Question

A student wants to prepare \(1.00 \mathrm{~L}\) of a \(1.00 \mathrm{M}\) solution of \(\mathrm{NaOH}\) (molar mass \(=40.00 \mathrm{~g} / \mathrm{mol}\) ). If solid \(\mathrm{NaOH}\) is available, how would the student prepare this solution? If \(2.00 \mathrm{M} \mathrm{NaOH}\) is available, how would the student prepare the solution? To help ensure three significant figures in the NaOH molarity, to how many significant figures should the volumes and mass be determined?

Short answer

To prepare 1.00 L of 1.00 M NaOH solution, a student would need 40.00 g of solid NaOH. The student should first dissolve the solid NaOH in approximately 800 mL of deionized water and then transfer the solution to a 1.00 L volumetric flask, adding water until reaching the 1 L mark. If given a 2.00 M NaOH stock solution, the student would need to dilute 0.500 L of the stock solution in a 1.00 L volumetric flask by adding deionized water until reaching the 1 L mark. To ensure three significant figures in the NaOH molarity, the mass of NaOH for the solid should be measured to the nearest 0.001 g, and the volume measurements for dilution should be to the nearest 0.001 L.

Step-by-step solution

Determine the amount of NaOH needed for 1.00 L of 1.00 M NaOH solution

To calculate the amount of NaOH needed, we can use the formula: Amount of solute (g) = volume of solution (L) × molarity (M) × molar mass (g/mol) The volume of the solution needs to be 1.00 L, and the molarity needs to be 1.00 M. The molar mass of NaOH is given as 40.00 g/mol. So, we need to multiply: Amount of solute (g) = 1.00 L × 1.00 M × 40.00 g/mol = 40.00 g Hence, 40.00 g of NaOH is required to prepare 1.00 L of 1.00 M NaOH solution.

Preparing the solution from solid NaOH

To prepare 1.00 L of 1.00 M NaOH solution using solid NaOH: 1. Weigh out 40.00 g of solid NaOH using a balance. 2. Slowly dissolve the solid NaOH in approximately 800 mL of deionized water in a beaker, stirring until the solid NaOH is completely dissolved. 3. Transfer the solution to a volumetric flask and add deionized water until the volume reaches the 1.00 L mark on the flask. 4. Stir the solution well in the flask to ensure uniform concentration.

Preparing the solution from 2.00 M NaOH

To prepare 1.00 L of 1.00 M NaOH solution using 2.00 M NaOH solution, use the dilution formula: C₁V₁ = C₂V₂ where C₁ and V₁ are the initial concentration and volume of the stock NaOH solution, and C₂ and V₂ are the final concentration and volume of the diluted NaOH solution. Rearrange the equation to find V₁: V₁ = (C₂V₂) / C₁ = (1.00 M × 1.00 L) / 2.00 M = 0.500 L To prepare the solution: 1. Measure 0.500 L of the 2.00 M NaOH stock solution using a volumetric pipette. 2. Transfer this volume to a 1.00 L volumetric flask. 3. Add deionized water until the volume reaches the 1.00 L mark on the flask. 4. Stir the solution well in the flask to ensure uniform concentration.

Determining significant figures

Since we want to ensure three significant figures in the NaOH molarity, it's necessary to have at least three significant figures in the mass and volume measurements: 1. Mass of NaOH for solid should be weighed to the nearest 0.001 g (e.g., 40.000 g). 2. Volume measurements for dilution should be to the nearest 0.001 L (e.g., 0.500 L). By taking these measurements with the specified precision, we can ensure three significant figures in the concentration of the prepared NaOH solutions.

Key concepts

Molar Mass Calculation

Understanding molar mass is fundamental when preparing molar solutions, as it enables a scientist or student to convert between the mass of a substance and the number of moles. The molar mass of a substance is the weight in grams of one mole of that substance. To calculate molar mass, you sum the atomic masses of all the atoms in the molecule.

Let's consider our example of sodium hydroxide (NaOH). This compound comprises sodium (Na), oxygen (O), and hydrogen (H) atoms. The atomic mass of Na is approximately 23.00 g/mol, O is about 16.00 g/mol, and H is around 1.00 g/mol. By adding these values, the calculated molar mass of NaOH is 40.00 g/mol. This translates to any amount of NaOH in moles multiplied by 40.00 g/mol to give the corresponding mass in grams necessary for a solution.

Solution Dilution

Dilution is a process frequently used to achieve the desired concentration for a solution. It involves adding more solvent to reduce the solute's concentration to a lower, more required level. The key formula here is the dilution equation:
\[C_1V_1 = C_2V_2\]
This equation signifies that the concentration (C) times the volume (V) before dilution (indicated by subscript 1) is equal to the concentration times the volume after dilution (indicated by subscript 2). To put this into practice, identify the desired final concentration and volume, then use these values along with the known initial concentration to solve for the initial volume.

In our NaOH example, we wanted to dilute a 2.00 M solution to a 1.00 M solution. Using the formula, you'd find you needed to start with 0.500 L of the 2.00 M solution, and then add water until the total volume is 1.00 L. This approach ensures the final concentration is exactly as desired, potentially safeguarding against chemical reactions' strength or yield.

Significant Figures in Chemistry

The accurate report of measurements in chemistry hinges on the concept of significant figures, which reflects the precision of a measurement. Significant figures in a number include all the digits known with certainty plus one final digit, which is somewhat uncertain or is estimated.

When making calculations, like those of a molarity problem in chemistry, it's crucial to maintain precision. Therefore, maintaining three significant figures throughout calculations, as in the NaOH problem, means every step of the measurement process — weighing the solute, measuring the solvent volume — should match this degree of precision. For instance, measuring 40.00 g of NaOH ensures that the final solution's concentration can confidently be said to be 1.00 M to three significant figures. Not adhering to significant figures can lead to inaccuracies, which in chemistry could lead to ineffective or unsafe outcomes, such as imprecise dosages of medication or incorrect substance concentrations for research experiments.