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svlad2 [7]
3 years ago
7

Calculate the silver ion concentration in a saturated solution of silver(I) sulfate (K sp = 1.4 × 10 –5). 1.5 × 10–2 M 1.4 × 10–

5 M 3.0 × 10–2 M 2.4 × 10–2 M None of the above.
Chemistry
1 answer:
spin [16.1K]3 years ago
7 0

Answer:

3.0x10⁻²M

Explanation:

Silver sulfate, Ag₂SO₄, has a product constant solubility equilbrium of:

Ag₂SO₄(s) ⇄ 2Ag⁺ + SO₄²⁻

When an excess of silver sulfate is added, some Ag₂SO₄ will react producing Ag⁺ and SO₄²⁻ until reach the equilbrium determined for the formula:

ksp = 1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]

Assuming the Ag₂SO₄ that react until reach equilibrium is X, we can replace in Ksp expression:

1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]

1.4x10⁻⁵ = [2X]² [X]

1.4x10⁻⁵ = 4X³

3.5x10⁻⁶ = X³

0.015 = X

As [Ag⁺] is 2X:

[Ag⁺] = 0.030 = 3.0x10⁻²M

The answer is:

<h3>3.0x10⁻²M</h3>
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In the reaction Zn(s) + NO3- → (aq) Zn2+ (aq) + NH4+ (aq) which element is reduced?
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Explanation:

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<em>[Rule 3: The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion; Rule 6: The oxidation state of hydrogen in a compound is usually +1]</em>

<em />

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Therefore, Zn moves from oxidation state of 0 to +2 (oxidation), while N moves from +5 to -3 (reduction).

4 0
3 years ago
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Answer:

Explanation:

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Step 3

Concentration of HPO42- produces during the dissociation of H2PO4- can be determined

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