Problem 201 - Introductory Chemistry Atoms First

Question

How many grams of \(\mathrm{NaOH}\) are needed to prepare \(500.0 \mathrm{~mL}\) of \(0.300 \mathrm{M} \mathrm{NaOH}\) solution?

Step-by-Step Solution

Answer

To prepare 500.0 mL of 0.300 M NaOH solution, 6.0 g of NaOH are needed.

Step 1: Understand the Problem and Gather Information

The problem is essentially asking to find out how many grams of NaOH are needed to prepare a 0.300 M solution of NaOH in a volume of 500.0 mL. The molar mass of NaOH is calculated by adding the atomic masses of sodium (Na - 23 g/mol), oxygen (O - 16 g/mol), and hydrogen (H - 1 g/mol), which results in 40 g/mol.

Step 2: Convert Volume from mL to L

Since molarity is defined in terms of liters of solution, the given volume (500 ml) has to be converted to liters. This is done using the conversion factor of 1 L = 1000 ml, resulting in: \[ 500.0 mL * \frac{1 L}{1000 mL} = 0.5 L \]

Step 3: Use the Molarity Formula to Find Moles of NaOH

Using the formula of molarity from earlier: \[Molarity = \frac{Moles~of~Solute}{Liters~of~Solution}\] the moles of NaOH can be found by rearranging the formula as follows: \[Moles~of~Solute = Molarity * Liters~of~Solution\] Then substituting the given molarity (0.300 M) and the converted volume (0.5 L): \[Moles~of~NaOH = 0.300 M * 0.5 L = 0.15 mol\]

Step 4: Convert Moles to Grams using Molar Mass

The final step involves converting the moles of NaOH to grams. This is done by multiplying the moles of NaOH by the molar mass of NaOH (40 g/mol): \[Grams~of~NaOH = Moles~of~NaOH * Molar~Mass~of~NaOH = 0.15 mol * 40 g/mol = 6.0 g\] In conclusion, 6.0 g of NaOH are needed to prepare 500.0 mL of a 0.300 M NaOH solution.

Key Concepts

Sodium Hydroxide (NaOH)

Sodium hydroxide, commonly known as lye or caustic soda, is a powerful alkaline compound. It often appears as a white, crystalline, odorless solid that is highly soluble in water. NaOH is frequently used in various industrial processes including paper production, textile manufacturing, and soap making.
Additionally, in chemistry labs, NaOH is utilized to create solutions of known concentrations due to its reactive nature and strong base properties.

• NaOH fully dissociates in water, providing hydroxide ions (OH⁻) that react with acids to form water and salts.
• Handling NaOH requires caution as it's highly corrosive to organic tissues and can cause severe chemical burns.

Ensuring proper handling and safety measures, such as wearing gloves and eye protection, is essential when dealing with this compound.

Molecular Weight

The molecular weight of a compound is crucial for calculating the quantity needed to prepare a solution.
It helps in converting between moles—a common measure of substance amount in chemistry—and grams, which is a more practical weight measurement.
The molecular weight of sodium hydroxide (NaOH) is determined by the sum of the atomic weights of its constituent atoms:

• Sodium (Na): 23 g/mol
• Oxygen (O): 16 g/mol
• Hydrogen (H): 1 g/mol

By adding these atomic weights, we find that NaOH has a molecular weight of 40 g/mol.
This simple calculation helps chemists determine exactly how much NaOH is required for a reaction or solution preparation.

Solution Preparation

Preparing a solution of known molarity involves several careful steps to ensure accuracy and safety. To make a 0.300 M NaOH solution, follow these general guidelines:

• Calculate the needed mass of solute using its molarity and volume of solution.
• Convert the volume of the solution to liters if needed, since molarity is defined as moles per liter of solution.
• Dissolve the solute in a small amount of solvent first, then dilute it to the final desired volume.

In this scenario, we need to determine how many grams of NaOH are needed to make 500 mL, or 0.5 L, of a 0.300 M solution. The entire process relies on understanding and using the molarity formula effectively.

Chemistry Calculations

Chemistry calculations are essential in determining the precise amounts of reactants and products in a chemical reaction. They usually involve various formulas and concepts such as molarity, molecular weight, and conversion factors.
For this task, we specifically utilized the molarity formula:
\[Moles = Molarity imes Liters~of~Solution\]
With the provided molarity (0.300 M) and converted solution volume (0.5 L), we calculated the moles of NaOH necessary for the solution.
Following this, the moles were converted to grams using the formula:
\[Grams = Moles imes Molar~Mass\]
Precision in these calculations ensures that the correct amount of solute is used, which is vital for the success of any experiment or chemical process.