Question

You wish to prepare \(0.12 \mathrm{M} \mathrm{HNO}_{3}\) from a stock solution of nitric acid that is \(15.8 M\). How many milliliters of the stock solution do you require to make up \(1.00 \mathrm{~L}\) of \(0.12 \mathrm{M} \mathrm{HNO}_{3} ?\)

Short answer

You need approximately 7.59 mL of the stock solution.

Step-by-step solution

Understand the problem

We want to dilute a concentrated solution of nitric acid from a molarity of \(15.8\, M\) to \(0.12\, M\). We need to find out how many milliliters of the concentrated solution are needed to make \(1.00\, L\) of the diluted solution.

Use the dilution formula

The dilution formula is \(C_1V_1 = C_2V_2\), where \(C_1\) and \(V_1\) are the concentration and volume of the initial solution, and \(C_2\) and \(V_2\) are the concentration and volume of the final solution. Here, \(C_1 = 15.8\, M\), \(C_2 = 0.12\, M\), and \(V_2 = 1.00\, L\). We need to find \(V_1\).

Substitute values into the equation

Plug the known values into the dilution equation: \(15.8\, M \times V_1 = 0.12\, M \times 1.00\, L\).

Solve for the initial volume \(V_1\)

Rearrange the equation to solve for \(V_1\): \(V_1 = \frac{0.12\, M \times 1.00\, L}{15.8\, M}\).

Perform the calculation

Calculate \(V_1\) using the equation: \(V_1 = \frac{0.12}{15.8} = 0.00759494\). To convert to milliliters, multiply by 1000, giving approximately \(7.59\, mL\).

Key concepts

Molarity

Molarity is a crucial concept in chemistry that deals with the concentration of a solution. It is defined as the number of moles of solute per liter of solution, typically expressed with the units "M" (moles per liter). Understanding molarity is essential because it allows chemists to accurately measure how concentrated a solution is.
For instance, if you have a 1 M solution of a substance, it means there is one mole of that substance in every liter of solution. When it comes to diluting solutions, as seen in our example, molarity helps determine how much of the stock solution is needed to achieve a desired concentration for the final solution.
To sum up, molarity provides a standardized way to express concentrations, making it easier to duplicate experiments and predict the outcomes of chemical reactions.

Solution Preparation

Preparing solutions accurately is a key skill in any chemistry lab. When making a solution, you start with a concentrated stock solution and dilute it to reach the desired concentration. This practice is common in laboratories because stock solutions are easy to store and readily adjustable to different concentrations.
Our example demands preparing a 0.12 M solution from a 15.8 M stock. This involves using a basic yet vital formula in chemistry: the dilution formula, \(C_1V_1 = C_2V_2\) where:

• \(C_1\) is the concentration of the stock solution.
• \(V_1\) is the volume of the stock solution required.
• \(C_2\) is the desired concentration of the final solution.
• \(V_2\) is the volume of the final solution.

The goal in solution preparation is to carefully measure and mix solutions to achieve the specific concentration needed for chemical reactions or tests. Proper preparation ensures accuracy in experiments and consistency in results.

Chemical Concentrations

Chemical concentrations tell us how much of a particular substance is present within a mixture or solution. Mastering this skill is a cornerstone of chemistry, enabling scientists to manage reactions and solutions efficiently.
By calculating chemical concentration, chemists can predict how substances will interact in a reaction. In our scenario, knowing the concentration of nitric acid allows us to dilute a concentrated solution to match an exact molarity, ensuring that subsequent experiments are accurate and reliable.
To calculate concentrations, you must first know the quantity of solute and the total volume of the solution, usually expressing it in terms of molarity (M). Accurately determining and manipulating chemical concentrations is vital whether you're in a lab setting or dealing with industrial applications, as even small errors can impact experimental results significantly.