a) Write the dissolution reaction for lead(II) iodide. How many grams of lead(II

Question

a) Write the dissolution reaction for lead(II) iodide. How many grams of lead(II) iodide will dissolve in 250.0 mL of water? What are the molarities of lead(II) ion and iodide ion? Neglect any volume change due to the lead(II) iodide.   Ksp of lead(II) iodide = 1.4 x 10-8

b) How many grams of lead(II) iodide will dissolve in 250.0 mL of 0.35 M HI?

c) Will the solubility of lead(II) iodide in 250.0 mL of 0.35 M HNO3 be the same as (a) or (b) ?

d ) How would the answers to (a) and (c) have compared if the salt under consideration were lead(II) fluoride, rather than lead(II) iodide? Ksp PbF2 = 7.1 x 10-7         KaHF = 7.2 x 10-4

e ) Lead(II) ion forms a 1:1 complex with a particular ligand, L (forms PbL2+) . Write the reaction for formation of the complex ion using “L” as the ligand. If the formation constant for this complex is 2.9 x 104 , calculate the solubility of lead(II) iodide in a 1.5 M solution of L.

Explanation / Answer

A: Dissolution reaction of lead iodide:

PbI2 <---> Pb2+ + 2I-

Ksp = [Pb2+] [I-]2

where, Ksp = solubility product = 1.4 *10-8

let the amount of PbI2 dissolve i.e solubility = x

concentration of Pb2+ = x

concentration if I- = 2x

1.4 *10-8 = x*(2x)2

4x3 =1.4 *10-8

x =0.0015 M

i.e concentration of Pb2+ = concentration of I- = 0.0015 M

conc. of I- = 2x = 2*0.0015 = 0.003 M

x= is the amount of PbI2 dissolved = 0.0015 M

to convert 0.0015 M into grams:

0.0015 M = 0.0015 mol /L

Also volume of solution = 250mL =0.025 L

0.0015 mol/L * 0.025 L * (461 g/mol of PbI2 / 1 mol of PbI2) = 0.017 g

grams of PbI2 dissolved in water = 0.017 g

B:  How many grams of lead(II) iodide will dissolve in 250.0 mL of 0.35 M HI?

Since, I- conc. initially is 0.35 M from HI

So. equilibrium is like;

PbI2 <---> Pb2+ + 2I-

initial conc of Pb2+ = 0 M

initial conc of I - = 0.35 M

final conc. of Pb2+ = x

final conc. of I- = 2x+0.35

Ksp = [Pb2+] [I-]2

1.4 x 10-8 = x*(2x+0.35)2

assume x<<0.35

1.4 x 10-8 = x*0.1225

x = 1.14 *10-7 M

Solubility in HI = x = 1.14 *10-7 M

convert Molarity into gram:

1.14 *10-7 M = 1.14 *10-7 mol/L

volume of HI = 250 mL = 0.025 L

1.14 *10-7 mol/L * 0.025 L * (461 g/mol of PbI2 / 1 mol of PbI2) = 1.3*10-6 g

grams of PbI2 dissolved in HI = 1.3*10-6 g

C :

Will the solubility of lead(II) iodide in 250.0 mL of 0.35 M HNO3 be the same as (a) or (b) ?

HNO3 have no ion in common to lead iodide, thus, HNO3 will not contribute to the solubility of PbI2 like HI

Hence, its solubility x will be same as that in water

D : How would the answers to (a) and (c) have compared if the salt under consideration were lead(II) fluoride, rather than lead(II) iodide? Ksp PbF2 = 7.1 x 10-7  

Since, Ksp = [Pb2+] [F-]2

Solvent water, HI, HNO3 has no ions common to Pb2+ and F- . Therefore the solubility product would be unaffected by all the solvents and solubility of lead flouride will be same in all the three solvents.

e) Lead(II) ion forms a 1:1 complex with a particular ligand, L (forms PbL2+) . Write the reaction for formation of the complex ion using “L” as the ligand. If the formation constant for this complex is 2.9 x 104 , calculate the solubility of lead(II) iodide in a 1.5 M solution of L.

PbI2 <---> Pb2+ + 2I- ; ksp = 1.4*10-8 ; ksp = [Pb2+ ] [I-]2

Complex formation :

PbI2 + L <---> PbL2+ ; Kf = 2.9*104

Kf*Ksp = K(equilibrium constant

K = [PbL2+] /L = a / (1.5-a)

Ksp *Kf = a  / (1.5-a)

1.4*10-8 * 2.9*104 = a / (1.5) where a<<<1.5

a =0.00061 = 6.1*10-4

i.e PbI2 dissolves in 1.5 M of L solution to produce 6.1*10-4 M of complex which is equal to the solubility of PbI2 in L

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