A variety of ions including #Na^+#, #K^+#, #NH_4^+#, #Pb^(2+)#, #Fe^(2+)#, #NO_3^-#, #Cl^-#, #Br^-#, #S^(2-)#, and #CO_3^(2-)# are mixed in a beaker of water. If you analyzed the precipitates which substances would you expect to find?

Answer 1

Well, you would expect to find FIRST of all, sulfide salts of the transition and main group metal ions.......

Precipitates......#FeS(s)#, #PbS(s)#, #PbCl_2(s)#, #PbBr_2(s)#.....
Possibly some #PbCO_3(s)#, but there might be proton transfer to form #HCO_3^(-)# ions.........(and bicarbonate salts tend to be soluble).

The solubilities of these salts are extensively tabulated; how else can we know these sulfides and halides are insoluble than by experimentation?

The following general guidelines can be used to evaluate solubilities in aqueous solution:

Ammonium and all of the alkali metal salts are soluble.

Each and every nitrate and perchlorate is soluble.

All halides are soluble EXCEPT for # AgX, Hg_2X_2, PbX_2"#.

All sulfates are soluble EXCEPT for #PbSO_4, BaSO_4, HgSO_4#.

Every sulfide is insoluble, as are all carbonates and hydroxides.

The given rules follow a hierarchy. Alkali metal and ammonium salts tend to be soluble in all circumstances. The one exception to this rule is #K^(+)BPh_4# and #NH_4^(+)BPh_4#, both of which are as soluble as bricks. #Na^(+)""^(-)BPh_4# is sold as #"kalignost"#
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Answer 2

Lead ions (Pb^(2+)) forming lead salts, iron ions (Fe^(2+)) forming iron salts, sulfide ions (S^(2-)) forming metal sulfides, and carbonate ions (CO_3^(2-)) forming metal carbonates would be expected precipitates if you examined the mixture of ions in the beaker.

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Answer 3

The substances that are likely to form precipitates when the ions are mixed in water depend on their solubility rules. Here are the likely precipitates:

  1. Lead(II) ion (Pb²⁺) can form a precipitate with chloride ion (Cl⁻) to produce lead(II) chloride (PbCl₂), which is insoluble.
  2. Silver(I) ion (Ag⁺) can form a precipitate with bromide ion (Br⁻) to produce silver bromide (AgBr), which is insoluble.
  3. Silver(I) ion (Ag⁺) can also form a precipitate with chloride ion (Cl⁻) to produce silver chloride (AgCl), which is insoluble.
  4. Silver(I) ion (Ag⁺) can further form a precipitate with carbonate ion (CO₃²⁻) to produce silver carbonate (Ag₂CO₃), which is insoluble.

Therefore, when the ions are mixed in water, the substances likely to form precipitates are lead(II) chloride (PbCl₂), silver bromide (AgBr), silver chloride (AgCl), and silver carbonate (Ag₂CO₃).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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