Determination of the Cations and Anions

Part 2: Identification of unknowns

1. Two compounds were identified through the observations of their solubility.
2. It should be sufficient to decide which of the three options on the label of the unknown was contained within.

Data/Results & Analysis:

Part 1: Observation and exploration of solubility rules

Table 1

Table 2

1. Identify any precipitates:-

                 AX and CX

2. And then any soluble compounds (including original solutions):

        BW, BZ and AY

Table 3

The cations that being used were Mg2+, Ni2+ and Na2+ and anions were NO3-, Cl-, SO42- and OH-. Matching of each letter ion in Table 3 with a specific ion from the list was made.

Based on observations and conclusions, some basic “solubility rules” were formulated.

Type 1 Cation:   All salts forms of the sodium ion were always soluble.

Type 2 Cation: Solubility of salt that forms from nickel ion and magnesium ion were following the solubility of their anions.

Type 1 Anion:  Salt forms from chloride ion, nitrite ion, and sulfide ion were generally soluble. Most salts of chloride ion were soluble except for AgCl and PbCl2, all salts of nitrite ion were soluble, and most salts of sulfide ion were soluble except for SrSO4, BaSO4, and PbSO4.

Type 2 Anion:    Most salts forms from hydroxide ion were generally insoluble except the salts forms with metal (Group I) cations and hydroxide anion.

Part 2: Identification of Unknowns.

Cation identification

Possible compounds (a) Pb(NO3)2   (b) Mg(NO3)2   (c) NaNO3 

Unknown X          Identification Pb(NO3)2.

Reason:-

Precipitate was formed on some mixture, so NaNO3 was excluded as unknown X. In the test, the compound that contained anion Cl- and SO42- formed precipitate in white colour. According to the theoretical results, only Pb2+ would react with Cl- and SO42- to form precipitate while Mg2+ formed soluble compound. Thus, the unknown X was Pb(NO3)2.

Anion identification

Possible compounds (a) Ni(NO3)2   (b) NiSO4   (c) NiCO3 

Unknown Y          Identification NiCO3.

Reason:-

When unknown Y was added into MgCl2 solution, precipitate with white in colour was formed. According to the solubility rules, only CO32- ion would form precipitate with Mg2+ ion. Thus, the unknown Y was NiCO3.

Result of testing substances X and Y

Theoretically for possible compound X

Result of testing substance X with other solutions:

NiCl2 (aq)         +         Pb(NO3)2 (aq)                         PbCl2 (s)        +        Ni(NO3)2 (aq)

(green solution)                (clear solution)                (white precipitate)        (green solution)

NiSO4 (aq)         +         Pb(NO3)2 (aq)                         PbSO4 (s)        +        Ni(NO3)2 (aq)

(green solution)                (clear solution)                (white precipitate)        (green solution)

NaNO3 (aq)         +         Pb(NO3)2 (aq)                         Pb(NO3)2 (aq)        +        NaNO3 (aq)

(clear solution)                (clear solution)                (clear solution)        (clear solution)

NaOH (aq)         +         Pb(NO3)2 (aq)                         Pb(OH)2 (s)        +        NaNO3 (aq)

(clear solution)                (clear solution)                (white precipitate)        (clear solution)

MgCl2 (aq)         +         Pb(NO3)2 (aq)                         PbCl2 (s)        +        Mg(NO3)2 (aq)

(clear solution)                (clear solution)                (white precipitate)        (clear solution)

Result of testing substance Y with other solutions

NiCl2 (aq)         +         NiCO3 (s)                         NiCO3 (s)        +        NiCl2 (aq)

(green solution)                (green precipitate)                (green precipitate)        (green solution)

NiSO4 (aq)         +         NiCO3 (s)                         NiCO3 (s)        +        NiSO4 (aq)

(green solution)                (green precipitate)                (green precipitate)        (green solution)

Na(OH) (aq)         +         NiCO3 (s)                         Na2CO3 (aq)        +        Ni(OH)2 (s)

(clear solution)                (green precipitate)                (clear solution)        (green precipitate)

