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siniylev [52]
3 years ago
10

A 25.00 ml solution of sulfuric acid H2SO4 is titrated to phenolphthalein end point with 27.00 ml of 1.700 M KOH

Chemistry
1 answer:
Nastasia [14]3 years ago
5 0
<h3>Answer:</h3>

0.918 M

<h3>Explanation:</h3>

Assuming the question requires we calculate the Molarity of sulfuric acid:

We are given:

  • Volume of the acid, H₂SO₄ = 25.00 ml
  • Volume of the base, KOH = 27.00 mL
  • Molarity of the base, KOH is 1.70 M

We can calculate the molarity of the acid using the following steps;

<h3>Step 1: Write the chemical equation for the reaction.</h3>

The reaction is an example of a neutralization reaction where a base reacts with an acid to form salt and water.

Therefore, the balanced equation will be;

H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l)

<h3>Step 2: Determine the moles of the base, KOH </h3>

When given molarity and the volume of a solution, the number of moles can be calculated by multiplying molarity with volume.

Number of moles = Molarity × Volume

                             = 1.700 M × 0.027 L

                              = 0.0459 moles

Thus, moles of KOH used is 0.0459 moles

<h3>Step 3: Determine the number of moles of the Acid, H₂SO₄</h3>

From the reaction, 1 mole of the acid reacts with 2 moles of KOH

Therefore, the mole ratio of H₂SO₄ to KOH is 1 : 2

Thus, moles of H₂SO₄ = Moles of KOH ÷ 2

                                     = 0.0459 moles ÷ 2

                                     = 0.02295 moles

<h3>Step 4: Calculate the molarity of the Acid </h3>

Molarity is the concentration of a solution in moles per liter

Molarity = Moles ÷ Volume

Molarity of the acid = 0.02295 moles ÷ 0.025 L

                                = 0.918 M

Thus, the molarity of the acid, H₂SO₄ is 0.918 M

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Answer:

18,1 mL of a 0,304M HCl solution.

Explanation:

The neutralization reaction of Ba(OH)₂ with HCl is:

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The moles of 17,1 mL≡0,0171L of a 0,161M Ba(OH)₂ solution are:

0,0171L*\frac{0,161moles}{L} = 2,7531x10⁻³moles of Ba(OH)₂

By the neutralization reaction you can see that 2 moles of HCl reacts with 1 mole of Ba(OH)₂. For a complete reaction of 2,7531x10⁻³moles of Ba(OH)₂ you need:

2,7531x10^{-3}molBa(OH)_{2}*\frac{2molHCl}{1molBa(OH)_{2}} = 5,5062x10⁻³moles of HCl.

The volume of a 0,304M HCl solution for a complete neutralization is:

5,5062x10^{-3}molHCl*\frac{1L}{0,304mol} = 0,0181L≡18,1mL

I hope it helps!

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