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#1
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I saw you guys were able to help a friend of mine understand some concepts better, maybe someone can help me with this week's lab.. eek!!
We were to titrate 1.7g of KIO3 in Na2S2O3 solution. Added the excess I with KI, starch, HCl, etc. Then we had to find the M of K2S2O3 with this information. HOW?! The second part, we titraded 10mL of Ca(IO3)2 in the same Na2S2O3 solution. Added the KI, starch, HCl once again. We need to find the M of IO3- and the solubility of it, so we can calculate the Ksp of Ca(IO3)2 using all this information. HOW?! I'm having a hard time seeing how all of this falls into place. I can get through Organic no problem, but I stumble through regular chemistry lab :oops: All, and any, help would be greatly appreciate it. ~Milla |
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#2
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You need two reactions:
IO3(-) + I- -> I2 to balance the equation remember that the it takes place in acidic solution. S2O3(2-) + I2 -> S4O6(2-) + I- this one is much easier to balance. Once you have balanced equations you should be able to do all the calculations without problems. Best, Borek |
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#3
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Dang! I hate that lab...I have to do it too...Good Luck to you :wink:
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