentration thresholds without causing adverse effects on health and wildlife. For example, the EPA has identified the maximum safe level of fluoride ion in tap water to be 4 ppm. Inline water filters are designed to reduce the concentration of fluoride and several other trace-level contaminants in tap water.

When reporting contaminants like lead in drinking water, ppm and ppb concentrations are often reported in mixed unit values of mass/volume. This can be very useful as it is easier for us to think about water in terms of its volume, rather than by its mass. In addition, the density of water is 1.0 g/mL or 1.0 mg/0.001 mL which makes the conversion between the two units easier. For example, if we find that there is lead contamination in water of 4 ppm, this would mean that there are:

Equivalents

Concentrations of ionic solutes are occasionally expressed in units called equivalents (Eq). One equivalent equals 1 mol of positive or negative charge. Thus, 1 mol/L of Na⁺(aq) is also 1 Eq/L because sodium has a 1+ charge.

A 1 mol/L solution of Ca²⁺(aq) ions has a concentration of 2 Eq/L because calcium has a 2+ charge. Dilute solutions may be expressed in milliequivalents (mEq)—for example, human blood plasma has a total concentration of about 150 mEq/L.

In a more formal definition, the equivalent is the amount of a substance needed to do one of the following:

react with or supply one mole of hydrogen ions (H⁺) in an acid–base reaction
react with or supply one mole of electrons in a redox reaction.

By this definition, an equivalent is the number of moles of an ion in a solution, multiplied by the valence of that ion. If 1 mol of NaCl and 1 mol of CaCl₂ dissolve in a solution, there is 1 equiv Na, 2 equiv Ca, and 3 equiv Cl in that solution. (The valence of calcium is 2, so for that ion you have 1 mole and 2 equivalents.)

Dilutions

A solution of a desired concentration can also be prepared by diluting a small volume of a more concentrated solution with additional solvent. A stock solution, which is a prepared solution of known concentration, is often used for this purpose. D

iluting a stock solution is preferred when making solutions of very weak concentrations, because the alternative method, weighing out tiny amounts of solute, can be difficult to carry out with a high degree of accuracy.

Dilution is also used to prepare solutions from substances that are sold as concentrated aqueous solutions, such as strong acids.

The procedure for preparing a solution of known concentration from a stock solution requires calculating the amount of solute desired in the final volume of the more dilute solution and then calculating the volume of the stock solution that contains this amount of solute.

Remember that diluting a given quantity of stock solution with solvent does not change the amount of solute present, only the volume of the solution is changing. The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution can therefore be expressed mathematically as:

Where M_s is the concentration of the stock solution, V_s is the volume of the stock solution, M_d is the concentration of the diluted solution, and V_d is the volume of the diluted solution.

Example of Dilution Calculations

What volume of a 3.00 M glucose stock solution is necessary to prepare 2500 mL of 0.400 M solution?

Given: volume and molarity of dilute solution, and molarity of stock solution

Asked for: volume of stock solution

Strategy and Solution:

For Dilution problems, as long as you know 3 of the variables, you can solve for the 4th variable.

Start by rearranging the equation to solve for the variable that you want to find. In this case, you want to find the volume of the stock solution, V_s

Next, check to make sure that like terms have the same units. For example, Md and Ms are both concentrations, thus, to be able to perform the calculations, they should be in the same unit (in this case they are both listed in Molarity). If the concentrations were different, say one was given in Molarity and the other in percent or one was in Molarity and the other was in Millimolarity, one of the terms would need to be converted so that they match. That way, the units will cancel out and leave you with units of volume, in this case.

Finally, fill in the equation with known values and calculate the final answer.

Note that if 333mL of stock solution is needed, that you can also calculate the amount of solvent needed to make the final dilution. (Total volume – volume of stock solution = volume of solvent needed for the final dilution. In this case 2,500 mL – 333 mL = 2,167 mL of water needed to make the final dilution (this should be done in a graduated cylinder or volumetric flask).

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