Na(NO3) (aq)         +         NiCO3 (s)                         Na2CO3 (aq)        +        Ni(NO3)2 (aq)

(clear solution)                (green precipitate)                (clear solution)        (green solution)

MgCl2 (aq)         +         NiCO3 (s)                         MgCO3 (aq)        +        NiCl2 (aq)

(clear solution)                (green precipitate)                (white precipitate)        (green solution)

Discussion:

        In part 1, only a small amount of solution was needed to observe whether a precipitate was formed. We must assume that only one precipitate was formed in this reaction. Firstly, small but equal quantities of the two named solution in Table 2 were mixed together. From Table 2, we could observe that not every product formed a precipitate. When observing precipitates, it was noted whether the precipitate thick or thin, its colour and how long the precipitate took to show up. All the reaction produced precipitate except for the reaction between AW and BY. Following were the products formed in this experiment:

                AW  +  BX   →   AX  +  BW             (green precipitate formed)

                CW  +  BX   →   CX  +  BW             (white precipitate formed)

                AW  +  BY   →   AY  +  BW      (no precipitate formed)

                AZ   +  BX   →   AX  +  BZ             (light green precipitate formed)

        The reaction of AW and BX produced an insoluble green product known as precipitation reaction. A precipitate is an insoluble solid formed by a reaction in solution. Precipitation reactions occur when certain pairs of oppositely charged ions attract each other so strongly that they form an insoluble solid. Precipitates often appear as milky in the solution, or they may form only a slight cloudiness, or may be black or brown in colour. This precipitation reaction also occur when CW with BX and AZ with BX were mixed together. Notice the reaction above that the cations in the two reactants exchange anions A ends up with X and B ends up with W. The chemical formulas of the products are based on the charges of the ions. One A ion was needed to give a neutral compound with X, as well as the formation of BW. It is only after the chemical formulas of the products are determined that the equation can be balanced. Such reactions are known as exchange reactions or metathesis reactions.

                

        

        First, we have to identify the precipitate between the two product formed in a reaction. After certain comparison, we can conclude that the precipitates formed in this whole experiment are AX and CX. AZ was chosen as the precipitate instead of BW or BZ in each reaction because all the reaction that produced AZ has precipitate as well. CX was considered as the precipitate instead of BW because not all the reaction that produced BW has precipitate. Thus, BW was a soluble compound. Since the reaction of CW and BX produced precipitate, then CW was the precipitate compound. Since only AW and CW were the precipitate, the rest of the product produced and the original solutions are soluble compound. The original solutions seem to have precipitates when observation was made. This statement was proved untrue because only when two or more compounds are added together only precipitate will formed. It is actually due to the supersaturation of the original solution.

        It can be concluded that the rest of the solution including the original solution were soluble compounds such as AW, AY, AZ, BW, BX, BY, BZ and CW. The solubility of a substance is the amount of that substance that can be dissolved in a given quantity of solvent. So, any substance with solubility less than 0.01 mol/L will be referred as insoluble. In those cases the attraction between the oppositely charged ions in the solid is too great for the water molecules to separates them to any significant extent, and the substance remains largely undissolved

        To complete Table 3, the ion type for each of the ions was decided. The cations used were either Mg2+, Ni2+ and Na+ and anions were NO3-, CI-, SO4- and OH-. The reason why this experiment conducted was to classify a number of ions on the basis of their general solubility. Thus, this classification would then be the conclusion drawn from the observation of the number of precipitation reactions. In this experiment, two unknown compound of its aqueous solution would be mixed together and observation was made whether or not a precipitate was formed. Unfortunately, there are no rules based on simple physical properties such as ionic charge to guide us in predicting whether a particular ionic compound will be soluble or not. Experimental observations, however have led to guidelines for predicting solubility for ionic compounds. From the observation, the cation and anion involved will be categorize using the following generic ‘solubility rules’

        In this reaction of Table 2, we can assume that made represent the general behavior of the ions. Thus, this will enables us to focus on the logical connection between observations and experimental conclusions. It is also to investigate a far broader range of ions for their precipitation behaviour, however, these observations are summarized in the well-accepted solubility rules. The aim of this section is to explore and understand these rules, not to reinvent them.

