Respuesta :
when Ag₂S dissolves, it dissociates as follows;
Ag₂S ---> 2Ag⁺ + S²⁻
First we need to calculate molar solubility which gives the number of moles dissolved in 1 L of solution.
If molar solubility of Ag₂S is y, then molar solubility of Ag²⁺ and S²⁻ is 2y and y respectively.
ksp gives the solubility constant
ksp = [Ag⁺]²[S²⁻]
ksp = [2y]²[y]
4y³ = 8.00 x 10⁻⁵¹
y³ = 2 x 10⁻⁵¹
y = 1.26 x 10⁻¹⁷ mol/L
molar mass = 247.8 g/mol
solubility of Ag₂S = 1.26 x 10⁻¹⁷ mol/L x 247.8 g/mol = 3.12 x 10⁻¹⁵ g/L
Solubility of Ag₂S = 3.12 x 10⁻¹⁵ g/L
Ag₂S ---> 2Ag⁺ + S²⁻
First we need to calculate molar solubility which gives the number of moles dissolved in 1 L of solution.
If molar solubility of Ag₂S is y, then molar solubility of Ag²⁺ and S²⁻ is 2y and y respectively.
ksp gives the solubility constant
ksp = [Ag⁺]²[S²⁻]
ksp = [2y]²[y]
4y³ = 8.00 x 10⁻⁵¹
y³ = 2 x 10⁻⁵¹
y = 1.26 x 10⁻¹⁷ mol/L
molar mass = 247.8 g/mol
solubility of Ag₂S = 1.26 x 10⁻¹⁷ mol/L x 247.8 g/mol = 3.12 x 10⁻¹⁵ g/L
Solubility of Ag₂S = 3.12 x 10⁻¹⁵ g/L
The solubility of Ag₂S : 3,125 .10⁻¹⁵ g/L
Further explanation
Solubility (s) is the maximum amount of a substance that can dissolve in some solvents.
Ksp is the product of ions in an equilibrium saturated state
Solubility (s) and solubility constants (Ksp) of the AxBa solution can be stated as follows.
AₓBₐ (s) ← ⎯⎯⎯⎯ → aAᵃ⁺ (aq) + b Bᵇ⁻ (aq)
s as bs
Ksp = [Aᵃ⁺] ᵃ [Bᵇ⁻] ᵇ
Ksp = (as) ᵃ (bs) ᵇ
Solubility units in the form of mol/liter or gram/liter
Ksp for Silver sulfide,Ag₂S, is 8.00 × 10⁻⁵¹
Ag₂S decomposition reaction :
Ag2S ⇒ 2Ag⁺+ S²⁻
s 2s s
Ksp = [Ag⁺]² [S²⁻]
Ksp = (2s)². s
Ksp = 4s³
8.00 × 10⁻⁵¹ = 4s³
s³ = 2.10⁻⁵¹
s = 1,259.10⁻¹⁷ mol / L
Molar mass Ag₂S = 248 g/mole
s = 1,259.10⁻¹⁷ x 248 = 3,125 .10⁻¹⁵ g/L
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