Problem 37 - Chemistry The Molecular Nature of Matter

Question

When a solution labeled \(0.50 \mathrm{M} \mathrm{HNO}_{3}\) is diluted with water to give \(0.25 \mathrm{M} \mathrm{HNO}_{3}\), what happens to the number of moles of \(\mathrm{HNO}_{3}\) in the solution?

Step-by-Step Solution

Answer

The number of moles of HNO3 remains unchanged after dilution.

Step 1: Understanding the Concept of Dilution

Dilution is the process of reducing the concentration of a solute in a solution, usually by adding more solvent. The number of moles of solute present before and after dilution remains the same because no solute is added or removed during the process.

Step 2: Applying the Dilution Formula

The dilution formula relates the concentrations and volumes before and after dilution. It can be expressed as M1 * V1 = M2 * V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

Step 3: Inferring the Change in the Number of Moles

Since the dilution formula shows that the product of the concentration and the volume is constant, and the concentration is halved from 0.50 M to 0.25 M, the volume must be doubled to keep the product constant. This implies that the number of moles of HNO3 remains the same before and after dilution.

Key Concepts

Molarity

Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute divided by the volume of the solution in liters. Mathematically, molarity can be expressed as:
\[ Molarity (M) = \frac{moles \: of \: solute}{volume \: of \: solution \: in \: liters} \]
The molarity of a solution tells us how 'strong' or 'concentrated' the solution is. For example, a 1 M solution of sodium chloride (NaCl) contains one mole of sodium chloride dissolved in one liter of solution. When dealing with dilutions, understanding molarity is crucial as it assists in determining the amount of solvent needed to achieve a desired concentration.

Dilution Formula

The dilution formula is a critical expression used to calculate the new concentration of a solute in a solution after adding more solvent. It helps in ensuring that the final solution has the desired molarity for chemical reactions or other scientific applications. The formula is:
\[ M_1 \times V_1 = M_2 \times V_2 \]
where:

• \(M_1\) is the initial molarity of the solution before dilution,
• \(V_1\) is the initial volume of the solution before dilution,
• \(M_2\) is the final molarity of the solution after dilution, and
• \(V_2\) is the final volume of the solution after dilution.

This formula is derived from the conservation of the number of moles of solute before and after dilution, presuming no chemical reaction occurs that would consume the solute. When using the formula, it's important to ensure that the volume measurements are in the same units, typically liters.

Concentration of a Solute

The concentration of a solute in a solution quantifies how much of the solute is present in a given volume of solvent. It is a ratio that can be expressed in various units, such as molarity, mass percent, or parts per million (ppm). Concentration can affect many properties of a solution, including boiling point, freezing point, and osmotic pressure.

In the context of the exercise, when diluting a 0.50 M nitric acid (\(HNO_3\)) solution to a 0.25 M solution, we are reducing the concentration of the solute, which is the nitric acid in this case. The number of moles of \(HNO_3\) does not change, but the volume of the solution increases to achieve the lower concentration. This is important in chemical processes where specific concentrations are necessary for reactions to proceed under controlled conditions.