        For the cation, compound A is under type 2 cation according to this generic ‘solubility rules’ because it sometimes soluble and sometimes don’t. It depends on the anion that present in the solution as it may form precipitate with the ‘insoluble anion’ such as the ‘insoluble anion’ X formed precipitate with the A compound. For compound A, we identified it as Ni2+. We choose Ni2+ instead of Mg2+ because the precipitate formed in this reaction was green colour. Ni2+ ion gives green colour to the precipitate. For compound B, it does not form any precipitate at any reaction and all of the compound that contain B were soluble. Thus, we identified compound B as Na+ because according to the ‘solubility rules’, all salts of Na+ are soluble compound and this made them under type 1.  While for compound C, we predicted it was under type 2, same as compound A. Since the precipitate formed contain compound C was white in colour, we identified it as Mg2+. On the other hand, for anion identification, W is CI-, X was OH-, Y was NO3- and Z was SO4-. All of these anions are under type 1 of the generic ‘solubility rules’ except for OH-, which is under type 2.

        

        Brieftly, according to the basic “solubility rules” that have formulate from observations, we can conclude that for type 1 cation, all salts forms from sodium ion was always soluble. While for type 2 cation, solubility of salt that formed from nickel ion and magnesium ion are depend on the solubility of their anions. For type 1 anion, it can be concluded that, salt forms from chloride ion, nitrite ion, and sulfide ion are generally soluble. Most salts of ion chloride are soluble except for AgCl and PbCl2; all salts of ion nitrite are soluble, most salts of ion sulfide are soluble except for SrSO4, Ba SO4, and Pb SO4. However, in short, for type 2 anion, most salts forms from ion hydroxide are insoluble except the salts forms with metal (Group I) cations and hydroxide anion.

        In part two of this experiment, given three option of the unknown X solution which are Pb(NO3)2, Mg(NO3)2,  and NaNO3. Solutions from the part one, NiCl2 (AW), NiSO4 (AZ), NaOH (BX), and MgCl2 (CW) were used to identify the unknown X solution through their solubility. Each solution was allowed to react with unknown solution in a small amount in order to observe whether a precipitate was formed and observe the colour of the mixture solution. Based on physical properties of each original solution such as colour, it is sufficiently helping in deciding which of the three options the unknown solution is. The actual colour precipitate that formed was observed careful as not to confuse with the colour when one of the original solutions is colored.

Based from the observations, when unknown X reacts with NiCl2, a white precipitate in green solution obtained. As we know, all salts of Cl- are soluble compounds except for halide of Pb+ and Ag+. Therefore, in this case the PbCl2 was insoluble which form the white precipitate when Pb(NO3)2  react with NiCl2. The reaction equation as below:-

Pb(NO3)2 + NiCl2 → PbCl2 + Ni(NO3)2

        From equation, the PbCl2 is the white precipitate formed in Ni(NO3)2 which was the green solution. Moreover, all salts of NO3- are soluble compounds; so the green solution obtained is the Ni(NO3)2 compound. Another two option compounds which are Mg(NO3)2, and NaNO3 were not the identification of unknown X solution because the MgCl2 and NiCl2 are soluble compounds and will not form precipitate. Thus, it could be concluded that the unknown X solution was identified as Pb(NO3)2.

Further test was made, the unknown X solution reacts with NiSO4 and a white precipitate in green solution obtained. Yet, again the test had prove that the unknown X solution was identified as Pb(NO3)2.  The reaction equation as follows:-

Pb(NO3)2 + NiSO4→ PbSO4 + Ni(NO3)2

        From equation, the PbSO4 was the white precipitate formed in Ni(NO3)2 which was the green solution. By following solubility rule table provided all salts of SO42-are soluble except for sulfates of Sr2+, Ba2+ and Pb2+. Therefore, the white precipitate; PbSO4 is insoluble compound and the green solution of Ni(NO3)2 was a soluble compound. In this case, the MgSO4 and NiSO4 were soluble compounds which only form green solution without any precipitate formed.

