Asked by sana
) Which of the following should dissolve the smallest amount of silver sulfide per liter, assuming no complex formation? A) 0.1 M HNO3
B) 0.1 M Na2S
C) 0.1 M AgNO3
D) 0.10 M NaNO3
E) pure water
B) 0.1 M Na2S
C) 0.1 M AgNO3
D) 0.10 M NaNO3
E) pure water
Answers
Answered by
DrBob222
You can calculate the value.
Ag2S ==> 2Ag^+ + S^-2
Ksp = (Ag^+)^2(S^-2)
Plug in Ksp and 0.1 for sulfide and calculate solubility.
Now do the same for Ksp and 0.1 for Ag^+ and calculate solubility. Which is smaller? (HNO3 you know isn't the one because HNO3 will dissolve all of the Ag2S, NaNO3 and pure water dissolve the usual amount of Ag2S (which you can calculate if you wish from Ksp.)) However, you know that EITHER B or C will be the smallest amount because both B and C contain a common ion which will shift the solubility equilibrium to the left, thus reducing the solubility of Ag2S.
Ag2S ==> 2Ag^+ + S^-2
Ksp = (Ag^+)^2(S^-2)
Plug in Ksp and 0.1 for sulfide and calculate solubility.
Now do the same for Ksp and 0.1 for Ag^+ and calculate solubility. Which is smaller? (HNO3 you know isn't the one because HNO3 will dissolve all of the Ag2S, NaNO3 and pure water dissolve the usual amount of Ag2S (which you can calculate if you wish from Ksp.)) However, you know that EITHER B or C will be the smallest amount because both B and C contain a common ion which will shift the solubility equilibrium to the left, thus reducing the solubility of Ag2S.
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