Part D on Experiment #15, Week 2. For the sulfide solids that were formed, how w

Question

Part D on Experiment #15, Week 2.

For the sulfide solids that were formed, how would the addition of 5 drops of 3 M nitric acid affect the solid’s solubility?

D. Sulfide Precipitation Precipitation of Sulfides. We know that sodium sulfide is a strong electrolyte, so it ionizes completely. Na2S aq) 2Na (aq) S2 (aq) (R18) Cu Gaq) will form a sulfide precipitate, Cus (s) with the [S21under appropriate conditions. Important note about iron: The addition of S reduces iron(III) to iron(ID: (R19) 2Fes (aq) 3 (aq) The black iron(ID sulfide is soluble in dilute nitric acid, which allows the white-yellow sulfur precipitate to be visible.

Explanation / Answer

Group III cations include Zn2+, Al3+, Ni2+, Co2+, Mn2+, Cr3+ and Fe3+ ions. These cations produce slightly soluble sulfides and have the Ksp values more than 10-20. Hence they require high quantities of basic H2S solution to precipitate corresponding sulphide salts. The sulfides of Mn2+ and Fe2+ can be redissolvedin 3M HNO3 producing free ions and elemental sulfur. The sulfides of Ni2+ and Co2+ are soluble in a mixture of conc. HNO3 and conc. HCl. The sulfides of Cu2+ and are soluble in concentrated HNO3 (16 M).

BaS releases H2S gas when it come in contact with water. So the H2S gas will be liberated from the reaction tube way earlier than the addition of the acid and there will be no precipitate of BaS to start with. K2S is highly basic and reacts with H2O to give mixture of KOH and KSH (potassium hydrosulfide). Both KOH and KSH are soluble in water. Hence like BaS, there would be no precipitate of K2S to react with HNO3.

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