Respuesta :
the volume of I- is missing in your question by assuming it = 1L
moles I- = molarity * volume
= 0.00237 * 1 L
= 0.00237 mol
[I-] = moles / total volume
= 0.00237 / 1.625L
= 0.00146 M
moles Pb2+ = molarity * volume
= 0.00785 * 0.625 L
= 0.0049 mol
[Pb2+] = 0.0049 / 1.625L
= 0.003 M
when PbI2(s) ↔ Pb2+(aq) + 2I-(aq)
when Q = [Pb2+][I-]^2 and we neglect [PbI2] as it is solid
∴ Q = 0.003 * (0.00146)^2
= 6.4 x 10^-9
by comparing the value of Q with Ksp value we will found that:
Q < Ksp which mean that more solid will dissolve, and this is an unsaturated solution which has ion concentrations < equilibrium concentrations, so the reaction will go forward until achieving equilibrium.
moles I- = molarity * volume
= 0.00237 * 1 L
= 0.00237 mol
[I-] = moles / total volume
= 0.00237 / 1.625L
= 0.00146 M
moles Pb2+ = molarity * volume
= 0.00785 * 0.625 L
= 0.0049 mol
[Pb2+] = 0.0049 / 1.625L
= 0.003 M
when PbI2(s) ↔ Pb2+(aq) + 2I-(aq)
when Q = [Pb2+][I-]^2 and we neglect [PbI2] as it is solid
∴ Q = 0.003 * (0.00146)^2
= 6.4 x 10^-9
by comparing the value of Q with Ksp value we will found that:
Q < Ksp which mean that more solid will dissolve, and this is an unsaturated solution which has ion concentrations < equilibrium concentrations, so the reaction will go forward until achieving equilibrium.
The value of Ksp of the given reaction is more so that precipitate will not form in the solution.
How do we know precipitate will form?
Precipitate will form in the reaction if the value of Qsp > Ksp and continue will form till Qsp = Ksp, and if the value of Qsp < Ksp then precipitate will not form in the solution.
Value of Qsp for PbI₂ will be calculated as:
Qsp = [Pb²⁺][l⁻]² / [PbI₂]
For this first we have to calculate the concentration of Pb²⁺ & l⁻ and we have to know the total volume of the solution and will be calculated as:
M₁V₁ = M₂V₂
M₁ = molarity of NaI = 0.00237M
V₁ = volume of NaI = ?
M₂ = molarity of Pb(NO₃)₂ = 0.00785M
V₂ = volume of Pb(NO₃)₂ = 625mL = 0.625L
V₁ = (0.00785)(0.625) / (0.00237) = 2.07 L
Total volume of the solution = 2.07 + 0.625 = 2.695 L
Given chemical reaction is:
2NaI(aq) + Pb(NO₃)₂ → PbI₂ + 2NaNO₃
NaI & Pb(NO₃)₂ both are strong electrolytes and fully dissociates into their ions, so their concentration is equal to their ions.
Moles of NaI = (0.00237)(2.07) = 0.0049059 moles
Concentration of NaI in the solution = 0.0049/2.695 = 0.00182
Moles of Pb(NO₃)₂ = (0.00785)(0.625) = 0.004906 moles
Concentration of Pb(NO₃)₂ in the solution = 0.004906/2.695 = 0.00182
Value of Qsp = [0.00182][0.00182]² / 1
Qsp = 6.02 × 10⁻⁹
Given value of Ksp = 1.40 × 10⁻⁸
Hence, valiue of Qsp < Ksp.
To know more about Qsp, visit the below link:
https://brainly.com/question/16053682
