How would you prepare exactly \(3.0\) litre of \(1.0 \mathrm{M} \mathrm{NaOH}\) by mixing proportions of stock solutions of \(2.50 \mathrm{M} \mathrm{NaOH}\) and \(0.40 \mathrm{M} \mathrm{NaOH}\). No water is to be used.

Understanding molarity and solution dilution is essential in preparing desired concentrations in chemistry. Molarity is a measure of concentration represented as moles of solute divided by liters of solution, commonly expressed as moles per liter (M or mol/L). To dilute a solution, one can mix a more concentrated solution (the stock solution) with a less concentrated one. Dilution does not involve adding water exclusively; it can be done by mixing different concentrations of the same solution to achieve a desired molarity.

For example, if a chemist needs to prepare a specific volume of a solution at a given molarity, they can use the dilution equation, which is a form of the conservation of moles concept. The equation, often written as \(C_1V_1 = C_2V_2\), relates the concentrations \(C_1\) and \(C_2\) and volumes \(V_1\) and \(V_2\) of two solutions before and after dilution. What's important is that the number of moles of solute remains the same before and after dilution; hence, the product of concentration and volume is constant. Popularly termed as a dilution equation, it is a valuable tool for the dilution of a concentrated solution to a lower concentration without specifying the amount of solvent used.

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. In solutions, stoichiometry involves calculations based on molar concentrations and volumes. A stoichiometric calculation might involve figuring out how different volumes of solutions will react based on their molar ratios. This process hinges on the concept that the final solution must contain the same total number of moles as the individual amounts combined.

An application of solution stoichiometry is when mixing different solutions to get a final combined solution of a target concentration. Here, understanding the mole concept, molarity, and volume is paramount. Careful calculation ensures the correct proportions are mixed, so that the essentially conserved properties—moles of solute and total volume—result in the desired concentration. These calculations are fundamental, notably in fields like pharmacy, where precise medication dosage is critical, and in research laboratories, where accurate reagent preparation can make or break an experiment.

Solving problems involving chemical solutions requires a systematic approach. First, clearly define what is sought—be it final volume, concentration, or amount of each solution to mix. Next, jot down what is known such as stock solutions' concentrations and any volume constraints. Then determine a strategy using mathematical relationships. Often in chemistry, algebraic equations like the dilution equation \(C_1V_1 = C_2V_2\) prove invaluable. With this, you can set up an equation system to solve for unknowns.

Take the initial example from the exercise: wanting to prepare a certain volume of a new concentration from two different stock solutions—no water added. By defining variables for the volumes of each stock solution and understanding that the total volume is fixed, one sets up two equations that can be simultaneously solved. Often, it involves algebra, potentially substituting one variable from one equation into the other, and then isolating the remaining variable. After calculating, always cross-verify the solution by checking that your results are sensible and they adhere to the conservation of mass. This thorough approach is a cornerstone in learning how to address chemical solution problems confidently.