Gravimetric analysis
Chemistry 321, Summer 2014
Analytical scheme:
A (aq) +
analyte
R (aq)
reagent
AR (s)
neutral complex
precipitate (ppt)
Goal: Isolate and purify ppt, before
drying and weighing. Use the mass of ppt
and rxn stoichiometry to determine the
mass of A in sample.
Analytical scheme:
Kf
A (aq) +
1
K f = [A] [R]
R (aq)
= formation constant
AR (s)
Kf needs to be large for rxn
to go to completion
Examples
Note: you must control
solution pH!
nickel ion
(analyte)
dimethylglyoxime (DMG)
(reagent)
silver ion
(reagent)
chloride ion
(analyte)
Ni(DMG)2 (solid precipitate)
silver chloride
(solid precipitate)
Can get good accuracy and precision
by gravimetric analysis
Analytic balance precision is 0.1 mg in 100 g of sample
0.1 × 10–3/100 = 10–6 uncertainty
Note that the formation constant is the
reciprocal of the solubility product, Ksp
Ksp
where Ksp = [Ag+] [Cl–]
Why is [AgCl (s)] left out?
Consider the actual equation from which Ksp is derived:
K°sp is the thermodynamic equilibrium
constant
o
sp
K = aAg+ aCl –
aAgCl
where ai = the activity of species i = fi [i]
and fi is the activity coefficient for i and [i] is measured
in molar
For a solid or a pure liquid at standard state, a = 1
+
o
sp
–
K = aAg+ aCl – = fAg+ [Ag ] fCl – [Cl ]
+
–
K sp = [Ag ] [Cl ] = K
o
sp
f Ag+ fCl –
The K°sp are tabulated at zero ionic strength (μ) where
fi = 1 so K°sp = Ksp
Conclusion: Use tabulated K°sp to perform
calculations…generally it’s okay to do so even if f is not
exactly 1; located in Table C3, pp. 806–807.
General Expression for Salt AxBy
Ksp
AxBy (s)
x Ay+ (aq) + y Bx– (aq)
Ksp = [Ay+]x [Bx–]y
Precipitation occurs from an aqueous solution
when [Ay+]x [Bx–]y ≥ Ksp
Solubility
Solid AgCl is added to pure H2O and the resulting
solution is allowed to come to equilibrium
Ksp = [Ag+] [Cl–] = 1 × 10–10 (from table)
Set up an ICE table (recall general chemistry
equilibrium) for this system
and define s = solubility of AgCl (M)
AgCl (s)
Ag+ (aq)
Cl– (aq)
initial (M)
—
0
0
change (M)
—
s
s
equil (M)
—
s
s
Ksp = s × s = s2 = 1 × 10–10
so s = [Ag+] = [Cl–] = 1 × 10–5 M
Common ion effect
A ppt is more soluble in pure water than in a solution
containing one or more of the ions in the ppt.
Example: To the saturated AgCl solution from the
previous slide, add 0.001 M AgNO3 and recalculate s
AgCl (s)
Ag+ (aq)
Cl– (aq)
initial (M)
—
0.001
0
change (M)
—
s
s
equil (M)
—
0.001 + s
s
Ksp = (0.001+ s) (s) = 1 × 10–10
(recall that Ksp is a constant!)
To avoid using the quadratic formula to solve for s,
make a simplifying assumption: s << 0.001 M (we
shall have to check this assumption)
so 0.001 + s ≈ 0.001, and (0.001) (s) ≈ 1 × 10–10
so s = [Cl–] = 1 × 10–7 M
check assumption: 1 × 10–7 M << 0.001 M
Adding Ag+ results in a significant
reduction in Cl– solubility
In 1 mM Ag+, the chloride ion is 100 times less
soluble than in pure water
Practical consequence: By adding
excess Ag+, Cl– is more effectively
precipitated, which helps reduce
measurement error
Challenge problem (next Wednesday)
The solubility of Cl– is 1 × 10–5 M with AgCl in
otherwise pure H2O.
How can gravimetry be used to determine [Cl–] in
aqueous samples if [Cl–] is thought to be ≈ 10–6 M?
Hint: The goal is to get 99.9% of the chloride ion
to precipitate (in other words, 0.1% Cl– left)
[Cl – ]left
0.1% = 0.001 = fractional error = –
[Cl ]initial sample
Apply the common ion effect – what would you add
to the sample to meet the goal?