_Manganese_ does not directly interfere, but the precipitated hydrate, which rapidly darkens through atmospheric oxidation, obscures the end reaction. It may be removed by passing sulphuretted hydrogen through the filtrate from the acetate separation: sulphides of nickel, cobalt and zinc will be precipitated, whilst manganese remains in solution: the addition of more sodium acetate may assist the precipitation. The precipitate must be filtered off and dissolved in nitric acid: the solution should be evaporated to dryness. The filtrate may retain a little nickel; if so, add ammonia till alkaline, then acidify with acetic acid and again filter; any small precipitate obtained here should be added to that first obtained. It is only when _cobalt_ is present that any further separation is required. Cobalt hydrate takes up oxygen from the air, and on adding potassium cyanide some may refuse to dissolve; and the solution itself acquires a brown colour, which becomes deeper on standing. At this stage the cobalt is easily separated. The solution containing the nickel and cobalt with no great excess of acid, is made alkaline by adding 20 c.c. of soda exactly as in preparing for a titration. So, too, the solution of cyanide is added so as to have an excess of 20 or 30 c.c.; the solution may have a brown colour, but if it is not quite clear it _must_ be filtered. Then warm (boiling is not needed) and add from 50 to 100 c.c. of bromine water. This throws down all the nickel as black peroxide in a condition easy to filter. Filter it off and wash with water. The precipitate can be dissolved off the filter with the greatest ease by a little warm sulphurous acid. The filtrate and washings, boiled till free from sulphurous acid, yield the nickel as sulphate in a clean condition. ~Determination of Nickel in Nickel Sulphate Crystals.~--Take 0.5 gram of the salt, dissolve in 50 c.c. of water and add 25 c.c. of solution of soda. Run in from a burette, say, 60 c.c. "cyanide." Add a few drops of potassium iodide and titrate back with "silver nitrate." Suppose 15.5 c.c. of the latter is required. Then 15.5 c.c. subtracted from 60 c.c. leaves 44.5 c.c., and since 100 c.c. = 0.25 gram of nickel, 44.5 c.c. will equal 0.11125 gram of nickel. This in 0.5 gram of the salt equals 22.25 per cent.