Question

Dr. Bunsen has a problem—he stored a rack of five bottles filled with clear colorless liquids too close to a heat vent and the labels fell off of each of the bottles. He needs your help in order to correctly identify each of the solutions. Below is a list of the solutions:

1. 0.10 M NaOH
2. 0.20 M BaCl2
3. 0.50 M H2SO4
4. 0.10 M AgNO3
5. 0.50 M Na2CO3

The bottles are now labeled ―A, B, C, D, and E‖. Match the letter on the bottle to the correct formula for the solute.
Phenolphthalein will be available for testing

***Please help me figure out a procedure for identifying which solution is which. Please include procedure steps and suggest tables needed for data collection.

Answer 1

Solution-

Given-

1. 0.10 M NaOH
2. 0.20 M BaCl2
3. 0.50 M H2SO4
4. 0.10 M AgNO3
5. 0.50 M Na2CO3

Phenolphthalein indicator is used in acid–base titrations. It becomes colourless in acidic solutions and pink in basic solutions.

We can identified the unknown solution by following number of test

Test-1 -We can add phenolphthalein to each solution (Labeled-A, B, C, D & EII) respectively.

Two solution of them becomes pink

Lets say those solution were A and D

                                       A                   B                     C                     D                 EII

Phenolphthalein       pink           Colourless   Colourless    pink       Colourless

(Indicator)

From above solution (1, 2, 3, 4, 5) two of them are basic. Then we know solution A and D are 1 and 5. But we don’t know which is 1 & 5.
Test-2- To a solution of A and D, was added a few drops of B, C & EII. Effervescence of CO2 was observes only in acid and carbonate. When acid and carbonate react to each other they produce carbon dioxide.

We can assume A and B produce the effervescence of CO2.

A was basic, then it must be the carbonate, and B must be the acid

Now we distinguish between A, B and D solution.
A = Na2CO3…………………5

B = H2SO4……………………3

D = NaOH………………….....1

Test-3- The remaining solution C and EII should be 2 & 4

We know the test of AgNO3

AgNO3 + NaOH → Brown precipitate of silver oxide

Then D was added to C and EII. One which is form brown precipitate was 4 (AgNO3) let’s say it was C and remaining will be EII

In this way we can identified labelled bottle (A, B, C, D, & EII)

A = Na2CO3…………………5

B = H2SO4……………………3

C = AgNO3…………………...4

D = NaOH………………….....1

EII = BaCl2………………………2

Test-4- Confirmatory test for solution.

i)BaSo4 is insoluble

e.g. BaCl2 + H2SO4 → White precipitate

        (EII)            (B)

ii) AgCl is insoluble

AgNO3 + BaCl2 → White precipitate

(C)           (EII)

iii) H2SO4 + NaOH + phenolphthalein → pink colour disappear at end point