Question

Three colorless solutions in test tubes, with no labels, are in a test tube rack on the laboratory bench. Lying beside the tests tubes are three labels : 0.10 M Na2CO3, 0.10 M HCL, and 0.10 M KOH. You are to place the labels on the test tubes using only the three solutions present. Here are your tests:

A few drops of the solutions from test tube 1 added to a similar volume of the solution in test tube 2 produces no visible reaction but the solution becomes warm.
A few drops of the solution from test tube 1 added to a similar volume of the solution in test tube 3 produces carbon dioxide gas.
Identify the labels for test tubes 1, 2, and 3

Answer 1

Test tube 1 0.10 M HCL

Test tube 2 0.10 M KOH

Test tube 3 0.10 M Na2CO3

Step-by-step explanation:

From the question we are told that

A few drops of the solutions from test tube 1 added to a similar volume of the solution in test tube 2 produces no visible reaction but the solution becomes warm

Generally this warmth is as a result of a reaction between an acid and a base and the acid is 0.10 M HCL and the base is 0.10 M KOH , the heat generated is know as the heat of neutralization,

The reaction is

`HCl_((aq)) + KOH_((aq)) \rightarrow KCl_((aq)) + H_2O_((l)) + \Delta H`

We are also told from the reaction that

A few drops of the solution from test tube 1 added to a similar volume of the solution in test tube 3 produces carbon dioxide gas.

Generally carbon dioxide gas is produced is as a result of a reaction between the acid HCl and Na2CO3.

The reaction is

`2HCl -{(aq)} + Na_2 CO_3_((aq)) \rightarrow 2 NaCl _((aq)) + CO_2_((g)) + H_2O_((l))`

Hence from this explanation above we see that the solution in test tube 1 is 0.10 M HCL while solution in test tube 2 is 0.10 M KOH and then solution in test tube three is 0.10 M Na2CO3