Calculating the Volume Required for Sodium Carbonate Solution Preparation

In this guide, we will walk through the process of calculating the volume of a sodium carbonate (Na2CO3) solution needed to prepare a specific concentration of decimolar solution. This is a common task in laboratory settings and helps in ensuring the accurate preparation of chemical solutions as required.

Understanding Sodium Carbonate Solutions

Sodium carbonate (Na2CO3) is a versatile chemical compound that is widely used in various industrial applications. It is a strong base and is commonly used in water treatment, glass manufacturing, and as a pH buffer. The density of 20% (m/m) sodium carbonate solution is 1.2086 g/mL at 30°C, making it an important parameter when dealing with its solution preparation.

Decimolar Solution and Molarity

A decimolar solution is a solution with a molar concentration of 0.1 mol/L (molarity, M). Molarity is defined as the number of moles of solute per liter of solution. The process of preparing such a solution involves knowing the required amount of solute and the final volume of the solution.

Calculating the Required Sodium Carbonate Mass

The first step in preparing a decimolar solution of sodium carbonate is to calculate the mass of sodium carbonate needed. The chemical formula for sodium carbonate is Na2CO3, and its molar mass (M) is approximately 106.0 g/mol. For a 1L decimolar (0.1 mol/L) solution of Na2CO3:

Number of moles of Na2CO3 required 0.1 mol/L * 1 L 0.1 mol

Mass of Na2CO3 number of moles * molar mass

Mass of Na2CO3 0.1 mol * 106.0 g/mol 10.6 g

Determining the Volume of 20% Sodium Carbonate Solution

Given the density of the 20% (m/m) sodium carbonate solution, we can calculate the volume of this solution that contains the required 10.6 g of Na2CO3. The density formula is: density mass/volume, which can be rearranged to solve for volume (V): V mass/density.

Here, the mass of sodium carbonate needed is 10.6 g and the density of the solution is 1.2086 g/mL. Therefore:

V 10.6 g / 1.2086 g/mL 8.79 mL

However, the calculated volume (8.79 mL) is for the pure sodium carbonate content within the 20% (m/m) solution. To obtain the total volume of the 20% solution, we need to account for the fact that 20% of the solution's mass is sodium carbonate. Let’s denote the total volume of the 20% solution as Vtotal.

Vtotal * 20% 8.79 mL

Vtotal 8.79 mL / 20% 8.79 mL / 0.20 43.85 mL

Thus, to prepare 1L of a decimolar (0.1 mol/L) sodium carbonate solution, you need 43.85 mL of 20% (m/m) sodium carbonate solution.

Conclusion

Accurately preparing chemical solutions requires a solid understanding of molarity, density, and the relationships between different solution concentrations. By using the density and concentration of a given solution, you can calculate the precise volume needed to achieve the desired molarity. This method ensures that the laboratory or industrial processes using these solutions are correctly conducted.

Related Keywords

* sodium carbonate: The specific chemical compound

* molar concentration: The measure of concentration of a solution, in moles of solute per liter of solution

* solution preparation: The process of creating chemical solutions with specific concentrations