A student mixed 5.00 mL of 0.0120 M Pb(NO3)2 with 5.00 mL of 0.0300 M KI and obs

Question

A student mixed 5.00 mL of 0.0120 M Pb(NO3)2 with 5.00 mL of 0.0300 M KI and observed a yellow precipitate.

a. What is the molecular formula of the precipitate? PbI2

b. How many moles of Pb2+ are present initially? ___

c. How many moles of I- are present initially? ___ The concentration of I- at equilibrium is experimentally determined to be 8.0*10^-3 M.

d. How many moles of I- are present in the solution (10mL)? ___

e. How many moles of I- precipitated? ___

f. How many moles of Pb2+ remain in solution? ___

g. What is the concentration of Pb2+ in the equilibrium solution? ____ M (Equilibrium solution is still 10mL)

h. Determine Ksp of PbI2 from this data. ____

Explanation / Answer

a) the reaction is given by

Pb(N03)2 + 2KI ---> PbI2 (s) + 2KN03

here

PbI2 is the precipitate formed

b)

we know that

moles = molarity x volume (L)

so

moles of Pb+2 = 0.012 x 5 x 10-3 = 6 x 10-5

c)

now

initially

moles of I- = 0.03 x 5 x 10-3 = 1.5 x 10-4

d)

now

at equilibrium

[I-] = 8 x 10-3

so

8 x 10-3 = moles of I- x 1000 / 10

moles of I- = 8 x 10-5

e)

moles of I- precipitated = intial moles - final moles

moles of I- precipitated = 1.5 x 10-4 - ( 8 x 10-5)

moles of I- precipitated = 7 x 10-5

f)

Pb+2 + 2I- --> PbI2

we can see that

moles of Pb+2 precipitated = 0.5 x moles of I- precipitated

so

moles of Pb+2 preciptated = 0.5 x 7 x 10-5 = 3.5 x 10-5

so

moles of Pb+2 remaining = initial - moles precipitated

moles of Pb+2 remaingin = 6 x 10-5 - ( 3.5 x 10-5)

moles of Pb+2 remaining = 2.5 x 10-5

g)

now

[Pb+2] = 2.5 x 10-5 x 1000 / 10

[Pb+2] = 2.5 x 10-3

h)

now

PbI2 --> Pb+2 + 2I-

Ksp = [Pb+2] [I-]^2

Ksp = [ 2.5 x 10-3 ] [8 x 10-3]^2

Ksp = 1.6 x 10-7

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.