Experiment 10
Topic : Ionic equilibrium
Purpose : To determine the dissociation constant of a weak acid using a pH meter.
Apparatus : (a) One 10.0cm³ pipette and pipette filler.
(b) One 50.0cm³ burette.
(c) Four 100.0cm³ beakers.
(d) Two beakers.
(e) One white tile.
(f) A retort stands with clamp.
Materials : KA 1 is a 1.0 mol dm־³ solution of a weak acid.
Distilled water.
Buffer solution.
Procedure : (a) Prepare four 100.0cm³ beakers labeled W, X, Y, and Z. Pipette 20.0cm³
of KA 1 into beaker W. Then pipette 2.0cm³ of KA 1 from beaker W
into beaker X. Run 18.0cm³ of distilled water from the burette into the
beaker X. Stir the solution. Pipette 2.0cm³ of the solution from beaker
X into beaker Y. Add 18.0cm³ of distilled water into beaker Y. Repeat
the same procedure for beaker Z using 2.0cm of solution from beaker
Y, and then dilute. Stir the solution produced.
Using the pH meter, determine the pH value of each of the solutions
produced in beakers W, X, Y, and Z. Record the pH readings and
completed the table below.
(Note: The final volume in beakers W, X, and Y is 18.0cm³ whereas it
is 20.0cm³ in beakers Z.)
Results : (b) Record and complete your readings in the table below.
Weak acid | Molarity/ mol dm ־³ | pH | [H+] | Degree of dissociation | Dissociation constant |
W | 1.000 | 2.4 | 3.981 × 10־ | 3.981 × 10־ | 1.585 × 10־ |
X | 0.100 | 2.9 | 1.259 × 10־ | 1.259 × 10־ | 1.585 × 10־ |
Y | 0.010 | 3.4 | 3.981 × 10־ | 3.981 × 10־ | 1.585 × 10־ |
Z | 0.001 | 3.9 | 1.259 × 10־ | 1.259 × 10־ | 1.585 × 10־ |
Question :
(c) Plot a graph of pH against the molarity of the solution.
(d) What is the effect of dilution on the pH of the solution?
Since [H+] = √Ka × c , when concentration of solution decrease, the
concentration of hydrogen ion [H+], will decrease. Hence pH are higher in very
dilute and less acidic because pH = - log10 [H+].
(e) What is the effect of dilution on the degree of dissociation?
Since Ka = ca² and Ka is a constant at constant temperature, the degree of
dissociation, a, increase when the concentration decrease. This means that a
weak acid dissociates rather well in very dilute solutions.
(f) Calculate the value of the dissociation constant for each of acid solutions KA 1 of
different concentrations.
Solution W
2.4 = ־log10 [H+] , pH = ־log10 [H+]
[H+] = 10(־2.4)
= 3.981 × 10־³ mol dm־³
3.981 × 10־³ = √Ka × 1.000 , [H+] = √Ka × c
Ka = (3.981 × 10־³)² × 1.000
= 1.585 × 10־ mol dm־³
Solution X
2.9 = ־log10 [H+] , pH = ־log10 [H+]
[H+] = 10(־2.9)
= 1.259 × 10־ mol dm־³
1.259 × 10־ = √Ka × 0.100 , [H+] = √Ka × c
Ka = (1.259 × 10־)² × 0.100
= 1.585 × 10־ mol dm־³
Solution Y
3.4 = ־log10 [H+] , pH = ־log10 [H+]
[H+] = 10(־3.4)
= 3.981 × 10־ mol dm־³
3.981 × 10־ = √Ka × 0.010 , [H+] = √Ka × c
Ka = (3.981 × 10־)² × 0.010
= 1.585 × 10־ mol dm־³
Solution Z
3.9 = ־log10 [H+] , pH = ־log10 [H+]
[H+] = 10(־3.9)
= 1.259 × 10־ mol dm־³
1.259 × 10־ = √Ka × 0.001 , [H+] = √Ka × c
Ka = (1.259 × 10־)² × 0.001
= 1.585 × 10־ mol dm־³
(g) What is the effect of dilution on the dissociation constant of the acid?
The dissociation constant of the acid, Ka, is a constant at constant temperature.
Thus, the dissociation constant of the acid, Ka, is not affected by the dilution of
the acid.
Comments : The reading is slightly different with the theory because there has
parallax error in the titration technique when the experiment is carried on.
Conclution : The dissociation constant of the weak acid is 1.585 × 10־ mol dm־³.
KA 1 solution is considered as acetic acid which has the same dissociation
constant value.
1 comment:
Thanks alot! =D
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