Concentration and dilution play vital roles in everyday life, from fine-tuning the strength of your morning coffee to ensuring the correct potency of medications. When mixing paints, artists rely on these concepts to achieve the perfect hue, while in the kitchen, chefs adjust the concentration of flavours to create culinary masterpieces. Understanding these principles is key to navigating and manipulating the mixtures that we encounter in our daily routines.
Expressing concentrations
Concentration (\(c\)) refers to the amount of solute present in a given quantity (known volume) of a solution. It is common to encounter concentration of a solute expressed using \(c(\textrm{chemical formula})\) or \(\textrm{[chemical formula]}\). For example, the concentration of \(\ce{NaCl}\) can be expressed as \(c(\ce{NaCl}\)) or \([\ce{NaCl}]\).
Concentration can be represented using a range of units:
percent concentration (%)
parts per million (ppm)
parts per billion (ppb)
molarity (M).
Percent concentration
Concentration can be represented as percentage. There are three ways to do this, depending on how the solution is prepared.
When you dissolve a mass of solute in a mass of solvent, concentration can be expressed using percentage by mass or mass/mass, \(\%(\textrm{m/m}\)). This is calculated using:
\[ \frac{\textrm{Mass of solute (g)}}{\textrm{Mass of solution (g)}}\times100\% \]
When you dissolve a volume of solute in a volume of solvent, concentration can be expressed using percentage by volume or volume/volume, \(\%(\textrm{v/v}\)). This is calculated using:
\[ \frac{\textrm{Volume of solute (mL)}}{\textrm{Volume of solution (mL)}}\times100\% \]
When you dissolve a mass of solute in a volume of solvent, concentration can be expressed using percentage mass/volume, \(\%(\textrm{m/v}\)). This is calculated using:
\[ \frac{\textrm{Mass of solute (g)}}{\textrm{Volume of solution (mL)}}\times100\% \]
Parts per million and parts per billion
Parts per million is the number of parts (mass or volume) present in one million parts (mass or volume). Similarly, parts per billion is the number of parts (mass or volume) present in one billion parts (mass or volume). For example, \(1\textrm{ ppm}\) solute could indicate that \(1\textrm{ g}\) of solute is present in \(1000000\textrm{ mL}\) of solution. This is the same as \(1\textrm{ mg}\) of solute in \(1\textrm{ L}\) of solvent.
Concentration in \(\textrm{ppm}\) is calculated for solutes and solvents expressed in the same units using:
\[ \frac{\textrm{Parts of solute (g)}}{\textrm{Parts of solvent (g)}}\times1000000 \]
In the same way, concentration in \(\textrm{ppb}\) is calculated using:
\[ \frac{\textrm{Parts of solute (g)}}{\textrm{Parts of solvent (g)}}\times1000000000 \]
Molarity
Molarity or molar concentration (\(M\)) is the amount of solute in moles present in one litre of a solution. It is calculated using:
\[ \textrm{Molarity (M)}=\frac{\textrm{Amount of solute (mol)}}{\textrm{Volume of solution (L)}} \]
From this, we can derive a formula that relates molar concentration (\(c\)) with amount of solute in moles (\(n\)) nand volume of solute in litres (\(V\)):
\[ \textrm{c}=\frac{n}{V} \]
Dilutions
Dilution is a process used to lower the concentration of the original solution by adding more solvent. The concentrated solution is known as the stock solution. During this process, the amount of solute present in the solution remains constant before and after the dilution; only the volume changes.
The following equation shows the relationship between concentration and volume before and after dilution. This formula is built upon the fact that the same amount of solute is present in the solution before (stock solution) and after (diluted solution) the dilution.
\[ c_{1}V_{1}=c_{2}V_{2} \]
In this formula,
\(c_{1}\)= concentration of the stock solution
\(V_{1}\)= volume of the stock solution
\(c_{2}\)= concentration of the diluted solution
\(V_{2}\)= volume of the diluted solution
Example 1 – representing concentration using percent concentration
Calculate the concentrations of the following solutions and express them in the units indicated.
A sodium chloride solution is prepared by dissolving \(3.25\textrm{ g}\) of sodium chloride in \(42.50\textrm{ g}\) of water. Express the concentration of the solution using percentage mass/mass.
\(8.00\textrm{ mL}\) of ethanol is dissolved in enough water to give \(80.00\textrm{ mL}\) of solution. Calculate the percent by volume concentration of ethanol in the resulting solution.
If \(5.0\textrm{ g}\) of \(\ce{MgCl2}\) is dissolved in enough water to give \(200\textrm{ mL}\) of solution, calculate the concentration of \(\ce{MgCl2}\) in \(\%(\textrm{m/v}\)).
