Solutions and Concentration
~8 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable calculating moles from mass using molar mass and understanding the concept of a limiting reagent in solid-phase reactions.
Why this matters
Chemical reactions rarely occur in dry powder; they happen in liquid, where concentration dictates the rate and extent of reaction.
In the laboratory, we almost never measure out individual atoms. Instead, we measure volumes of liquids. To predict how much product forms when two solutions are mixed, we need a bridge between the volume of liquid we pipette and the number of particles reacting. This lecture introduces molarity, the standard unit of concentration, and shows how it allows us to apply the stoichiometric rules you already know to reactions in solution. It also covers the practical skill of dilution, which is fundamental to preparing reagents of precise strength.
Counting by Volume
Imagine you have a bucket of water and you add a spoonful of salt. The salt dissolves, and the water becomes salty. If you add another spoonful, it gets saltier. If you add a bucket of fresh water, it gets less salty. In chemistry, we need to quantify "salty" precisely. We cannot just say "a lot" or "a little." We need to know exactly how many salt particles are present in a specific amount of water. This is where the mole comes in. We count the salt in moles, not grams, because chemical equations are written in moles. However, we measure the water in litres, not moles. Therefore, we need a unit that combines these two ideas: moles of substance per litre of solution. This unit is called molarity, denoted by the symbol \(c\). It tells us how "concentrated" the solution is. A high molarity means there are many particles in a small volume. A low molarity means there are few particles in the same volume. This concept is crucial because when you mix two solutions, the amount of reaction depends on how many particles are available to collide. If you know the molarity and the volume you use, you can calculate the exact number of moles of reactant you are adding. This allows you to predict the outcome of the reaction using the same stoichiometry you use for solids and gases. It turns a liquid mixture into a countable quantity, making the invisible world of atoms measurable and predictable in the lab.
Defining Molarity
Molarity is defined as the number of moles of solute divided by the volume of the solution in litres. It is important to distinguish between the volume of the solute and the volume of the solution. When you dissolve a solid in water, the final volume is not simply the volume of the water plus the volume of the solid. The particles fit into the spaces between water molecules, and the total volume changes. Therefore, molarity is always calculated using the final volume of the mixture. The formula is straightforward: \(c = n/V\). Here, \(c\) is the molarity in \(\mathrm{mol\,dm^{-3}}\) (which is equivalent to \(\mathrm{mol\,L^{-1}}\)), \(n\) is the amount of substance in moles, and \(V\) is the volume in litres. To calculate molarity, you first convert the mass of the solute to moles using its molar mass. Then, you divide by the volume of the solution. For example, if you dissolve \(58.5\ \mathrm{g}\) of sodium chloride (\(\mathrm{NaCl}\)) in water and make up the solution to a final volume of \(1.00\ \mathrm{L}\), the molarity is \(1.00\ \mathrm{mol\,L^{-1}}\). This is because the molar mass of \(\mathrm{NaCl}\) is approximately \(58.5\ \mathrm{g\,mol^{-1}}\), so you have exactly one mole of \(\mathrm{NaCl}\) in one litre of solution. This unit is central to all solution chemistry because it directly links the macroscopic volume you measure to the microscopic number of particles involved in the reaction.
The Dilution Principle
In the laboratory, it is often more convenient to prepare a stock solution of high concentration and then dilute it to the desired lower concentration. Dilution involves adding solvent (usually water) to a solution. When you add solvent, the volume of the solution increases, but the amount of solute (the number of moles) remains exactly the same. You have not added or removed any solute particles; you have just spread them out over a larger volume. This conservation of moles is the key to the dilution equation. If \(n_1\) is the moles of solute before dilution and \(n_2\) is the moles after dilution, then \(n_1 = n_2\). Since \(n = cV\), we can substitute to get \(c_1V_1 = c_2V_2\). Here, \(c_1\) and \(V_1\) are the concentration and volume of the initial stock solution, and \(c_2\) and \(V_2\) are the concentration and volume of the final diluted solution. This equation allows you to calculate any one of these four variables if you know the other three. It is a powerful tool for preparing solutions. For instance, if you have a \(10.0\ \mathrm{mol\,L^{-1}}\) stock solution and you need \(250\ \mathrm{cm^3}\) of a \(0.500\ \mathrm{mol\,L^{-1}}\) solution, you can use this equation to find out how much stock solution to measure. The principle relies on the fact that the identity of the solute does not change during dilution, only its spatial distribution.
