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gizmo_the_mogwai [7]
3 years ago
14

Question 9

Chemistry
1 answer:
Shkiper50 [21]3 years ago
7 0

Answer: There are 20 milliliters of 1.0 M HCl is required to be diluted to make 200 mL of a 0.1 M solution.

Explanation:

Given: M_{1} = 1.0 M,    V_{1} = ?

M_{2} = 0.1 M,    V_{2} = 200 mL

Formula used is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute values into the above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\1.0 M \times V_{1} = 0.1 M \times 200 mL\\V_{1} = \frac{0.1 M \times 200 mL}{1.0 M}\\= 20 mL

Thus, we can conclude that there are 20 milliliters of 1.0 M HCl is required to be diluted to make 200 mL of a 0.1 M solution.

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Give the complete ionic equation for the reaction (if any) that occurs when aqueous solutions of MgSO3 and HI are mixed. Give th
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Answer:

Mg²⁺(aq) + SO₃²⁻(aq) + 2 H⁺(aq) + 2 I⁻(aq) ⇄ Mg²⁺(aq) + 2I⁻(aq) + H₂O(l) + SO₂(g)

Explanation:

<em>Give the complete ionic equation for the reaction (if any) that occurs when aqueous solutions of MgSO₃ and HI are mixed.</em>

When MgSO₃ reacts with HI they experience a double displacement reaction, in which the cations and anions of each compound are exchanged, forming H₂SO₃ and MgI₂. At the same time, H₂SO₃ tends to decompose to H₂O and SO₂. The complete molecular equation is:

MgSO₃(aq) + 2 HI(aq) ⇄ MgI₂(aq) + H₂O(l) + SO₂(g)

In the complete ionic equation, species with ionic bonds dissociate into ions.

Mg²⁺(aq) + SO₃²⁻(aq) + 2 H⁺(aq) + 2 I⁻(aq) ⇄ Mg²⁺(aq) + 2I⁻(aq) + H₂O(l) + SO₂(g)

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Conduct metric Titration of H_2(SO_4) and Ba(OH)_2 Write an equation (including states of matter) for the reaction between H_2(S
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Answer:

a) H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)

b) H₂SO₄, H⁺, HSO₄⁻, SO₄²⁻. H₂O, H⁺, OH⁻.

c) H⁺, HSO₄⁻, SO₄²⁻

d) As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. The conducting species are Ba²⁺, SO₄²⁻, H⁺ and OH⁻.

f) After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

Explanation:

a) Write an equation (including states of matter) for the reaction between H₂SO₄ and Ba(OH)₂.

The <em>balanced equation</em> is:

H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)   [1]

b) At the very start of the titration, before any titrant has been added to the beaker, what is present in the solution?

In the beginning there is H₂SO₄ and the ions that come from its <em>dissociation reactions</em>: H⁺, HSO₄⁻, SO₄²⁻. There is also H₂O and a very small amount of H⁺ and OH⁻ coming from its <em>ionization</em>.

H₂SO₄(aq) ⇄ H⁺(aq) + HSO₄⁻(aq)

HSO₄⁻(aq) ⇄ H⁺(aq) + SO₄²⁻(aq)

H₂O(l)  ⇄ H⁺(aq) + OH⁻(aq)

c) What is the conducting species in this initial solution?

The main responsible for conductivity are the <em>ions</em> coming from H₂SO₄: H⁺, HSO₄⁻, SO₄²⁻.

d) Describe what happens as titrant is added to the beaker. Why does the conductivity of the solution decrease? What is the identity of the solid formed? What is the conducting species present in the beaker?

As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) What happens when the conductivity value reaches its minimum value (which is designated as the equivalence point for this type of titration)? What is the conducting species in the beaker?

At the <em>equivalence point</em> equivalent amounts of H₂SO₄ and Ba(OH)₂ react. Only BaSO₄ and H₂O are present, and since they are <em>weak electrolytes</em>, there is a small amount of ions to conduct electricity. The conducting species are Ba²⁺ and SO₄²⁻ coming from BaSO₄ and H⁺ and OH⁻ coming from H₂O.

f) Describe what happens at additional titrant is added past the equivalence point. Why does the conductivity of the solution increase? What is the conducting species present in the beaker?

After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

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