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Mekhanik [1.2K]
3 years ago
15

HELP PLS HELP i rlly need help omg

Chemistry
1 answer:
sesenic [268]3 years ago
3 0
I think g (sorry if it’s wrong)
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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–
spin [16.1K]

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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