Question

The value of ksp for silver sulfide, ag2s, is 8.00×10−51. calculate the solubility of ag2s in grams per liter.

Answer 1

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 

Answer 2

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

Learn more

the rate of solubility

brainly.com/question/9551583

influencing factors are both solubility and the rate of dissolution

brainly.com/question/2393178

The solubility of a substance

brainly.com/question/2847814