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Answer is: Ksp for silver sulfide is 8.00·10⁻⁴⁸.
Reaction of dissociation: Ag₂S(s) → 2Ag⁺(aq) + S²⁻(aq).
s(Ag₂S) = s(S²⁻) = 1.26·10⁻¹⁶ M.
s(Ag⁺) = 2s(Ag₂S) = 2.52·10⁻¹⁶ M; equilibrium concentration of silver cations.
Ksp = s(Ag⁺)² · s(S²⁻).
Ksp = (2.52·10⁻¹⁶ M)² · 1.26·10⁻¹⁶ M.
Ksp = 6.35·10⁻³² M² · 1.26·10⁻¹⁶ M.
Ksp = 8.00·10⁻⁴⁸ M³.

The value of solubility product constant for the Ag₂S is 8.00·10⁻⁴⁸ M³.

How we calculate solubility product constant?

Solubility product constant (Ksp) of any reaction is define as the product of the concentration of ions in the dynamic solubility equilibrium condition.

Given chemical reaction is:

Ag₂S(s) ⇄ 2Ag⁺(aq) + S²⁻(aq)

In the question given that,
Solubility concentration of ions is always equal to the concentration of their given compound. From the stoichiometry of the reaction, it is clear that:

1 mole of Ag₂S = produce 1 mole of S²⁻ ion

1 mole of Ag₂S = produce 2 mole of Ag⁺ ion

Given solubility concentration of Ag₂S = 1.26 × 10⁻¹⁶ M

So, solubility concentration of S²⁻ ion = 1.26 × 10⁻¹⁶ M

Solubility concentration of Ag⁺ ion = 2 × 1.26×10⁻¹⁶ = 2.52 × 10⁻¹⁶ M

Ksp For the reaction is calculated as:

Ksp = [Ag⁺]²[S²⁻]

On putting all values in the above equation, we gwt

Ksp = (2.52 × 10⁻¹⁶)² × (1.26 × 10⁻¹⁶)

Ksp = 8.00·10⁻⁴⁸ M³

Hence, 8.00·10⁻⁴⁸ M³ is the value of Ksp.

To know more about solubility product constant, visit the below link:

https://brainly.com/question/9807304

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