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alexandr1967 [171]
3 years ago
10

5.0 mL of 0.050 M Ba(NO3)2 M and 100.0 mL of 0.10 M KIO3 are added together in a 250 mL erlenmeyer flask. In this problem, ignor

e ionic strength effects and neglect the possible presence of any complex ions in solution. a. Will Ba(IO3)2 precipitate out of solution? Support your answer with appropriate calculations. b. What are the concentrations of Ba2+ and 103 in the solution? Report your answer using 2 significant figures.
Chemistry
1 answer:
DiKsa [7]3 years ago
7 0

Answer:

The Ba(IO₃)₂ will precipitate out in the solution.

Explanation:

The solubility of Ba(IO₃)₂ is:

Ba(IO₃)₂(s) → Ba²⁺(aq) + 2 IO₃⁻(aq)

Where ksp, is defined as:

ksp =  [Ba²⁺] [IO₃⁻]² = 1.5x10⁻⁹

If the multiplication of  [Ba²⁺] [IO₃⁻] is higher than ksp, the reaction will produce Ba(IO₃)₂, that means the solid will precipitate.

In solution, concentration of Ba²⁺ and  IO₃⁻ are:

[Ba²⁺] = 0.050M × (5mL / 105mL) = <em>0.00238M</em>

[IO₃⁻] = 0.10M × (100mL / 105mL) = <em>0.0952M</em>

Replacing in ksp formula:

[0.00238] [0.0952]² = 2.16x10⁻⁵. As the result is higher than ksp, <em>The Ba(IO₃)₂ will precipitate out in the solution.</em>

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Answer:

Cathode: Hydrogen gas

Anode: Chlorine gas

Explanation:

① Write down the ions present in the electrolyte

Cations: Na⁺, H⁺

Anions: Cl⁻ , OH⁻

② Decide which ions are preferentially discharged.

These are the factors:

For the discharge of cations,

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(The lower the position of the cation in the reactivity series, the easier it is to be discharged)

For the discharge of anions,

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(The more concentrated the ion, the easier for it to be discharged)

- If solution is not concentrated (dilute), look at the position of the anion on the electrochemical series.

(The lower the position of the anion on the electrochemical series, the easier of it to be discharged)

In this case:

For cations, H⁺ ions are selectively discharged at the cathode as its position is lower than Na⁺ in the reactivity series.

Anion: Cl⁻ ions, being more concentrated, are selectively discharged at the anode.

☆For electrolysis,

Cation at the cathode (-ve terminal)

Anion at the anode (+ve terminal)

In summary, here's what happened at each electrode:

<u>C</u><u>athode</u>

- H⁺ selectively discharged

- ionic half equation: 2H⁺ (aq) +2e⁻ → H₂ (g)

<u>Anode</u>

- Cl⁻ ions selectively discharged

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<em>Answer:</em>

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They have general formula as follow

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Answer:

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Explanation:

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Its atomic mass is 75 amu.

Its symbol is As.

It is usually present in combine with sulfur and metals.

it is used in bronzing.

It is also used for hardening.

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