Step 1: For part a, calculate the concentration using the formula for \(\%(\textrm{m/m}\)).
\[\begin{align*}
c(\ce{NaCl}) & = \frac{\textrm{Mass of solute (g)}}{\textrm{Mass of solution (g)}}\times100\textrm{%}\\
& = \frac{3.25\textrm{ g}}{42.50\textrm{ g}}\times100\%\\
& = 7.65\%(\textrm{m/m})
\end{align*}\]
Step 2: For part b, calculate the concentration using the formula for \(\%(\textrm{v/v}\)).
\[\begin{align*}
c(\textrm{Ethanol}) & = \frac{\textrm{Volume of solute (mL)}}{\textrm{Volume of solution (mL)}}\times100\textrm{%}\\
& = \frac{8.00\textrm{ mL}}{80.00\textrm{ mL}}\times100\%\\
& = 10.0\%(\textrm{v/v})
\end{align*}\]
Step 3: For part c, calculate the concentration using the formula for \(\%(\textrm{m/v}\)).
\[\begin{align*}
c(\ce{MgCl2}) & = \frac{\textrm{Mass of solute (g)}}{\textrm{Volume of solution (mL)}}\times100\%\\
& = \frac{5.0\textrm{ g}}{200\textrm{ mL}}\times100\%\\
& = 2.5\%(\textrm{m/v})
\end{align*}\]
Example 2 – representing concentration using parts per million
Drinking water contains \(0.002\textrm{ g}\) of \(\ce{Mg}^{+}\) ions in \(100\textrm{ g}\) of water. Calculate the concentration in parts per million.
Step 1: Calculate the concentration using the formula for \(\textrm{ppm}\).
Example 3 – representing concentration using molarity
\(2.50\textrm{ mol}\) of \(\ce{CuSO4}\) is dissolved in enough water to give a \(500.0\textrm{ mL}\) solution. Calculate the molar concentration of the resulting solution.
Step 1: Calculate the concentration using the formula for \(M\). Remember that the volume needs to be converted to litres.
Example 4 – calculating the quantity of solute using concentration
Calculate the mass of glucose present in \(250\textrm{ mL}\) of \(1.50\textrm{M}\) glucose solution. The molar mass of glucose is \(180.15\textrm{ g mol}^{-1}\).
Step 1: Calculate the amount in moles using the formula for \(M\). Remember that the volume needs to be converted to litres.
Step 2: Calculate the mass in grams using \(M=\frac{m}{n}\).
\[\begin{align*}
M & = \frac{m}{n}\\
m & = n \times M\\
& = 0.375 \textrm{ mol} \times 180.15\textrm{ g mol}^{-1}\\
& = 67.6 \textrm{ g}
\end{align*} \]
Calculate the volume (in \(\textrm{mL}\)) of water required to prepare a \(1.00\textrm{M}\) \(\ce{HCl}\) solution from \(7.3\textrm{ g}\) of \(\ce{HCl}\). The molar mass of \(\ce{HCl}\) is \(36.458\textrm{ g mol}^{-1}\).
Step 1: Calculate the amount in moles using \(M=\frac{m}{n}\).
\[\begin{align*}
M & = \frac{m}{n}\\
n & = \frac{m}{M}\\
& = \frac{7.3\textrm{ g}}{36.458\textrm{ g mol}^{-1}}\\
& = 0.20\textrm{ mol}
\end{align*} \]
Step 2: Calculate the volume using the formula for \(M\). Remember that the calculated volume will be in litres.
\[\begin{align*}
c & = \frac{n}{V}\\
V & = \frac{n}{c}\\
& = \frac{0.20\textrm{ mol}}{1.00\textrm{M}}\\
& = 0.20\textrm{ L}
\end{align*} \]
A chemist prepared \(10\textrm{ mL}\) of a \(1.00\textrm{M}\) \(\ce{HCl}\) solution by adding water to the \(3.0\textrm{M}\) stock solution of \(\ce{HCl}\). Determine the volume of stock solution that should be taken for the dilution.
Step 1: Identify the known variables and the unknown variable.
We are given the concentration of the diluted solution (\(c_{2}=1.00\textrm{M}\)), the volume of the diluted solution (\(V_{2}=10\textrm{ mL}\)), and the concentration of the stock solution (\(c_{1}=3.0\textrm{M}\)). The unknown variable is the volume of the undiluted stock solution (\(V_{1}\)).
Step 2: Calculate the volume of the stock solution using the dilution formula. It is helpful to rearrange the formula first. Make sure all concentration values are in the same units, and all volume values are in the same units.
