Chemistry Experiment (Module: Shift of Equilibrium Position of a Weak Acid)
IJSO Training: Chemistry Experiment
Experiment: Shift of Equilibrium Position of a Weak Acid
Purpose:
To study the shift of equilibrium position of a weak acid:
HA(aq)
A(aq) + H+(aq)
Ka =
[A(aq)][H+(aq)]
[HA(aq)]
Principle:
In this experiment, we study the shift of equilibrium position of a weak acid by (a)
adding acid; (b) adding alkali; and (c) dilution. To observe the results, we use pH
indicators as the weak acids here. Indicator molecule has different colors for the
neutral form, HA(aq), and for the anionic form, A(aq).
Safety
Handle all chemicals with great care. Avoid direct contact
of chemicals with skin. Dispose of chemical waste, broken
glassware and excess materials according to your teacher’s
instruction.
Further information on the chemicals used in the experiment
can be found in the Material Safety Data Sheet (MSDS).
Consult your teacher for details.
EYE PROTECTION
MUST BE WORN
Procedure:
Perform the following:
(a) adding acid; (b) adding alkali; and (c) dilution
to the two indicator solutions and observe the color changes. If time allows, study the
UV-visible spectra for the solutions.
Chemistry Experiment (Module: Shift of Equilibrium Position of a Weak Acid)
Results:
Indicator #1: bromocresol green (Ka ~ 1.9 x 105 M)
Color:
Acidic (small pH)
Slightly acidic
(dissolved in water)
Adding
acid/alkali
Alkaline (large pH)
Dilution:
-----------------------
-----------------------
Further
dilution:
-----------------------
-----------------------
Indicator #2: methyl orange (Ka ~ 3.5 x 104 M)
Color:
Acidic (small pH)
Slightly acidic
(dissolved in water)
Adding
acid/alkali
Alkaline (large pH)
Dilution:
-----------------------
-----------------------
Further
dilution:
-----------------------
-----------------------
If time allows, obtain the UV-visible spectra for the solutions. Sketch the spectra
below. Write the peak positions and absorbances.
Indicator #1: bromocresol green (Ka ~ 1.9 x 105 M)
Acidic
(small pH)
Alkaline
(large pH)
Chemistry Experiment (Module: Shift of Equilibrium Position of a Weak Acid)
Slightly
acidic
(dissolved
in water)
Dilution:
Further
dilution:
Indicator #2: Methyl Orange (Ka ~ 3.5 x 104 M)
Acidic
(small pH)
Alkaline
(large pH)
Slightly
acidic
(dissolved
in water)
Dilution:
Further
dilution:
Chemistry Experiment (Module: Shift of Equilibrium Position of a Weak Acid)
Discussions:
(1)
Adding acid to a weak acid solution causes its equilibrium position to shift to
the left/right and adding alkali causes the position to shift to the left/right.
(2)
Dilution of the solution of Indicator #1 results in color change. But dilution of
the solution of Indicator #2 does not result in color change. (Color is lighter in
both cases because of dilution.)
(a)
Dilution causes equilibrium position to shift to the right. It can be
rationalized in the following way. Suppose the solution is diluted by a
factor of 2. Immediately after dilution, [HA(aq)], [A(aq)], and [H+(aq)]
are halved. The fraction, [A(aq)][H+(aq)]/[HA(aq)] becomes smaller.
The equilibrium position will shifts in such a way to make this fraction
equal to Ka again, i.e., to make this fraction increase. Therefore, the
product-side is favored to increase both [A(aq)] and [H+(aq)]; and to
decrease [HA(aq)]. In brief, the lower the concentration, the more
favorable the product-side is.
(b)
Mathematical treatment:
HA(aq)
Initial:
x0
Final:
(x0  x)
A(aq)
0
x
Ka =
+
H+(aq)
0
x
[A(aq)][H+(aq)]
[HA(aq)]
=
x2
(x0  x)
We have the quadratic equation: x2 + Ka x  Ka x0 = 0.
The (positive) solution is: x =
 Ka +  (Ka2 + 4 Ka x0)
2
This formula allows one to calculate [HA(aq)], [A(aq)], and [H+(aq)].
For example, for a 0.01 M acetic acid solution (Ka for acetic acid = 1.8 x
105 M), [H+(aq)] = 4.2 x 104 M, corresponding to pH = _____.
(c)
Using the above formula, we can study how [HA(aq)] and [A(aq)]
changes upon dilution for the 3 cases: (i) x0  Ka; (ii) x0 ~ Ka; and (iii) x0
 Ka with some numerical values.
(i) x0  Ka
x0 (M)
1 x 102
5 x 103
2 x 103
Ka (M)
1 x 106
[A(aq)]/x0
0.01
0.01
0.02
[HA(aq)]/x0 Conclusion
0.99
Mostly HA(aq)
0.99
in solution
whatever x0 is
0.98
Chemistry Experiment (Module: Shift of Equilibrium Position of a Weak Acid)
1 x 104
(ii) x0 ~ Ka
5 x 105
2 x 105
1 x 105
(iii) x0  Ka
1 x 105
5 x 106
2 x 106
5x
104
0.36
0.64
0.46
0.54
0.73
0.27
0.98
0.99
1.00
0.02
0.01
0.00
x0 for bromocresol green: ~ 5 x 105 M
x0 for methyl orange:
~ 1 x 105 M
Relative
amounts of
HA(aq) and
A(aq) change
upon dilution
Mostly A(aq)
in solution
whatever x0 is
(Ka ~ 1.9 x 105 M)
(Ka ~ 3.5 x 104 M)
Therefore, we observe color change upon dilution for bromocresol green
[similar to Case (ii)] but not for methyl orange [similar to Case (iii)].
Exercise: (if you are interested in algebra)
(a)
Determine x0 (in terms of Ka) if [HA(aq)] = [A(aq)].
(b)
Show the following:
(i) if x0  Ka, then [A(aq)]  (Ka x0)1/2 and [HA(aq)]  x0.
(ii) if x0  Ka, then [A(aq)]  x0 and [HA(aq)]  (x02 / Ka).