Question

With respect to the dissolution of PbI₂ determine:

a) the reactions involved in the dissolution of PbI₂ in water

b) Generate the expression to determine the solubility of PbI₂.

c) Calculate the maximum solubility of PbI₂ in water at 25 C since its k_sp is 7.9 x10^-9.

d) Calculate the maximum solubility of PbI₂ (ksp is 7.9 x10^-9) at 0.500 M NaI at 25 C.

e) Determine the percent difference between the solubility calculated in step d and that calculated in step c where is the result of reference.

Answer 1

ANSWER

a) Dissolution of PbI2 in water:- PbI2 may be insoluble with water at room temperature, it's solubility increasing marginally with temperature. Simply, when the ionic compound dissolve in water they dissociate into there component ions. This dissolution can either given or take out energy from the surrounding. In case of PbI2 it dissociate into Pb^2+ and 2I^- ion. This take energy with temperature to promote dissociate of lead(2) iodide.

b) PbI2(s) + H2O(l) ===$>$ Pb^2+(aq) + 2I^-(aq)

Ksp = [Pb^2+][I^-]^2 = S × S^2 = S^3 (S - Solubility)

  S = CubeRoot( Ksp) ....................... (1)

c) ​​​​​​ if at 25°C Ksp =7.9×10^-9 then S=?

From eqn 1

S = Cuberoot( Ksp) = Cuberoot (7.9 × 10^-9 )= 1.99 × 10^-3

S = 1.99 × 10^-3

d) when 0.5M NaI added in PbI2

Here, S is much less than 0.5

by PLI2 + H2o - pbet +27 S s Her Matti- 0.5 0.5 0.5 Due to common in effect the addition Nei I . Kep = 1 pet] [ 5 ] ² (s) (s+0.5)? x10.5]? • ksp - sx 0.25 с – - К. 0.25 2-3 х 202 0.25 S = 31,6 x 10 M

e) 63 × 10^-10 % difference of step c