# Solubility Product Constant

Length: 490 words (1.4 double-spaced pages)

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Solubility Product Constant

Objective: To determine in the laboratory the solubility constant
(Ksp) of Calcium Sulfate(CaSO4)

Materials and Instruments

* Burette

* Pipette

* Clamp

* Glass rod

* Digital balance

* Teflon stopper

* Stopcock

* Sheet of black paper

* 0.050 M CaCl2

* 0.050 M Na2SO4

* Distilled water

Procedure:

1. With the Molarity formulua, calculate how many gr of CaCl2 are
needed in order to form a 0.050 M CaCl2 solution with 500 mL of
water

2. Repeat the calculations with the right numbers for Na2SO4

3. Weight the compounds, separately, and get the amount you need
in order to prepare your solution

4. Prepare both solutions. Mix the indicated weight of each
compound with 500 mL of water. Use different flasks for each
compound

5. With the graduated cylinder measure 100 mL of 0.050 CaCl2. Pour

6. Fill a clean, dry burette with 0.050 M Na2SO4. Drain burette
until no bubbles are left in the tip. Record the initial volume
of Na2SO4 of the burette in a data table

7. Lay a sheet of black paper beneath the burette, place the flask
with the CaCl2 solution on top

8. Add Na2SO4 to the flask from the burette and gently swirl the

MLA Citation:
"Solubility Product Constant." 123HelpMe.com. 23 Apr 2019
<https://www.123helpme.com/view.asp?id=150027>.

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### Popular Essays

Close stopcock when the solution in the flask begins to
form a white precipitate.

9. Record the final volume of Na2SO4 in the burette.

10. Repeat stedp 3 to 6 for a second trial.

Calculations:

0.050 M solution Na2SO4

0.050 M solution CaCl2

CaSO4 â†” (Ca2+) + (SO42-)

Ksp = (Ca2+)(SO42-)

CaCl2 + Na2SO4â†” CaSO4 + NaCl Equilibriumâ†’ 1st Cristal of CaSO4 formed

(Tritation)

Solubility CaSO4 = 0.209 g/100mL

Molarity: moles solote

Liters solution

0.050 M = moles solote

0.500 L solution

Moles solote = 0.025 m

0.025 mol NaSO4 x 142 gr Na2SO4 = 3.55 gr Na2SO4

1 mol Na2SO4

0.025 mol CaCl2 âˆ™ 2H2O x 146 gr CaCl2 âˆ™ 2H2O = 3.65 gr CaCl2 âˆ™ 2H2O

1 mol CaCl2 âˆ™ 2H2O

Trial 1

Trial 2

Inicial vol Na2SO4 (mL)

100

100

Final vol Na2SO4 (mL)

61

60

Volume Na2SO4 used (mL)

39

40

Total Volume of solution

139

140

Average

139,5 mL

Molarity = moles solote

Liters of solution

0.050 M = moles solute (Na2SO4)

0.0395 liters solution

Moles solute (Na2SO4) = 1.98 x 10 -3

Molarity = 1.98 x 10 -3 moles

0.1395 liters solution

Molarity = 0.014193548 moles CaCl2

M = 1.42 x 10 -2

Ksp = (Ca+)(SO4)

Ksp = )1.42 x 10 -2)(1.42 x 10 -2)

Ksp = 2.0164 x 10 -4

Ksp = 2.02 x 10 -4

Sources of error:

Â· Burette: not volumetric, not precise measure

Â· Material could have been lost during the transfer of the compound

Â· Because of the minimal reaction, almost unnoticeable, the endpoint
of the experiment was very difficult to visualize

Â· A non-volumetric flask was used to measure CaCl2

Â· Mostly non-volumetric glasses were used throughout the experiment

Conclusion

The precipitate formed as a resultant of the experiment was very
difficult to visualize, it was barely seen, making the experiment hard
to prove. Sources of error were present through out the experiment,
and influenced the final solubility constant, by our calculations, the
solubility constant was 2.02 x 10 -4, when it should have been 2.4 x
10 -5. Considering the conditions in which the experiment was made and
the materials used, the results were satisfactory.