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Chapter 6: Problem 18

A student wants to prepare \(1.00 \mathrm{L}\) of a \(1.00-M\) solution of NaOH (molar mass \(=40.00 \mathrm{g} / \mathrm{mol}\) ). If solid NaOH is available, how would the student prepare this solution? If \(2.00 \mathrm{M}\) 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

Expert verified
To prepare a 1.00 L of 1.00-M NaOH solution using solid NaOH, the student should weigh out 40.0 g of NaOH, dissolve it in approximately 800 mL of water, and then add more water until the total volume reaches 1.00 L. To prepare the solution using a 2.00 M NaOH solution, measure 0.500 L of the 2.00 M solution, mix it with water until the total volume is 1.00 L. The volumes and mass should be determined with at least three significant figures to ensure accurate molarity.

Step by step solution

01

1. Calculating the mass of solid NaOH required

To prepare a 1.00 L solution of 1.00-M NaOH, we first need to calculate the mass of solid NaOH required. Using the molar mass of NaOH (40.00 g/mol), we can find the mass as follows: Mass of NaOH = (Molarity × Volume) × Molar mass Mass of NaOH = (1.00 mol/L × 1.00 L) × 40.00 g/mol Mass of NaOH = 40.0 g
02

2. Preparing the 1.00-M NaOH solution from solid NaOH

To prepare the 1.00 L of 1.00-M NaOH solution using solid NaOH, the student should: 1. Measure out 40.0 g of solid NaOH. 2. Dissolve the 40.0 g of solid NaOH in approximately 800 mL of water in a 1 L volumetric flask or beaker. 3. Once the NaOH has dissolved, carefully add water to the flask or beaker until the 1 L mark is reached. 4. Stir the solution to ensure homogeneity.
03

3. Calculating the dilution ratio for preparing 1.00-M NaOH from 2.00-M NaOH solution

To prepare a 1.00 L of 1.00-M NaOH solution using the 2.00 M NaOH solution, we can use the formula: M1 × V1 = M2 × V2 where M1 is the initial molarity (2.00 M), V1 is the initial volume, M2 is the final molarity (1.00 M), and V2 is the final volume (1.00 L). Solving for V1, we get: V1 = M2 × V2 / M1 V1 = (1.00 mol/L × 1.00 L) / 2.00 mol/L V1 = 0.500 L
04

4. Preparing the 1.00-M NaOH solution from 2.00-M NaOH solution

To prepare the 1.00 L of 1.00-M NaOH solution using the 2.00 M NaOH solution, the student should: 1. Measure out 0.500 L of the 2.00 M NaOH solution. 2. Pour the 0.500 L of 2.00 M NaOH solution into a 1 L volumetric flask or beaker. 3. Carefully add water to the flask or beaker until the 1 L mark is reached. 4. Stir the solution to ensure homogeneity.
05

5. Determining the required number of significant figures for the volumes and mass

To help ensure three significant figures in the NaOH molarity, the volumes and mass should have at least three significant figures. In this case, the mass of solid NaOH is already measured with three significant figures (40.0 g). For volumes, we should measure the 0.500 L with three significant figures, meaning we should use a volumetric flask or graduated cylinder that has markings up to at least 0.001 L increments.

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