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該法的实质是:在酸性溶液中,亚硝酸将Co~(2+)离子氧化成Co~(3+)离子,Co~(3+)离子在由亚硝酸—亚硝酸鈉組成的緩冲溶液中(PH=3.8-4.0),在有过量亚硝酸鈉存在的情况下,絡化成Co(NO_2)_6~(-3)离子。然后再与溶液中的K~+离子作用生成深黄色的亚硝酸鈷鉀[K_3Co(NO_2)_6]。在沉淀过程中,鎳也生成亚硝酸鎳鉀的絡合物[K_4Ni(NO_2)_6],但是它溶于水(在100毫升水中能溶解46.56克),因而鎳仍然留在溶液內。过程可用如下的方程表示之: H_2SO_4+2NaNO_2=2HNO_2+Na_2SO_4, H_2SO_4+2CoSO_4+2HNO_2
The essence of this method is that in acid solution, nitrous acid oxidizes Co2 + ions into Co3 + ions, and the Co3 + ions in the buffer composed of nitrous acid-sodium nitrite Solution (pH = 3.8-4.0), in the presence of excess sodium nitrite, the formation of Co (NO_2) _6 ~ (-3) ions. And then with the solution of K ~ + ions to produce dark yellow potassium cobaltous nitrite [K_3Co (NO_2) _6]. Nickel also forms a complex of nickel and potassium nitrite [K_4Ni (NO_2) _6] during precipitation, but it dissolves in water (46.56 g in 100 ml of water), so nickel remains in the solution. The process can be expressed by the following equation: H_2SO_4 + 2NaNO_2 = 2HNO_2 + Na_2SO_4, H_2SO_4 + 2CoSO_4 + 2HNO_2