CHEMICAL REACTIONS:

Metathesis and Complexation Reactions

Tushar Khanna

Lab Instructors:

Zachary Strasser, Naurenn Rizvi

Purpose:

To observe the physical changes of chemical reactions and thus study their chemical behavior and property.

PART A1

PROCEDURE:

In a small test tube add drops of 0.1 M MgSO4 solution to drops of 0.1 M BaCl2 solution, mix the contents, and note what happens.

OBSERVATIONS:

A white colored precipitate of BaSO4 is formed.

EXPLANATIONS:

MgSO4 (aq)+BaCl2 (aq) -----( BaSO4 (s)+ MgCl2(aq)TYPE OF REACTION: Double Displacement (Precipitate)reaction

PART A2

PROCEDURE:

2 ml of saturated Na2CO3 solution was placed in a small test tube and five drops of 6M HCl were added.

OBSERVATIONS:

A brisk effervescence of Co2 is produced.

EXPLANATION:

Na2CO3 + HCl --( NaCl(aq) + H20 +CO2(g)TYPE OF REACTION:

Double displacement reaction

PART A3

PROCEDURE:

A 2 ml 0.15M solution of CH3COOH is taken in a test tube and a 2 ml solution of unknown molarity of NaOH is taken in another test tube

Ten drops of acetic acid solution were added in a third test tube in addition to two drops of phenolphthalein indicator.

To this solution NaOH was added drop by drop till there is a change in the color of the solution. The number of drops which were added are noted.

A similar experiment is now performed with NaOH instead, with NaOH now inside the fourth test tube with the indicator and acid added dropwise. Thus, 10 drops of unknown NaOH solution is placed along with two drops of phenolphthalein in the fourth test tube. To the NaOH solution standardized acetic acid acid is added drop wise until there is a color change.

OBSERVATIONS:

In the first case when the acid is in the test tube and base is added, the color of the solution changes from colorless to pink. 7 drops of NaOH were required to neutralize the acid.

In the second case when the base is in the test tube and acid is added, the color of the solution changes from pink to colorless. 13 drops of CH3COOH were used to neutralize the reaction

EXPLANATION:

NaOH + CH3COOH --( NaCH3COOH + H20

(base) (acid) (salt)

CALCULATIONS:

Mol (NaOH)= Mol (CH3COOH)

or MNaOH VCH3COOH=M CH3COOH VCH3COOHInstead of volume we use the number of drops (volume no. of drops added)

Hence,

In the first case, the number of drops used (for NaOH)= 7

Drops of CH3COOH= 10

Molarity of CH3COOH= 0.15 M

Hence the molarity of NaOH= =0.21M

In the second case, the number of drops used (for CH3COOH)= 14

Drops of NaOH = 10

Molarity of CH3COOH= 0.15 M

Hence molarity of NaOH= =0.21M

Thus by the two trials we find that the morality of NaOH is 0.21M

TYPE OF REACTION: Neutralization reaction

PART A4:

PROCEDURE:

5 drops of 0.1 M NaCl solution is added to 5 drops of 0.1 M AgNO3 solution. The contents were mixed by tapping the test tube.

OBSERVATIONS:

Solution changes from colorless to white with white precipitates of AgCl being formed.

EXPLANATION:

NaCl+ AgNO3--( NaNO3+AgCl

TYPE OF REACTION: Double Displacement precipitation reaction.

PART A5:

PROCEDURE:

STEP 1)To the AgCl prepared in part A4, a 6M NH3 was added dropwise, and the contents were mixed till the precipitate completely dissolved.

STEP 2) To this a 3 M HNO3 solution was added dropwise.

OBSERVATIONS:

STEP 1)The white precipitate becomes colorless

STEP2) The colorless solution becomes white again

EXPLANATION:

AgCl+NH3 ---( Ag(NH3)2+ + Cl-

( Colorless)

[Ag(NH3)2]Cl + HNO3 --( AgCl+TYPE OF REACTION: Complexation reaction

PART 6

PROCEDURE: To a 10 cm test tube containing 0.100 M CuSO4 solution, 10 drops of 0.100 M NaOH solution were added. The contents were mixed and observed after 90 seconds.

OBSERVATIONS: A crystalline(blue colored) solid (Cu(OH)2)is formed.

EXPLANATION: CuSO4+ NaOH----(Na2SO4+ Cu(OH)2TYPE OF REACTION: It is a double displacement type reaction

PART 7:

PROCEDURE: There are 3 steps to this experiment:

STEP 1: Two drops of 6M NH3 was added to the Cu(OH)2 precipitate obtained in the previous experiment and the contents were stirred till the precipitate dissolves completely.

STEP2: 4 drops of 0.5M HNO3 were added to the solution in the test tube and stirred. Next 2 drops of 0.5 M HNO3 solution were added and stirred again. Following this, about 10 drops of 0.5 M HNO3 was added till there was a color change.

STEP 3:A few more drops of the 0.5 M HNO3 solution was added until the precipitate dissolves and the solution becomes clear.

OBSERVATIONS:

STEP 1: A dark blue colored solution is formed.

STEP 2:A cloudy light blue colored solution is formed

STEP 3:A clear light blue colored solution is formed.

TYPE OF REACTION:

Complexation reaction.

Chemical Reactions:1.MgSO4 (aq)+BaCl2 (aq) -----( BaSO4 (s)+ MgCl2(aq)2Na2CO3 + HCl --( NaCl(aq) + H20 +CO2(g)3.NaOH + CH3COOH --( NaCH3COOH + H20

(base) (acid) (salt)4.NaCl+ AgNO3--( NaNO3+AgCl5. AgCl+NH3 ---( Ag(NH3)2+ + Cl- [Ag(NH3)2]Cl + HNO3 --( AgCl+6.CuSO4+ NaOH----(Na2SO4+ Cu(OH)2