1a) Ten drops of 1.0M Na 2 CO 3 are placed into each of two small test tubes. To

Question

1a) Ten drops of 1.0M Na2CO3 are placed into each of two small test tubes. To the first test tube, 2 drops of 0.1 M Ca(NO3)2 is added, and to the second test tube, 2 drops of 0.1M Mg(NO3)2 is added. Describe the likely observations for each test tube and state how these results could or could not be used to distinguish between the cations.

b) Six drops of 0.5M bromine in organic solvent is placed in each of two small test tubes. To the first test tube, 15 drops of 0.1NaBr is added, and to the second test tube, 15 drops of 0.1M NaI is added. Describe the likely observations for each test tube and state how these results could be used to distinguish between the anions.

Explanation / Answer

1a)

Na2CO3 = 2Na+ + CO3-2

Na+ = alkali metal ion

Ca(NO3)2 -- > Ca+2 + NO3-

Ca+2 = alkaline earth ion

note that there will be no reaction at all

Mg(NO3)2 -- > Mg+2 + NO3-

Mg+2 = alkaline earth ion

simlar to Ca+2, there is no reaction

there can't be seen any type of reaction

Q2

Bromine in organic solvent --> most likely Br2(l)

tube 1:

NaBr --> aqueous NaBr, implies Br-(aq) i.e. ionic Br-

there will be:

Br2(l) --> 2Br-(aq)

according to the Kd vlaue of extraction, some liquid forms Br- and some Br- forms Br2(l) according to the extraction coefficient

tube 2

2e- + Br2(l) --> 2Br-(aq)

2I-(aq) --> 2e- + I2(aq)

note that, Bromine has higher electronegativity, so some Br2(l) from th eorganic layer react swith I2- to form Br-(aq) ions

the I- ions form I2(aq) which is also I2(s)... will distribute according to I- and I2 trends on aqueous + organic layers

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