Stoichiometry in Solution
Once you have established the concentration of a solution, you can use it to perform stoichiometric calculations for reactions that occur in liquid phase. The process is identical to stoichiometry with solids, but the first step is slightly different. Instead of converting mass to moles, you convert volume and concentration to moles using \(n = cV\). Once you have the number of moles of each reactant, you proceed with the standard limiting reagent analysis. You compare the mole ratio from the balanced chemical equation with the actual mole ratio of the reactants you have. The reactant that is consumed first is the limiting reagent, and it determines the maximum amount of product that can form. For example, consider the reaction between hydrochloric acid and sodium hydroxide: \(\mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O}\). The stoichiometry is 1:1. If you mix \(50.0\ \mathrm{cm^3}\) of \(0.100\ \mathrm{mol\,L^{-1}}\) \(\mathrm{HCl}\) with \(50.0\ \mathrm{cm^3}\) of \(0.100\ \mathrm{mol\,L^{-1}}\) \(\mathrm{NaOH}\), you have \(0.00500\ \mathrm{mol}\) of each. Since the ratio is 1:1, they react completely, and there is no limiting reagent in the sense of excess; they are stoichiometric. However, if you used \(100\ \mathrm{cm^3}\) of the acid, you would have \(0.0100\ \mathrm{mol}\) of acid and \(0.00500\ \mathrm{mol}\) of base. The base would be the limiting reagent, and the acid would be in excess. This method allows chemists to design titrations and other quantitative analyses with precision.
The result
Concentration and dilution
Molarity \(c=n/V\) (moles of solute per litre of solution) is the standard measure of solution concentration. Diluting a solution by adding solvent changes its volume but not the moles of solute present, so \(c_1V_1=n=c_2V_2\) — the dilution equation — relates any solution's concentration and volume before and after dilution, and together with \(n=cV\), extends mole-based stoichiometry (mole-avogadro, limiting-reagent) to reactions carried out in solution.
A common misconception is that molarity does not change with temperature. In reality, the volume of a liquid expands or contracts with temperature changes, while the number of moles of solute remains constant. Therefore, the molarity of a solution changes slightly with temperature. For high-precision work, concentrations are often specified at a standard temperature (e.g., \(20\ \mathrm{^\circ C}\)). If you prepare a solution at \(25\ \mathrm{^\circ C}\) and then use it at \(10\ \mathrm{^\circ C}\), the volume will be slightly smaller, making the molarity slightly higher than calculated at \(25\ \mathrm{^\circ C}\). This effect is usually small for dilute aqueous solutions but can be significant in precise analytical chemistry or when working with organic solvents that have higher thermal expansion coefficients. C
Worked example
Problem. A chemist needs to prepare \(250.0\ \mathrm{cm^3}\) of a \(0.200\ \mathrm{mol\,L^{-1}}\) solution of sulfuric acid (\(\mathrm{H_2SO_4}\)). She has a stock solution of \(1.00\ \mathrm{mol\,L^{-1}}\) \(\mathrm{H_2SO_4}\). Calculate the volume of stock solution required.
- Identify the knowns: \(c_1 = 1.00\ \mathrm{mol\,L^{-1}}\), \(c_2 = 0.200\ \mathrm{mol\,L^{-1}}\), \(V_2 = 250.0\ \mathrm{cm^3} = 0.2500\ \mathrm{L}\).
- Identify the unknown: \(V_1\), the volume of stock solution needed.
- Rearrange the dilution equation \(c_1V_1 = c_2V_2\) to solve for \(V_1\): \(V_1 = \frac{c_2V_2}{c_1}\).
- Substitute the values: \(V_1 = \frac{0.200\ \mathrm{mol\,L^{-1}} \times 0.2500\ \mathrm{L}}{1.00\ \mathrm{mol\,L^{-1}}}\).
- Calculate the result: \(V_1 = 0.0500\ \mathrm{L}\).
- Convert to \(\mathrm{cm^3}\): \(0.0500\ \mathrm{L} \times 1000\ \mathrm{cm^3\,L^{-1}} = 50.0\ \mathrm{cm^3}\).
Units check. The units of concentration cancel out, leaving volume, which is consistent with the required answer.
Three quick tests
- What is the molarity of a solution containing \(0.500\ \mathrm{mol}\) of solute in \(2.00\ \mathrm{L}\) of solution?
Answer
\(c = n/V = 0.500/2.00 = 0.250\ \mathrm{mol\,L^{-1}}\). - If you dilute \(100\ \mathrm{cm^3}\) of a \(2.00\ \mathrm{mol\,L^{-1}}\) solution to a final volume of \(500\ \mathrm{cm^3}\), what is the new concentration?
Answer
\(c_1V_1 = c_2V_2 \Rightarrow 2.00 \times 100 = c_2 \times 500 \Rightarrow c_2 = 0.400\ \mathrm{mol\,L^{-1}}\). - How many moles of \(\mathrm{NaOH}\) are present in \(50.0\ \mathrm{cm^3}\) of a \(0.100\ \mathrm{mol\,L^{-1}}\) solution?
Answer
\(n = cV = 0.100 \times 0.0500 = 0.00500\ \mathrm{mol}\).
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Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Results cited: T-017 · Threads: structure