Syuzanna Abrahamyan Chem 102 11/29/18 How can you use pH to separate Ni(II) ions from Hg(II) ions? The solubility product of HgS is roughly 2 x 10-53 and that of NiS is 3 x 10-19. If we wanted to separate these two ions by a differential precipitation, how might I do that? Consider the use of hydrosulfuric acid as your precipitating agent. Hg(II) is a Group II metal ion, whereas Ni(II) is a Group IV metal ion. The Group II metal ions are precipitated as their sulphides by passing H2S in presence of diluted HCl. The addition of HCl before passing H2S is necessary to ensure that Group IV metal ions are not precipitated as their sulphides. H2S is a weak electrolyte H2S 2 H + + S 2− The presence of HCl produces a common ion H+ which suppresses the dissociation of H2S. This lowers the concentration of sulphide ions in the solution. Even with these low S2concentrations the ionic product exceeds the solubility product of Group II metal sulphides and thus they are precipitated. The amount of S2- is determined by the pH of the solution Ka for H2S is 10-22 [ H + ][ S 2− ] Ka = = 10−22 [H 2S ] Ka[ H 2 S ] 12 10−22  0.1 12 [H ] = ( ) =( ) = 0.32 [ S 2− ] 10−22 + pH = − log[ H + ] = − log 0.32 = 0.5 Therefore, saturated H2S solutions have a concentration of 0.1M A 𝑆 2− ion concentration of 10−22M would be enough to precipitate Group II cations At pH 0.5 Group II sulphides get precipitated. To precipitate out Group IV cations, (𝑁𝑖2+ ).(𝑆 2− ) will be 10−6 𝑀 So that precipitation occurs In this case 1 − 22 10  0.1 2 [H + ] = ( ) = 3.2 10−9 M 10−6 pH = − log[ H + ] = − log 3.2  10−9 = 8.5 The group reagent for Group IV precipitation is NH Cl + NH OH + H S 4 4 2