Problem 76 - Chemistry Matter and Change

Question

How much of 5.0 \(\mathrm{M}\) nitric acid (HNO \(_{3} ),\) in milliliters, is needed to make 225 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{HNO}_{3} ?\)

Step-by-Step Solution

Answer

You need 45 mL of 5.0 M nitric acid.

Step 1: Understand the Dilution Formula

We will use the dilution formula given as \( M_1V_1 = M_2V_2 \) where \( M_1 \) and \( V_1 \) are the molarity and volume of the concentrated solution, and \( M_2 \) and \( V_2 \) are the molarity and volume of the diluted solution.

Step 2: Identify Known Values

From the problem, we know \( M_1 = 5.0 \, \text{M} \), \( M_2 = 1.0 \, \text{M} \), and \( V_2 = 225 \, \text{mL} \). Our task is to find \( V_1 \), the volume of the concentrated solution required.

Step 3: Apply the Dilution Formula

Substitute the known values into the formula: \( 5.0 \, \text{M} \times V_1 = 1.0 \, \text{M} \times 225 \, \text{mL} \). Simplify to find \( V_1 \).

Step 4: Solve for \( V_1 \)

To isolate \( V_1 \), divide both sides by \( 5.0 \, \text{M} \): \( V_1 = \frac{1.0 \, \text{M} \times 225 \, \text{mL}}{5.0 \, \text{M}} \). Calculating gives \( V_1 = 45 \, \text{mL} \).

Key Concepts

Molarity

Molarity is a way to express the concentration of a chemical solution. It tells us how many moles of a solute are present in one liter of solution. Simply put, it tells us the 'strength' of a solution. More formalized, molarity is calculated using the formula:

• M (Molarity) = moles of solute / liters of solution

If you have a 1.0 Molar (1.0 M) nitric acid solution, it means there is 1 mole of nitric acid in every liter of the solution. Molarity is crucial in chemistry because it helps in knowing the precise amount of a substance participating in a reaction.
In dilution problems, like the one in the exercise, understanding molarity helps us achieve the desired concentration by adjusting how much solute is present in a given volume.

Volume

Volume is the amount of space that a substance or object occupies. In the context of solutions, volume is used to measure how much liquid we have. It is usually measured in liters or milliliters. Knowing the volume of solutions is vital in chemistry as it affects the concentration and the outcome of chemical reactions.
In dilution calculations, volume plays a significant role in determining how concentrated or diluted a solution will become. For instance, when mixing a concentrated solution with water, the final volume determines the new concentration according to its molarity.
Understanding volume is essential in practical chemistry tasks such as making solutions of a specific concentration by adding or mixing measured amounts of liquid.

Concentration

Concentration indicates how much of a solute is present in a solution relative to the volume of the solution. It tells us how 'strong' or 'weak' a solution is. In terms of everyday items, think of making a drink from a powdered mix; the more powder you add, the stronger the drink.

• Concentration is often expressed in molarity, but it can also be expressed in percentage or parts per million (ppm).

Understanding concentration is key in preparing solutions for experiments where precise chemical reactions are required. It lets you control the amount of active ingredient interacting in a particular reaction.
Chemists rely on the concentration of solutions to predict interaction speeds, the amount of product generated, and overall reaction behavior.

Dilution Formula

The dilution formula is a simple but powerful tool in chemistry. It helps calculate how to dilute a solution to achieve a desired molarity. The formula is:

• M₁V₁ = M₂V₂

- **M₁** and **V₁** represent the molarity and volume of the concentrated solution. - **M₂** and **V₂** represent the molarity and volume of the diluted solution. This formula shows that the product of the initial concentration and initial volume equals the product of final concentration and volume. It assumes no loss of substance during the mixing process and ensures that the amount of solute before and after dilution remains constant.
Using this formula allows for accurate preparation of solutions needed in various experiments, like achieving the right acidity for a reaction or making an exact solution for measured observation.