A solution may contain the following ions Ag⁺,Cu²⁺,Mn²⁺,Ca²⁺,and Na⁺.No precipitate formed when 0.10 M NaCl was added but a dark colored precipitate formed when H₂S was added to an acidic portion of the solution.After the removal of the solid the solution was made basic and more H₂S was added and a dark precipitate again formed.Treatment of the filtrate with (NH₄)₂CO₃ resulted in a white precipitate.If no further tests were made then what conclusions can you draw? A)possible ions present Cu²⁺,Mn²⁺,Na⁺ B)possible ions present Cu²⁺,Mn²⁺,Ca²⁺ C)possible ions present Cu²⁺,Mn²⁺,Ca²⁺,Na⁺ D)possible ions present Ag⁺,Cu²⁺,Mn²⁺,Ca²⁺,Na⁺

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The presence of Ag+ was not confirmed in this series of tests, so we can eliminate Choice D. The NaCl test indicates that no Cl- or Ag+ is present, so we can eliminate Choice A. The formation of a dark precipitation with H₂S in acidic solution suggests the presence of Cu2+ or Mn2+. The subsequent precipitation with the same reagent in basic solution confirms this, and suggests the presence of both Cu2+ and Mn2+ ions. The final test with (NH₄)₂CO₃ indicates the presence of Ca2+ ions. Na+ ions were present in the original solution and were not eliminated by any of the tests, so the possible ions present are: Cu2+, Mn2+, Ca2+, and Na+. Therefore, the best choice is C.

A Solution May Contain the Following Ions Ag+,Cu2+,Mn2+,Ca2+,and Na+

A Solution May Contain the Following Ions Ag⁺,Cu²⁺,Mn²⁺,Ca²⁺,and Na⁺