A Few Chemical Magic Tricks Based on the Clock Reaction
-+ -
The reactions discussed here are based on the famous clock reaction, which uses KI03 and NaHS03. If sufficient NaHS03 is present, the reaction will proceed as follows: 3NaHS03 KI03 3NaHSOh + KI. There will be no change in appearance. When little NaHSOa is present, the reaction proceeds thus: 5/2 NaHS03 K103 5/2 NaHSOl+ % Ip %KOH. In the above reactions, suitable concentrations of the solutions are 0.05M KI03 and 0.125 M NaHS03. These two solutions will be referred to as "A and "B" respectively. In the followingreactions, theseare diluted by about half. If A and B are equal in volume, the mole ratio will be 2 5 1 ; the subsequent reaction will be the clock reaction, and 1%will become visible in about one minute. If the volumeof the NaHSOx solution is 1.2 times that of the KIO~solution.the mole ratio becomes 31. and the first reaction shove will result. In practice, reliability is achieved by using a v&me of N ~ H S O solution ~ more than 1.3 times the volume of the KIO- solution. NOTE: The concentration should he those stated ahove unleas otherwise meeified.
+
+
This will be referred t o as solution "C". Trick-LMethod. Dilute 30 ml of A with 110ml of water,and add 6 ml of aO.l N solution of HgCI2. Then add 45 ml of B, and mix continuously by a magnetic stirrer. The solution will first appear colorless and transparent, but as I- is produced through reaction of A and B, gradually HgIz will appear as an orange suspension. As I- increases further, the Hg12 suspension will then change to the camolex ion HeIa2-and the solution will become colorless and transoarent as before. NOTE: When toomuch HeCI.,isadded. the suspension will not disappear completely. Caution is advised in the use of HgClz because of its toxicity. Alternate Method. To 30 ml of A, 100 ml of C and 7.5 ml of a 0.1 N solution of HC12, add 45 ml of B, and mix continuously by a magnetic stirrer. The solution, which is colorless and transparent a t first, will gradually become a muddy orange. After some time, it will suddenly turn a dark brown. If a suitable quantity of starch is added heforehand, it will turn a deep purplish blue. If the amount of Hg2+isless, the color of t h e 4 will appear after the suspension disappears. Triek-11. Method. Dilute 30 ml of A with 30 ml of water, and add 90 ml of a 0.1 N solution of Pb(N03)2. Then add 45 ml of B. A dense white suspension will form. Mix continuously by amagnetic stirrer. Since I-is produced by reactionof A and B, the superfluous Ph2+reacts with I- and becomes PbIz, and the suspension will thus turn yellow. As mixing is continued, Sod2-is yielded, and the PbI, will change to PhSOa, which is a less soluble substance than P b l ~and , the suspentions will turn white again. Note: When there is too little Ph2+, in spite of the presence of I- PbIz will not he formed, and the suspension will not turn yellow. On the other hand, if there is too much Pb2+, in spite of the presence of Sob2- all the PhIz will not turn into PbSOn,and the yellow color will not disappear completely. Trick-111. Method. Mix 30 ml of B (1 M), 45 ml of a 2 N solution of KI, and 75 ml of C. The resulting solution will be an almost colorless (slightly yellowish) solution. Let the solution stand far a while, and it willgradually change from yellow t o bright yellow. The color will subsequently fade away. If the solution is allowed to stand for a while longer, the solution will begin to turn yellow again and finally will become red-orange. The last change will take place more rapidly if thesolution is heated. The color changes occur when the mole ratio of HzS03 t o I- is in the range 1:4.1 wish to thank Dr. J. A. Campbell of Harvey Mudd College for his kind advice, and I thank Mr. Hill who kindly corrected my English. Shakujii Senior High School Tokyo, J a p a n
184 1 Journal of Chemical Education
Eiichi Shigematsu