        When the unknown X solution reacts with NaOH, a white precipitate in white solution obtained. According to the solubility rule table provided, most salts of OH- are insoluble compounds except for salts of NH4+ and alkali metal cations. Thus, the unknown X solution is Pb(NO3)2.  The reaction equation as below:-

Pb(NO3)2 + NaOH → Pb(OH)2 + NaNO3

        From the equation, the white precipitate formed was Pb(OH)2 and the white solution was NaNO3. The Pb(OH)2 obtained was insoluble compound because it formed white precipitate when Pb(NO3)2 reacts with NaOH. Whereas for Mg(NO3)2 and NaNO3 were not used for the identification of unknown X solution because no precipitate was obtained from the reaction with NaOH and the products formed were soluble compound.

        A white precipitate in colorless solution was obtained when the unknown X solution reacted with MgCl2. Based from the solubility rule table given; all salts of Cl- are soluble compounds except for halide of Pb+ and Ag+. Therefore, both compounds of Mg(NO3)2 and NaNO3 were not used to identify to the unknown X solution because the products formed were soluble compound which will not form any precipitate during the reaction with MgCl2.

        In this case, the PbCl2 was insoluble compound because it formed a white precipitate when Pb(NO3)2  react with MgCl2. The reaction equation shown as below:-

Pb(NO3)2 + MgCl2 → PbCl2 + Mg(NO3)

        From the equations, the Mg(NO3) is the product that form the colorless solution and it contain white precipitate which is the PbCl2 compound.

        From the reactions and observation above, it was strongly proven that unknown X solution was Pb(NO3)2 solution. The Pb(NO3)2 solution formed white precipitate when reacts with NiCl2, NiSO4, NaOH, and MgCl2. It also proven through the well-accepted solubility rule table that provided to guide in deciding which three option compound was the unknown X solution. As a conclusion, the unknown X solution was Pb(NO3)2 solution.

        In this experiment, the Na(NO3)2  which is ‘BY’ solution was not used in part two as to identify the unknown X solution. This was because, according to the solubility rule table provided all salts of NO3- are soluble compounds. Therefore, either three option solution which were Pb(NO3)2, Mg(NO3)2 or NaNO3 when react with Na(NO3)2 would form the same products. The products that formed were soluble compound that not forming any precipitate in the reaction.

        On the other hand, unknown Y was NiCO3. This was because when unknown Y was added into MgCl2 solution, precipitate with white in colour was formed. According to the solubility rules, only CO32- ion would form precipitate with Mg2+ ion. Thus, the unknown Y was NiCO3.

Precaution steps:

1. Handle the sodium hydroxide (NaOH) carefully as it is very caustic and may damage the skin.
2. Advisable to hold the test tube properly when placing it in test tube rack as to prevent the bottom of test tube to break.
3. Make sure that the test tube is totally dry and doesn’t contain any form of water because tap water contains a wide variety of ions and could result in misleading observations.
4. It is advisable not to put or to use the wrong dropper into a bottle because the entire contents of the bottle will become contaminated.
5. When mixing the solution in a test tube, it is very important that the solutions are stirred using a thin stirring rod so that the contents of the test tube do not overflow and the reactions can proceed to completion.

Conclusions:

1. Precipitation reaction is a reaction between 2 different solutions which produces an insoluble product known as precipitate.
2. Na+ is type 1 cation, where all the salts forms from this ion are always soluble.
3. Mg2+ and Ni2+ are type 2 cation, where the solubility of the salt formed from these ions are depend on its anions.
4. NO3-, CI- and SO4- are type 1 anion, where the salts formed from these ions are generally soluble.
5. OH- is type 2 anion, where the salts formed from these ions are generally insoluble.
6. Compound X is Pb(NO3)2 and compound Y is NiCO3.

References:

1. Brown, LeMay, Bursten, Chemistry- The Central Science, 9th edition, Prentice Hall, pages 117-119.
2. http://www.lpi.usra.edu/meetings/LPSC99/pdf/1922.pdf
6. http://en.wikipedia.org/wiki/Precipitation
7. http://en.wikipedia.org/wiki/Salt