Question

Start with a 1.0 L solution with a 0.40 M concentration of sulfuric acid. The above solution is divided into two equal parts of the same volume. To the first part, 0.50 L of water is added. To the second part 1.5 L of water are added. After the previous procedure, the two parts are mixed and 2.0 L of 0.1 M sulfuric acid are added to this new solution. Determine the final molar concentration.

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Answer 1

Explanation:Before dividing the initial solution:

Initial volume = 1.0 L

Initial concentration = 0.40 M

After dividing into two equal parts:

Each part has a volume of 0.5 L

The first part has a concentration of 0.40 M

The second part has a concentration of 0.20 M (diluted by 50% with 1.5 L of water)

When the two parts are mixed:

Total volume = 1.0 L + 0.5 L + 0.5 L + 1.5 L = 3.5 L

Total moles of sulfuric acid = (1.0 L x 0.40 M / 1000) + (0.5 L x 0.40 M / 1000) + (0.5 L x 0.20 M / 1000) + (1.5 L x 0 / 1000) = 0.5 mol

Final concentration before adding more sulfuric acid = 0.5 mol / 3.5 L = 0.14 M

When 2.0 L of 0.1 M sulfuric acid are added:

Total volume = 3.5 L + 2.0 L = 5.5 L

Total moles of sulfuric acid = 0.5 mol + (2.0 L x 0.1 M / 1000) = 0.7 mol

Final concentration = 0.7 mol / 5.5 L = 0.13 M

Therefore, the final molar concentration of the solution is 0.13 M.