How to Calculate the Volume of 20% HCl Needed for Dilution to Achieve 1 Molar Solution

Understanding the proper dilution process is crucial when working with chemical solutions, especially in laboratory settings and industrial applications. In this article, we will guide you through the steps to calculate the volume of a 20% by mass HCl solution needed to dilute into a higher molarity solution to achieve a specific desired concentration. We will use the example of adding 20% HCl to 4 liters of a 0.6 molar solution to obtain a 1 molar HCl solution.

Understanding the Concentration of HCl Solutions

Before we dive into the calculation, it's important to understand the concentration of the HCl solution. A 20% by mass HCl solution means that every 100 grams of the solution contains 20 grams of HCl. In terms of density, if the density (ρ) is 1.1 g/cm3, we can determine the concentration as follows:

1 cm3 1 mL

Molar mass of HCl 1 x 1.008 1 x 35.45 36.458 g/mol

The mass of HCl in 1000 g of solution: 1000 g / 1.1 g/cm3 909.1 mL or 0.9091 L

The mass of HCl in 1 L of solution: 200 g / 0.9091 L 220 g

The number of moles of HCl in 220 g: 220 g / 36.458 g/mol 6.03 mol

The concentration or Molarity of the HCl solution: 6.03 mol / 1 L 6.03 M

Dilution Calculation

Now, let's apply the dilution formula to calculate the volume of the 20% HCl needed. The formula used is:

C1 V1 C2 V2

Where:

C1 Initial concentration of HCl (6.03 M)

V1 Volume of HCl to be added (to be determined)

C2 Final concentration of HCl (1 M)

V2 Volume of the 0.6 M HCl solution (4 L)

V3 Final volume (V1 4 L)

Substituting the values into the formula:

6.03 x V1 0.6 x 4 1 x (V1 4)

Solving for V1 will give us the required volume of 20% HCl:

6.03V1 2.4 V1 4

5.03V1 1.6

V1 1.6 / 5.03 0.319 L

Converting L to mL: 0.319 L x 1000 mL/L 319 mL

Verification

To verify the solution, we can calculate the mass of the 20% HCl and confirm if the molarity matches:

Mass of 20% HCl in 319 mL: 0.319 L x 1100 g/L 350.9 g

Moles of HCl: 350.9 g / 36.458 g/mol 9.62 mol/L

Final solution: (9.62 mol/L x 0.319 L) (0.6 mol/L x 4 L) 1.32 2.4 3.72 mol/L

Dividing by the final volume (4.32 L): 3.72 mol/L / 4.32 L 1 M

This confirms that the solution is indeed 1 M.

Conclusion

Calculating the required volume of a 20% HCl solution is essential for achieving a desired molarity in the final solution. By understanding the concentration and applying the correct dilution formula, you can effectively prepare your solution.