A QUALITATIVE TEST FOR TIN IN THE PRESENCE OF ANTIMONY DONALD BOOK* Oberlin College, Oberlin, Ohio
EDITOR'SN ~ ~ ~ . q bfreshmen e r h are encouraged "to do a little inventing," as Professor Holmes puts it. The accomanying note is published, not only for its intrinsic interest, but as an example of the kind of work an interested and intelligent freshman can do.
SbOa does not reduce Bi(OH)a to free bismuth, if (2) NasSb03 does not interfere with Na2SnOz in the reduction of Bi(OH)S, if (3) the presence of other group IIB ions does not hinder the formation or the action of NazSnOl, then the reduction of Bi(OH)3by NazSnOe is a good test for tin ions. THE usual method of testing for tin, namely, the Experiments to test these conditions, (1) the effect reduction of HgCL to insoluble HgCl by divalent tin of NasSb03 upon Bi(OH)a, (2) the effect of a solution ions, is in many cases an unsatisfactory test because of NasSbOa and Na2SnOz upon Bi(OH)3, and (3) the the precipitate of HgCl is not always heavy enough to analysis of two known solutions, the first containing be conclusive. Furthermore, the original solution con- arsenic, mercury, and antimony, and the second containing the divalent tin is often somewhat cloudy. taining arsenic, mercury, antimony, and tin show the The reduction of Bi(OH)3to free bismuth by a solution conditions to be satisfied and therefore that the Biof NaSnO* seems to be a more conclusive test for tin (0H)~-NasSnOz reaction is a good test for tin ions. than any other. When NaOH is added to a solution In experiment (1) the Na3Sb03produced no effect upon containing both antimonous and stannous ions, the the Bi(OH)3. The appearance of unmistakable black following reactions occur. bismuth in experiment (2) also showed satisfactory results. Experiment (3) indicated particularly what SbCL + 3NaOH +s b ( 0 H ) ~+ 3NaCI impurities could be expected in the solution of SbClb Sn(OH)* + 2NaCI SnCb + 2NaOH+ and SnClr after Sb& and SnSz were dissolved in 9N Excess hydroxide gives HCl. In most cases when HgS is present, the solution Sb(OH)a 3NaOH +NasSbOs 3H2O of the antimony and tin salts is milky. This turbidity Sn(OH)% 2NaOH +N@SnO2+ 2H.O can be reasonably explained by the presence of some Addition of the solution of the antimonite and stan- HgCl because HgS is partially soluble even in cold nite to Bi(0H)a should give free bismuth if NaaSbOs 9N HCl. Some HgClz may also be present, but there is no other impurity of a noticeable amount in the does not interfere with the action of the Na2SnOn. antimony-tin solution. Those named above did not 2Bi(OH)8 + 3NanSnOl--+3NalSnOa + 3HeO + 2Bi affect the NasSb03 or the NasSnOz. In this test, as Providing Na3SbOadoes not also reduce Bi(OH)3, the in the more common test for tin, i t is necessary to reappearance of free bismuth when the solution of Nas- duce the SnC4 completely to SnCll with nascent SbOa and Na2SnOl is added to Bi(OH)3 is a good test HZ; otherwise Na2SnOawill be formed from the SnCl, for tin. We must also consider whether or not the and no reduction of Bi(OH)a can take place. analytical procedure will admit other ions, especially 2A1CIa + 3H2 (1) 2Al + 6HC1+ those of group IIB, to the antimony-tin solution which (2) SnCL + HZ--+ SnCb + 2HC1 may interfere with the formation or with the action (3) SnC4 + 6NaOH +NazSnOs + 4NaC1+ 3Ha0t of Na3Sn03. t Although in many reactions there is a marked similarity If, then, three conditions are satisfied, if (1) Na3- between the properties of tin and antimony, most probably
+ +
+
* During 1 9 3 H 6 Mr. Book was a student in the general chemistrv class at Oberlin Calleee. It is sienificant that a student h his first year of college :hemistry cokd work out the problem of this paper independently. He-was not aware of the fact that Vanino and Treubert had developed this test in 1898.
explained by the positions of the metals next to each ot&r in the periodic table, NasSbO. does not behave illogically in not reducine Bi(0HL. The oeriodic table shows that antimonv is sli~htryless m&llic than tin and, therefore, does not oxidize so easily tin. Consequently, a stronger oxidizing agent than Bi(0H). seems necessary to change NaBbOa to Na8Sb0,.
as
