1 through 9. not really sure how to do this. thank you. pUSSIDIE T8n is present,

Question

1 through 9. not really sure how to do this. thank you.

pUSSIDIE T8n is present, absent, or undetermined. 1. Test for Agt, Cu*, Fe Some 6 M HCl is added to a solution that may contain the three ions. A white precipitate forms. Ions present: Ions undetermined: 2. Test for Mn2, Fe*, and Cu* Some 3% H20, and 6 M NaOH are added to a pale blue solution. dissolves when 6 M H,S0, is added. Addition of solution containing a dark precipitate. The dark precipitate completely dissolves i A dark precipitate forms, which totally 6 M NH, to the solution until it is basic results in a deep blue n H,SO Ions present:-- Ions absent: Ions undetermined: 3. Test for Cu2-, Al3+, and Zn2 Addition of 6 M NH, to a solution known to contain one or more of the above ions causes the formation of a deep-blue colored solution containing a gelatinous precipitate. Ions absent: Ions undetermined: Ions present:

Explanation / Answer

Qualitative analysis of ions

1. with 6 M HCl forms precipitate of AgCl

ions present - Ag+

ions absent - Cu2+, Fe3+

ions undetermined - none

2. addition of 3%H2O2 and NaOH gives blue coloration - Cu2+ ions

forms dark precipitate with NH3 which dissolves in H2SO4 - Mn2+ ions

forms dark precipitate with NH3 which dissolves in H2SO4 - Cu2+ ions

ions present - Mn2+, Cu2+

ions absent - Fe3+

ions undetermined - none

3. 6M NH3 forms deep-blue solution with gelatinous precipitate

ions present - Cu2+, Al3+

ions absent - Zn2+

ions undetermined - Zn2+

4. 6M HCl gives bubbles

ions present - CO3^2-

ions absent - Cl-

ions undetermined - Cl-

5. acidified with KMnO4 and gave purple color to mineral oil

ions present - I-

ions absent - Br-

ions undetermined - Cl-

6. no precipitate with AgNO3 or BaCl2

ions present - NO3-

ions absent - Cl-, SO4^2-, Br-, I-, CO3^2-

ions undetermined - none

7. 2NaCl(aq) + BaCl2(aq) --> No reaction

Na2SO4(aq) + BaCl2(aq) --> BaSO4(s) + 2NaCl(aq)

8. NO3-(aq) + OH-(aq) + Al(s) ---> Al(NO3)3(aq) + CO2(g)

CO3^2-(aq) + 3OH-(aq) + Al(s) ----> Al(OH)3(s) + CO3^2-

9. K+(aq) + Cl-(aq) --> KCl(aq)

Ag+(aq) + Cl-(aq) --> AgCl(s)

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