An ore sample collected near the Orange river was treated so that the resulting 25.0 dm³ 0.00226 dm-3 of Ni²+ (aq) ions

[answerhappygod]

An Ore Sample Collected Near The Orange River Was Treated So That The Resulting 25 0 Dm 0 00226 Dm 3 Of Ni Aq Ions 1 (74.47 KiB) Viewed 68 times

An ore sample collected near the Orange river was treated so that the resulting 25.0 dm³ 0.00226 dm-3 of Ni²+ (aq) ions and contained mol solution 0.00125 mol dm-3 of Co²+ (aq) ions. The solution was kept saturated with an aqueous solution of 0.0250 mol dm-3 H₂S. The pH was then carefully adjusted selectively precipitate the first metal ion (as a metal sulphide) from the second. first precipitate was filtered off from the remaining solution, dried and reduced to pure metal form. The pH of the remaining solution was then carefully adjusted to. the second time until the entire concentration of the second metal ion, together with a trace concentration of the first metal ion, were co-precipitated as metal sulphides. This co-precipitate was also filtered off, dried and reduced to the metal form. Based upon this information and that in the data sheet, calculate: (12) 4.1 The pH at which maximum separation of the two metal ions was achieved. 4.2 The percentage mass impurity of the metal that was obtained from the reduction of the last precipitate. (8) 7/9