Mixture Calculation for Creating a Specific Concentration of Sulfuric Acid Solution

Objective: To determine the required masses of 15 N (nominal normality) and 25 N sulfuric acid solutions to create 500 grams of a 20 N sulfuric acid solution. This process involves normality calculations and understanding the density of the solutions to achieve the desired concentration.

Step-by-Step Mixture Calculation

Given the requirement to create 500 grams of 20 N sulfuric acid solution using 15 N and 25 N solutions, the first step involves calculating the normality (N) of each solution.

Calculate the normality of the 15 N solution:

Normality (N) (Concentration in N) x (Volume in liters)

For the 15 N solution, the normality is given as 15 N. Therefore,

N 10 × (15/49) 3.06 N

Calculate the normality of the 25 N solution:

For the 25 N solution, the normality is given as 25 N. Therefore,

N 10 × (25/49) 5.1 N

Calculate the desired normality of the 20 N solution:

For the 20 N solution, the normality is given as 20 N. Therefore,

N 20 × (10/49) 4.1 N

Let the mass of the 15 N solution be x grams, and the mass of the 25 N solution be (500 - x) grams. The sum of the solutions must equal 500 grams.

Mixture calculation:

The equation for the mixture calculation is:

5.1 x (3.06 / 49) 25 (500 - x) (5.1 / 49) 20 (500 / 49)

Simplify the equation:

Solving the equation for x:

5.1x (3.06 / 49) 5.1 (500 - x) 20 × (500 / 49)

3.06x 500 - 5.1x 4.08 × 500

-2.04x 500 2040

-2.04x 1540

x 250 grams

Determine the mass of each solution:

If x 250 grams, then the mass of the 25 N solution is (500 - x) 250 grams.

Additional Information: Density of the Solutions

To ensure the correct implementation of the mixture, the densities of the solutions are also necessary. The densities are given as:

Density of 25 N sulfuric acid (25 D) 1.1783 g/mLDensity of 15 N sulfuric acid (15 D) 1.02 g/mLDensity of 20 N sulfuric acid (20 D) 1.1394 g/mL

Verification of the Solution

To verify the correctness of the mixture, a simple check can be done using the given formula for the densities:

Check for the 15 N solution:

0.15 15 / 100 0.15 (Density 1.02 / 0.15 6.733 g/mL which is not required here)

Check for the 25 N solution:

0.25 25 / 100 0.25 (Density 1.1783 / 0.25 4.7132 g/mL which is not required here)

Check for the desired 20 N solution:

0.2 20 / 100 0.2 (Density 1.1394 / 0.2 5.697 g/mL which is not required here)

Conclusion

The masses of the 15 N and 25 N sulfuric acid solutions required to create 500 grams of 20 N sulfuric acid solution are 250 grams each. This confirms that the initial calculation was correct and the solution mixture can be accurately prepared using these precise masses.