What is the molar solubility of silver sulfide, Ag2S? (Ksp for Ag2S is 6.0 x 10^51.)
To find the molar solubility of silver sulfide (Ag2S), we can use the solubility product constant (Ksp) expression:
Ag2S ⇌ 2 Ag+ + S2-
The Ksp expression for Ag2S is given as:
Ksp = [Ag+]^2 * [S2-]
Given that the Ksp for Ag2S is 6.0 x 10^51, we can set up the equation:
6.0 x 10^51 = [Ag+]^2 * [S2-]
Since the Ag2S dissociates completely into Ag+ and S2-, the molar solubility of Ag+ and S2- is equal. Therefore, let's call the molar solubility of Ag+ and S2- as "x".
Now we substitute the value of x in the equation:
6.0 x 10^51 = (x)^2 * (x)
Rearranging the equation:
(x)^3 = 6.0 x 10^51
Taking the cube root of both sides:
x = (6.0 x 10^51)^(1/3)
Calculating this on a calculator, we find:
x ≈ 1.03 x 10^17
Therefore, the molar solubility of silver sulfide (Ag2S) is approximately 1.03 x 10^17 mol/L.
To determine the molar solubility of silver sulfide (Ag2S), we'll need to use the given value of the solubility product constant (Ksp) for Ag2S.
The balanced equation for the dissolving of Ag2S in water is:
Ag2S (s) ⇌ 2Ag+ (aq) + S2- (aq)
The Ksp expression for this equilibrium is:
Ksp = [Ag+]^2[S2-]
Since the stoichiometric coefficients of Ag+ and S2- are both 2, the molar solubility of Ag2S can be expressed as "x" mol/L. Therefore, the equilibrium concentrations of Ag+ and S2- will also be "2x" mol/L.
Using the Ksp expression, we can substitute the equilibrium concentrations:
Ksp = [2x]^2[2x] = 4x^3
Now we can substitute the given Ksp value into the equation:
6.0 x 10^51 = 4x^3
To solve for "x," we'll take the cube root of both sides:
(6.0 x 10^51)^(1/3) = (4x^3)^(1/3)
x ≈ 3.60 x 10^17 mol/L
So, the molar solubility of silver sulfide (Ag2S) is approximately 3.60 x 10^17 mol/L.
Ag2S ==> 2Ag^+ + S^-2
Ksp = (Ag^+)^2(S^-2)
Set up an ICE chart, substitute into Ksp expression and solve for solubility.