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Mars2501 [29]
3 years ago
13

The solubility of silver sulfate (Ag2SO4), in moles per liter, can be expressed in terms of the resulting ion concentrations. Wh

ich relationship is correct?
Chemistry
1 answer:
lisabon 2012 [21]3 years ago
8 0

Answer:

Solubility = [SO₄²⁻]

Explanation:

Solubility of Ag₂SO₄ can be determined by its ksp equilibrium:

Ag₂SO₄(s) ⇄ SO₄²⁻ + 2Ag⁺

<em>Where 1 mole of silver sulfate dissolves producing 1 mole of sulfate ion and 2 moles of silver ion.</em>

Solubility is defined as the amount of solid that can be dissolved per liter of solution.

The moles of Ag₂SO₄ dissolved are equal to moles of SO₄²⁻.

That means:

<h3>Solubility = [SO₄²⁻]</h3>
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Answer:

CrO₂ --------------------> Cr⁴⁺ and O²⁻

VCO₃ -------------------> V²⁺ and CO₃²⁻

Cr₂(SO₄)₃ -------------> Cr³⁺ and SO₄²⁻

(NH₄)₂S ----------------> NH₄⁺ and S²⁻

Explanation:

Within ionic compounds, the cation is listed first, followed by the anion. Some of the ions are polyatomic, meaning they are covalently bonded to other elements. Polyatomic ions always have a specific charge.

All of these ionic compounds have an overall charge of 0. As such, the charges of the cations and anions must balance out. In order to do so, there are some compounds which have more than one atom of each ion.

2.) CrO₂

------> Oxygen (O) always forms the anion, O²⁻.

------> Therefore, if there are 2 oxygen anions, the chromium (Cr) must have the cationic form of Cr⁴⁺.

------> +4 + (-2) + (-2) = 0

3.) VCO₃

------> Carbonate (CO₃), a polyatomic ion, always has the state CO₃²⁻.

------> If there is only one atom of each ion, the charges must perfectly balance, making vanadium (V) be the cation V²⁺.

------> +2 + (-2) = 0

4.) Cr₂(SO₄)₃

------> Sulfate (SO₄), a polyatomic ion, always has the state SO₄²⁻.

-------> The only way the charges could balance out is if the chromium (Cr) is in the cationic form Cr³⁺.

------> +3 + 3 + (-2) + (-2) + (-2) = 0

5.) (NH₄)₂S

------> Ammonium (NH₄), a polyatomic ion, always has the state NH₄⁺.

------> Sulfur (S) always forms the anion S²⁻.

------> +1 + 1 + (-2) = 0

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