the molar solubility of PbS, in a 0.150 M ammonium sulfide (NH4)2S solution is? The Ksp for PbS= 3.2E-28.
...........PbS ==> Pb^+2 + S^-2
initial.....0.......0......0
change......x.......+x.....+x
final.......x........x......x
Ksp = (Pb^+2)(S^-2) = 3.2E-28
..............(NH4)2S ==> 2NH4^+ + S^-2
initial.......0.15M........0........0
change.........-0.15......0.30......0.15
final...........0..........0.30.....0.15
Substitute into Ksp and solve for PbS.
We want solubility PbS; therefore, we want to solve for x.
PbS = x
Pb^+2 = x
S^-2 = x+0.15
3.2E-28 = (x)(x+0.15).
Solve for x.
To find the molar solubility of PbS in a 0.150 M ammonium sulfide solution, we need to use the concept of the solubility product constant (Ksp) and the common ion effect.
The balanced equation for the dissolution of PbS in water is:
PbS(s) ⇌ Pb²⁺(aq) + S²⁻(aq)
According to the Ksp expression for PbS, we have:
Ksp = [Pb²⁺][S²⁻]
Given that the Ksp for PbS is 3.2 x 10^(-28), we can use this value to calculate the molar solubility of PbS.
Let's assume that 'x' is the molar solubility of PbS. Since PbS dissociates to form 1 mol of lead ions (Pb²⁺) and 1 mol of sulfide ions (S²⁻), the concentrations of lead and sulfide ions will be 'x'.
Therefore, the Ksp expression becomes:
Ksp = [Pb²⁺][S²⁻] = (x)(x) = x²
Since we are given a 0.150 M concentration of ammonium sulfide ((NH4)2S), it means the sulfide ions (S²⁻) will initially be 0.150 M.
However, when PbS dissociates, it will consume some of the sulfide ions to achieve equilibrium. Let's assume the change in concentration of sulfide ions is represented by 'y'.
Thus, the final concentration of sulfide ions at equilibrium will be (0.150 - y).
Now, we can rewrite the Ksp expression with these concentrations:
Ksp = [Pb²⁺][S²⁻] = (x)(0.150 - y)
Since the stoichiometry of the balanced equation for the dissolution of PbS is 1:1, we can conclude that the concentration of lead ions ([Pb²⁺]) at equilibrium is also x.
Now, substitute these expressions into the Ksp expression:
3.2 x 10^(-28) = (x)(0.150 - y)
We are given that the initial concentration of ammonium sulfide is 0.150 M, and we can assume that all the ammonium sulfide will be converted to sulfide ions.
To solve for 'y', we can use the initial concentration of ammonium sulfide as a limiting factor:
0.150 M = y
Now, we can substitute the values into the equation:
3.2 x 10^(-28) = (x)(0.150 - 0.150)
Simplifying further gives:
3.2 x 10^(-28) = 0
Here, we see that 0 ≠ 3.2 x 10^(-28), which means that this system is not feasible. Thus, considering the Ksp value and initial concentration of ammonium sulfide, there is no molar solubility of PbS in a 0.150 M ammonium sulfide solution.