When 25.0 mL of sulfuric acid solution was completely neutralized in a titration with 0.05 M NaOH solution, it took 18.3 mL of the NaOH(aq) to complete the job. The reaction is:NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l)
What was the molarity of the sulfuric acid solution?

Answers

Answer 1

Answer: Balanced equation will be as follows

2NaOH + H2SO4=Na2SO4 + 2H2O
from MaVa/MbVb=Na/Nb where Ma is molarity of acid=?
Va is volume of acid=25.0 mL
Mb is molarity of base= 0.05 M
Vb is volume of base=18.3 mL
Nb number of moles of base= 2
Na is number of moles of acid =1

Ma*25 mL / 18.3mL*0.05M = 1/2
Ma=18.3*0.05M / 50 = 0.0183M

The molarity of sulphuric acid = 0.0183M

Answer 2

Answer:

Final answer:

The molarity of the sulfuric acid solution is found using the formula: M1V1/2 = M2V2 to account for the 2:1 ratio of NaOH to H2SO4 in the reaction. Substituting the given values yields the molarity of sulfuric acid solution as 0.0732M.

Explanation:

This problem involves a concept known as titration, specifically acid-base titration. In this process, one solution (in this case, the base NaOH) is added to another solution (the acid H2SO4) until the mixture is neutralized. The balanced chemical equation NaOH + H2SO4 -> Na2SO4 + 2H2O tells us that the reaction occurs in a 2:1 ratio of NaOH to H2SO4.

To find the molarity of the sulfuric acid, we use the formula: M1V1 = M2V2, where M1 is the molarity of the NaOH, V1 is the volume of the NaOH, M2 is the molarity of the H2SO4 (which we want to find), and V2 is the volume of the H2SO4. However, we need to adjust the equation to M1V1/2 = M2V2 because the reaction occurs in a 2:1 ratio of NaOH to H2SO4.

So substituting the given values into the equation, we have (0.05 M * 18.3 mL) / 2 = M2 * 25.0 mL. Solving for M2, the molarity of the sulfuric acid solution, yields M2 = 0.0732M.

Learn more about Acid-Base Titration here:

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The following three solutions are mixed: 100.0mL of 0.100M Na₂SO₄, 50.0mL of 0.300M ZnCl₂, and 100.0mL of 0.200M Ba(CN)₂. Cyanide (CN-) solubilities are not in most tables, but Ba(CN)₂ is soluble, Zn(CN)₂ is not. a. What ionic compounds will precipitate out of solution? b. What is the molarity of each ion remaining in the solution, assuming complete precipitation of all insoluble compounds, and assuming that volumes are additive?

Answers

a. To determine the ionic compounds that will precipitate out of solution, we need to consider the solubility rules. According to the solubility rules:

1. All sodium (Na+) salts are soluble, so Na₂SO₄ will remain in solution.
2. Zinc (Zn2+) salts are generally soluble, except for zinc sulfide (ZnS) and zinc hydroxide (Zn(OH)₂). However, in this case, we are adding ZnCl₂ to the solution, which contains chloride (Cl-) ions. Chloride ions form soluble salts with most cations, including Zn2+. Therefore, ZnCl₂ will remain in solution.
3. Barium (Ba2+) salts are generally soluble, except for barium sulfate (BaSO₄) and barium carbonate (BaCO₃). However, in this case, we are adding Ba(CN)₂ to the solution, which contains cyanide (CN-) ions. Cyanide ions form insoluble salts with most cations, including Ba2+. Therefore, Ba(CN)₂ will precipitate out of solution as Ba(CN)₂ is not soluble.

b. Assuming complete precipitation of all insoluble compounds and that volumes are additive, we can calculate the molarity of each ion remaining in the solution.

For Na₂SO₄:
- Sodium (Na+) ion concentration: 2 * 0.100 M = 0.200 M
- Sulfate (SO₄2-) ion concentration: 0.100 M

For ZnCl₂:
- Zinc (Zn2+) ion concentration: 0.300 M
- Chloride (Cl-) ion concentration: 2 * 0.300 M = 0.600 M

For Ba(CN)₂:
- Barium (Ba2+) ion concentration: 0.200 M
- Cyanide (CN-) ion concentration: 2 * 0.200 M = 0.400 M

Therefore, the molarity of each ion remaining in the solution, assuming complete precipitation of all insoluble compounds and that volumes are additive, are as follows:
- Sodium (Na+) ion: 0.200 M
- Sulfate (SO₄2-) ion: 0.100 M
- Zinc (Zn2+) ion: 0.300 M
- Chloride (Cl-) ion: 0.600 M
- Barium (Ba2+) ion: 0.200 M
- Cyanide (CN-) ion: 0.400 M

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Answers

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