The titration is performed in the usual way, running the standard solution of silver nitrate into a solution containing a known weight or volume of the material containing the cyanide. The _finishing point_ is determined in one of two ways, both of which are largely used. In the first place, as long as there remains any free cyanide in the solution the silver nitrate will combine with it forming the double cyanide and yielding a clear solution; but as soon as all the free cyanide is used up the silver nitrate will react with the double cyanide[41] forming silver cyanide, which separates as a white precipitate and renders the solution turbid. But, in the second place, if potassium iodide is present in the solution the excess of silver nitrate will react with it,[42] rather than with the double cyanide; and silver iodide will separate as a yellowish turbidity which is easily recognised. In working with pure solutions, the two finishing points give the same results; and this is true even when there is much difference in the degree of dilution. The finishing point with the iodide, however, has an advantage in precision. Moreover, it is but little affected by variations in alkalinity, which render the other finishing point quite useless. The great difference between the two is shown when zinc is present in the solution. In this case, when working without the iodide, the first appearance of a turbidity is less distinct; the turbidity increases on standing and as a finishing point is unsatisfactory. It can be determined with precision only by very systematic working and after some experience. The turbidity is due to the separation of an insoluble zinc compound. A most important point (to which reference has already been made) is that less silver nitrate is required to give this turbidity and, consequently, a lower strength in cyanide is reported. On the other hand, as much silver nitrate is required to give the yellow turbidity due to silver iodide as would be required if no zinc were present.