Qualitative Analysis: Charting Help! use a flow chart (Fig. 1) to help you separ

ID: 877969 • Letter: Q

Question

Qualitative Analysis: Charting Help!

use a flow chart (Fig. 1) to help you separate and identify the cations in your system. The first part of the chart has been left blank. Using the information in the paragraphs below, propose the steps to fill in the flow chart to isolate the chloride group (Ag+, Hg22+ and Pb2+) from a mixture, separate each ion from the others and confirm the presence of each ion.

Background:

Silver, mercury(I) and lead(II) are often called the "chloride" group because they form sparingly soluble to insoluble precipitates with chloride ions. All three solids are white. The first step in isolating these ions from a solution is to add HCl to form the chloride precipitates. Silver and mercury(I) chlorides are much less soluble (Ksp values of 1.8 x 10-10 and 1.3 x 10-18, respectively) than lead(II) chloride (Ksp of 1.6 x 10-5). If solid PbCl2 is heated in water to 100 oC for a few minutes, it will dissolve. The other two chlorides will not. Lead(II) in solution will form an insoluble white precipitate when allowed to react with sulfuric acid (Ksp = 6.3x10-7). The addition of ammonia to solid silver chloride causes the formation of a colorless silver-ammonia complex ([Ag(NH3)2]+, Kf = 1.7 x 107). The addition of nitric acid will cause the equilibrium to shift to free the silver which ca

n then react with the chloride again. Mercury(I) chloride reacts with ammonia to form Hg (metallic liquid), HgNH2Cl (s, white), and Hg2O (s, black). The solid mixture will have an overall grayish color.

Cation Determination

Before examining an unknown mixture it is helpful to observe the behavior of known ions in a mixture. You will separately analyze two known mixtures of cations using the procedures outlined in Fig. 1. Mixture A contains silver(I) (Ag+), mercury(I) (Hg22+), aluminum (Al3+), barium (Ba2+) and potassium (K+) ions. Mixture B contains lead(II) (Pb2+), iron(III) (Fe3+), nickel(II) (Ni2+), magnesium (Mg2+) and zinc (Zn2+) ions. There will also be solutions available that contain the individual ions that you will be analyzing. You can use these solutions to confirm the behavior of the ions in your mixture.

For each sample, you should record every step of the analysis and your observations as you proceed in a table similar to the one shown as Table 1. You should be as specific as possible when describing your observations of the known mixtures so that you can use those observations to identify your unknowns.

Table 1. Sample data table for recording the results of each experimental step.

The first steps in the cation procedure are to identify and remove the chloride group as described above. After these have been isolated, the remaining cations will be separated based on their reactions with hydroxide. The reactions of hydroxide ions with cations are very interesting. By carefully controlling the pH of the solution, only certain metal hydroxides can be caused to precipitate from solution or form soluble complexes. After the chloride group ions are precipitated with hydrochloric acid, the solution will be acidic. An ammonia/ammonium buffer is then created in order to make it just neutral, which will shift the hydroxide concentration of the solution causing the precipitation of only the most highly insoluble hydroxides, Fe(OH)3 (Ksp = 1.6 x 10-39) and Al(OH)3 (Ksp = 3 x 10-34). It is important that the solution is not made overly basic as the additional hydroxide will cause the [Al(OH)4]- complex to form too soon. The remaining ions will either form soluble complex ions with the added ammonia or remain dissolved in solution. After separation from the supernatant, the aluminum hydroxide can be re-dissolved by increasing the concentration of hydroxide ions with the addition of sodium hydroxide. This addition will favor the formation of the complex, [Al(OH)4]- (Kf = 2.0 x 1033). However, the iron(III) hydroxide will not re-dissolve. Aluminum can be confirmed by adding aluminon, a dye, and then making the solution alkaline with concentrated ammonia. The presence of a pink lake (dyed precipitate) suspended in solution confirms the presence of aluminum. Make sure that it is not the solution itself that is pink by centrifuging. The presence of Fe3+can be confirmed in two ways. Iron forms a red complex with SCN- and a blue solid, KFe[Fe(CN)6], upon the addition of K4Fe(CN)6.

After the iron and aluminum are removed from the solution, the hydroxide concentration can be manipulated, again, to selectively precipitate two cations. An increase in the concentration of hydroxide ions with the direct addition of aqueous sodium hydroxide will cause the precipitation of nickel hydroxide (Ksp = 6 x 10-16) and magnesium hydroxide (Ksp = 6 x 10-10). The other cations will remain in solution; zinc as the hydroxide complex [Zn(OH)4]- and Ba2+ and K+ as the solvated ions. Like most hydroxides, magnesium and nickel hydroxide can be dissolved in a solution that is acidified and warmed. After re-establishing the ammonia buffer, the addition of Na2HPO4 will cause the magnesium to slowly precipitate as MgNH4PO4. Nickel will remain in solution in the form of a nickel ammonia complex. The magnesium can be redissolved in hydrochloric acid. The magnesium will form a blue lake in an alkaline solution containing the organic compound 4-(p-nitrophenylazo)-resorcinol. Dissolved nickel ions will form a deep pink precipitate upon the addition of another organic compound, dimethylglyoxime.

The only ions remaining in solution after the hydroxide concentration is raised are zinc, barium and potassium. Barium forms an insoluble precipitate with sulfate ions (Ksp = 1.1 x 10-10). The barium can be further confirmed by the presence of a persistent green flame test. Zinc can be precipitated by the addition of phosphoric acid, H3PO4 (Ksp for Zn3(PO4)2 is 5 x 10-36).

At this point, the only unknown ion remaining in solution will be potassium. Potassium forms very few insoluble precipitates. The simplest way to identify it is by a flame test after other ions are removed. The flame will turn a fleeting violet color when exposed to potassium ions. Because this color may be masked by the orange flame of sodium ions, the flame should be viewed through a thickness of cobalt blue glass.

Step
Number Procedure Observations Add 0.5 mL 20% NH4Cl and neutralize with 6 M NH,, Heat in bath for 3 minutes Add 5 drops of6 M HCI Cool in ice for 1 minute Fill in the flow chart for the chloride group efore coming to lab Supernatant All other ions Fe(OH)3. A(OH)2 Rusty, white Supernatant All other ions Wash precipitate 2 | Add 5 drops of 6 M NaOH. Fe(OH)3 Al(OH)4 Add 1-2 drops aluminon Add 5-10 drops of 6 M HCI and warm in bath (if needed) until solid dissolves. Make alkaline with concentrated NH Fe3+ Centrifuge to confirm pink lake Dilute with 1-2 mL of water and divide into two portions Add several drops of 0.25 M K,Fe(CN)s- Add several drops of 0.1 M KSCN. Fe(SCN)2. Red Blue

Explanation / Answer

Answer:

Reaction with HCl

Reaction with NH3 followed by heat

On reaction with HCl only Ag, Hg and Pb forms metal chlorides seperated out.

Fe and Al forms precipitate gets separated out.

Which on warm gets Al separated while Fe warm HCl gets seperated

Ni, Mg forms precipitate gets separated.

On heating withNa2HPO4 Mg gets separated

Reaction with H2SO4

Followed by heated up with H3PO4

Ba forms precipitate gets separated.

Zn forms precipitate gets separated.

Step Procedure Observations 1