In: Chemistry
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. ____
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