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Prelab for Determination of an Ionic Compound SD

front 1

What famil(y)ies of the periodic table is/are being investigated in this experiment?

back 1

Halogens AND Alkaline earth metals

front 2

Use the following chemical equation to answer the following question

X2 + 2Y 1 minus ----> 2X 1 minus + Y2

What species is being reduced?

back 2

X2

front 3

In solubility test section of the experiment you will be determining the solubility of various alkaline earth salts. What anions will you be using for your solubility test?

back 3

SO4 2 minus, IO3 minus, C2O4 2 minus and CO3 2 minus

front 4

In a test tube you add 2 mL of saturated chlorine water (Cl2) followed by about 1 mL of an organic solvent such as heptane (or cyclohexane). The test tube is shaken and you notice that the heptane layer has a yellow color. To this mixture you then add 1 mL of 0.1 M N a I. After shaking you notice a color change from yellow to purple in the heptane layer. What can you conclude about Cl and I from this experiment?

Information: Cl2 is pale yellow; I2 is purple; I minus (N a I) and Cl minus (N a Cl) are colorless.

Hint: For A2 + 2 B 1 minus ---> B2 + 2 A 1 minus

If a reaction occurs, A is more active than B.

If a reaction does not occur, B is more active than A.

back 4

Cl is more active than I because Cl2 reacted with I minus to form I2 and Cl minus

Ex.

By sholanejo on Brainly, https://brainly.com/question/15720257

The organic compounds (heptane/cyclohexane) will get dissolve in Cl₂ water forming a yellow colored solution.

When NaI is added into it, the following reaction between NaI and Cl2 takes place;

Cl₂ + NaI ⇒ I₂ + NaCl

In above reaction, oxidation state of Cl is getting reduced from 0 in Cl2 to -1 in NaCl.

Therefore, Cl2 is getting reduced and acting as an oxidizing agent in the reaction. While I⁻ is getting oxidised annd acting as a reducing agent.