Laporan Praktikum Kimia-Reaksi Pengendapan Ca, Sr, Ba

CHEMISTRY LAB REPORT

“DEPOSITION REACTION OF COMPOUNDS

OF CALCIUM, STRONTIUM AND BARIUM”

A.    Purpose:

To study the deposition of compounds of calcium, stronsium and barium.

B.     Basic Theory:

The relationship between KSP and deposition that is when the two kinds of electrolytes are mixed. Then, there will be three possibilities

1.      The result of multiplying the concentration of ions in solution is smaller than the solubility product of the electrolyte, the deposition does not occur.

2.      The result of multiplying the concentration of ions in solution is the same with the solubility product of the electrolyte, then exact solution is saturated and the deposition is started.

3.      The result of multiplying the concentration of ions in solution is larger than the solubility product of the electrolyte, the deposition is occur.

So, an electrolyte can be deposited from the solution by enlarge the concentration of electrolyte solution that will deposited, so the result of multiplying the concentration of ions there is in solution is greater than its KSP price.

In this case, the deposition is the process of sediment formation in liquid.

The following table of KSP prices:                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                 

Compounds	KSP
CACO₃	2,3 . 10⁻⁹
SrCO₃	5,4 . 10⁻⁰
BaCO₃	2,6 . 10⁻⁹
CaSO₄	2,4 . 10⁻⁵
SrSO₄	2,8 . 10⁻⁷
BaSO₄	1,08 . 10⁻⁰
CaC₂O₄	2,3 . 10⁻⁹
SrC₂O₄	3,6 . 10⁻⁵
BaC₂O₄	1,1 . 10⁻⁷
CaCrO₄	7,1 . 10⁻
BaCrO₄	2 . 10⁻⁰

•         n Na₂CO₃ = m . V = 5.10⁻ . 5.10⁻ = 25.10⁻⁵

        Na₂CO₃ ↔ 2Na⁺ + CO₃⁻

        25.10⁻⁵                25.10⁻⁵

        [CO₃⁻] =  =  = 25.10⁻ mol/L

•        n CaCl₂ = m . V = 5.10⁻ . 5.10⁻ = 25.10⁻⁵

CaCl₂ ↔ Ca⁺ + 2Cl⁻

       25.10⁻⁵            25.10⁻⁵

      [Ca⁺] =  =  = 25.10⁻ mol/L

•        CaCO₃ ↔  Ca⁺ + CO₃⁻

      Q CaCO₃ = [Ca⁺] [CO₃⁻] = 25.10⁻ . 25.10⁻ = 625.10⁻6

       KSP CaCO₃ = 4,8 . 10⁻⁹

       Q > KSP → Deposition occur. Because, the volume and all the solution is the same.      Then Q price all the solution also the same.  

C.     Tools and Materials:

1.      Measuring glass

2.      Reaction tube

3.      Rack of reaction  tube

4.      Dropped pipette

5.      Na₂CO₃ solution

6.      CaCl₂ solution

7.      SrCl₂ solution

8.      BaCl₂ solution

9.      Na₂SO₄ solution

10.  Na₂C₂O₄ solution

11.  K₂CrO₄ solution

D.    Working procedures:

1.      Prepare all the tools and materials.

2.      React 5ml 0,05 M Na₂CO₃ solution with 5ml 0,05 M CaCl₂ solution, 5ml 0,05 M SrCl₂ solution and 5ml 0,05 M BaCl₂ solution.

3.      Repeat the above experiment by replacing the Na₂CO₃ solution with Na₂SO₄ solution, Na₂C₂O₄ solution, and K₂CrO₄ solution.

4.      Observe and write down the observation result on the table!

E.     Observation result:

(5ml 0,05 M)	5ml solution 0,05 M
	CaCl₂	SrCl₂	BaCl₂
Na₂CO₃	Deposite ( white )	Deposite ( white )	Deposite ( white )
Na₂SO₄	No deposite	No deposite	Deposite ( white )
Na₂C₂O₄	Deposite ( white )	Deposite ( white )	Deposite ( white )
K₂CrO₄	Deposite ( white )	No deposite	Deposite ( yellow )

F.      Questions:

1.      Fill the following table with the deposite + colour / no deposite!

Answer:

(5ml 0,05 M)	5ml solution 0,05 M
	CaCl₂	SrCl₂	BaCl₂
Na₂CO₃	Deposite ( white )	Deposite ( white )	Deposite ( white )
Na₂SO₄	No deposite	No deposite	Deposite ( white )
Na₂C₂O₄	Deposite ( white )	Deposite ( white )	Deposite ( white )
K₂CrO₄	Deposite ( white )	No deposite	Deposite ( yellow )

2.      Explain with the matter whether deposition occur if 50cm 0,1 M SrCl₂ solution react with 50cm 0,1 M Na₂SO₄. KSP SrSO₄ = 3.10⁻⁷.

Answer:

3.      What the reagent can be used to distinguish the presence of:

a.       Ba⁺ ion with Ca⁺ and Sr⁺. Explain!

b.      Ca⁺ ion with Sr⁺ and Ba⁺. Explain!

Answer:

a.       Reagent that are used to distinguish Ba⁺ ion with Ca⁺ is Na₂SO₄. Because, Ba⁺ white deposite and Ca⁺ no deposite.

Reagent that are used to distinguish Ba⁺ ion with Sr⁺ is K₂CrO₄.

G.    Conclusion:

“ Based on the above experiment, we can get the conclusion that is:

• Barium if reacted with Na₂CO₃,  Na₂SO₄, Na₂C₂O₄, and K₂CrO₄ all are deposite.

• Strontium all are deposite except if reacted with K₂CrO₄ solution.

            • Calcium if reacted with Na₂CO₃,  Na₂C₂O₄ are deposite. But, if reacted with Na₂SO₄ and K₂CrO₄.