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Flauer [41]
3 years ago
15

Solid sodium iodide is slowly added to a solution that is 0.0050 M Pb 2+ and 0.0050 M Ag +. [K sp (PbI 2) = 1.4 × 10 –8; K sp (A

gI) = 8.3 × 10 –17] Calculate the Ag + concentration when PbI 2 just begins to precipitate.
Chemistry
1 answer:
UkoKoshka [18]3 years ago
6 0

Answer:

[Ag⁺] = 5.0x10⁻¹⁴M

Explanation:

The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:

Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸

Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷

The PbI₂ <em>just begin to precipitate when the product  [Pb²⁺] [I⁻]² = 1.4x10⁻⁸</em>

<em />

As the initial [Pb²⁺] = 0.0050M:

[Pb²⁺] [I⁻]² = 1.4x10⁻⁸

[0.0050] [I⁻]² = 1.4x10⁻⁸

[I⁻]² = 1.4x10⁻⁸ / 0.0050

[I⁻]² = 2.8x10⁻⁶

<h3>[I⁻] = 1.67x10⁻³</h3><h3 />

So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:

[Ag⁺] [I⁻] = 8.3x10⁻¹⁷

[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷

<h3>[Ag⁺] = 5.0x10⁻¹⁴M</h3>

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