Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

Example 4- E:

Equilibrium concentration after precipitation of lead nitrate (PbNO 3 )

We wish to determine the equilibrium concentrations of Pb2+ and I- ions formed after precipitation of PbI 2

in a solution formed by mixing 100.0 mL of a 0.0500 M Pb(NO 3 ) 2 solution and 200.0 mL of a 0.100 M

NaI solution.

Initially it must be determined whether or not solid PbI 2 is formed during the mixing of the solutions. The

solubility product Ksp for PbI 2 is 1.4 · 10-8 M^3. The concentrations of Pb2+ and I- after the mixing is easily

achieved as:

 M

L

L mol L

Pb^201. 67103

0. 300

^0.^1000.^0500 /

and

 M

L

I L mol L^2

(^00). 300 6.^6710
0.^2000.^100 /
Hereby the ion product for PbI 2 is determined as:
Q = [Pb2+] 0 · [I-] 02 = 1.67 · 10-3 M · (6.67 · 10-2 M)^2 = 7.43 · 10-5M^3
As in this case Q is larger than Ksp precipitation of PbI 2 (s) will happen. In order to determine the
concentrations of Pb2+ and I- ions after precipitations, we set up initially a before and an after scheme on
the basis of moles following the principles below:
Pb2+(aq) + 2 I-(aq)  PbI 2 (s)
Before
precipitation:
0.100 L · 0.0500
mol/L = 5.0 · 10-3
mol
0.200 L · 0.100 mol/L =
2.0 · 10-2 mol
After
precipitation:
0 mol 2.0 · 10-2 mol – 5.0 · 10-3 mol
= 1.0 · 10-2 mol
The amount
of precipitated
matter will
not affect the
equilibrium
Now we let the equilibrium take place. At equilibrium the concentration of [Pb2+] is not zero as indicated
above because the precipitation reaction does not proceed until all the lead ions are used. One may
imagine that when PbI 2 precipitated a small amount of PbI 2 is simultaneously dissolved such that an
equilibrium is set up. As there is an excess of I- ions the small amount of re-dissolved PbI 2 is dissolved
into a solution containing 1.0 · 10-2 mol I- in 300.0 mL solution. This gives a concentration of I- of:
 M
L
mol L
I^2
2
3. 3310
0. 300
1.^010 /
Hereafter our problem may be formulated as: What is the solubility of solid PbI 2 in a 3.33 · 10-2 M NaI
solution? As earlier examples we start by writing the initial concentrations:
Equilibrium