Determination of the Solubility Product Constant of a Saltby njose528
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General
Determination
Date
Chemistry
of
the
Performed:
II
Solubility
Lab
Product
March
(CHEM
Constant
of
1,
a
1106)
Salt
2011
INTRODUCTION
If solid KHC4H4O6 is added to a beaker of water, the salt will begin to dissolve. The amount of
solid diminishes, and the concentrations of K+(aq) and HC4H4O6-(aq) in the solution increase.
When no KHC4H4O6 dissolves, the concentrations of K+(aq) and HC4H4O6-(aq) will not
increase further and any additional KHC4H4O6 added after this point will remain as a solid. At
this point, the solution is said to be saturated. The rates at which KHC4H4O6 is dissolving and
reprecipitating are equal in a saturated solution, so that no net change is observed. This process
is another example of dynamic equilibrium. The equation for potassium hydrogen tartrate
dissolving in water is: KHC4H4O6 (s) K+(aq) + HC4H4O6-(aq). For the generalized chemical
reaction aA + bB cC + dD, the equilibrium constant expression is keq = [C]c[D]d/[A]a[B]b. The
value of the constant that reflects the solubility of a compound is referred to as the solubility
product constant, or the ksp. The solubility product constant can be calculated using the
equation: Ksp = [k+][HC4H4O6-]. On the other hand, the solubility of a compound refers to the
concentration of a solute in equilibrium with undissolved solute in a saturated solution. In this
experiment, the solubility and the solubility product constant in water of KHT was determined.
The concentration of HC4H4O6- in solution can be determined by the direct titration of a
filtered solution with a standard sodium hydroxide solution, since it is a slightly soluble salt.
Because one mole of sodium hydroxide neutralizes one mole of HT- ions, the number of moles
of sodium hydroxide used in the titration will equal the number of moles of HT-. The
concentration of HT- can be determined by dividing its moles by its volume. The concentration
of HT- will be equal to the concentration of K+, which in turn will be equal to the solubility of
KHT.
From
there,
the
Ksp
can
be
determined.
PROCEDURE
1.
10
mL
of
filtered
KHT
was
measured
in
a
beaker.
2. 50 mL of distilled water was added to the KHT and placed in a beaker. 3. 2 drops of
phenolpthalein was added to the mixture in the beaker. 4. A buret was then filled with 0.096M
NaOH.
5. NaOH solution was then used to titrate the KHT mixture, until the first faint, but permanent,
pink colored appeared. 6. A second determination by titration, using the same exact directions,
as above, was performed. 7. Finally, the average ksp of KHT in distilled water, and the average
solubility
of
KHT
was
determined.
DATA
AND
CALCULATIONS
TABLE
I
Temperature of filtered KHT solution, 25ºC | Trial 1 | Trial 2 | Concentration of NaOH solution,
mol L-1 | .096 M | .096M | NaOH solution Final buret reading, mL | 19.6 | 22.7|
Initial
buret
reading,
mL
|
16|
19|
Volume
of
NaOH
used,
mL
|
3.6|
3.7|
Number of moles of NaOH used | 3.6 x 10-4 | 3.6 x 10-4|
Number of moles of HT- titrated | 3.6 x 10-4| 3.6 x 10-4| [HT] in KHT solution, mol L-1| 3.6 x
10-2|
3.6
x
10-2|
[K+]
in
KHT
solution,
mol
L-1|
3.6
x
10-2|
3.6
x
10-2|
Solubility
of
KHT,
mol
L-1|
3.6
x
10-2|
3.6
x
10-2|
Average
solubility
of
KHT,
mol
L-1|
3.6
x
10-2|
Ksp|
1.30
x
10-3|
1.30
x
10-3|
Average
Ksp
|
1.30
x
10-3|
DISCUSSION
As already mentioned, the equation for KHT dissolving in water obeys the law of equilibrium
and can be written as: KHC4H4O6 (s) K+(aq) + HC4H4O6-(aq). From the calculations above, the
average solubility product constant value was determined to be 1.30 x 10-3M. The solubility
product constant pinpoints the extent to which a compound dissolves. A very small solubility
product value indicates that the compound is insoluble. Similarly, a very small solubility product
constant value indicates that the compound will form a precipitate. Conversely, a very large
solubility product constant value indicates that the compound is very soluble and will not form
a precipitate. Here, the value indicates that potassium hydrogen tartrate is slightly soluble. The
solubility of potassium hydrogen tartrate is .677 grams per 100 mL. The literature value for its
solubility is .650 grams per 100 mL. There is a 4.15% error difference, which indicates that the
values are fairly correct. The percentage error could be attributed to inaccurately reading the
pipette during titration. Furthermore, if any of the instruments were not thoroughly cleaned
this
could
have
also
added
to
the
experimental
error.
CONCLUSION
The solubility product constant is the equilibrium constant for solids dissolving in a solvent. In
this experiment because of stoichiometry, one mole of sodium hydroxide neutralizes one mole
of HT- ions. The number of moles of sodium hydroxide used in the titration was equal to the
number of moles of HT-. Thus, the concentration of HT- was equal to the concentration of K+,
which in turn was equal to the solubility of KHT. Using the concentrations of K+ and HC4H4O6-,
the
ksp
for
KHT
was
determined.
Bibliography
* "Determination of the Solubility Product of a Salt" Signature Labs Series. Mason: Cengage
Learning, 2009. 143-46. * Kotz, Treichel, and Townsen. "Solubility of Salts." General Chemistry.
Brooks/Cole. 634-635. Print.