How much of a \(12.0 \mathrm{M} \mathrm{HNO}_{3}\) solution should you use to make \(850.0 \mathrm{~mL}\) of a \(0.250 \mathrm{M} \mathrm{HNO}_{3}\) solution?

Making sense of molarity is essential for anyone dabbling in the science of solutions. In essence, molarity (\(M\)) expresses the concentration of a solution, measuring how much of a substance, known as the solute, is dissolved in a certain volume of solvent, typically water. The formula is profoundly simple: take the number of moles of solute and divide it by the volume of the solution in liters. For example, if you have a 1.0 M solution of sodium chloride, this means there's 1 mole of sodium chloride in every liter of the solution.

In the exercise we're exploring, molarity plays a crucial role when both mixing and diluting solutions. Understanding how to manipulate molarity is key to accurately preparing a dilute solution from a more concentrated one without altering the amount of solute present.

Stoichiometry is the cornerstone of reacting chemicals by quantifying the relationships between reactants and products. With stoichiometry, you can predict how much of each substance is needed or produced in a chemical reaction. Within the context of our example, stoichiometry takes the spotlight when we talk about the relationship between the volume and concentration of solutions before and after dilution.

The equation you apply here is the dilution equation, \(M1V1 = M2V2\), which ties together the initial and final states of a solution. This equality demonstrates stoichiometrically that the amount of solute remains unchanged during dilution; what changes is its concentration and the volume of the solvent.

Solution preparation is a fundamental procedure in chemistry that involves dissolving a solute in a solvent to achieve a desired molarity. This process requires precision and attention to detail to ensure the resulting chemical properties are achieved. When preparing a diluted solution from a concentrated stock, you begin by calculating the number of moles in the final solution, using the formula provided in the molarity section. Then, using the formula from the stoichiometry section, you can ascertain the volume of stock solution needed.

Practical Tip

Always add the concentrated solution to water, not the other way around, to avoid a thermal reaction or splashing that could occur if water is added to a concentrated acid or base.

In the realm of scientific measurements, significant figures are the digits that carry meaning towards the precision of those measurements. When calculations are completed, it's key to round your answer to the least number of significant figures used in the given values to maintain the same level of precision. This means, quite simply, that our answer must mirror the precision of the least precisely known number.

In our textbook example, the least precise value is the molarity of the dilute solution, 0.250 M, which has three significant figures. Thus, when we determine the volume of the concentrated solution that we need, our final answer should also be expressed with three significant figures to ensure that we are not implying greater precision than our measurements can support.