4
$\begingroup$

I am running phosphate by the ascorbic acid method (SM 4500-PE, PDF), and making the antimony potassium tartrate reagent. Standard methods uses $\pu{1.3715 g}$ of $\ce{K(SbO)C4H4O6 * 1/2 H2O}$ per $\pu{500 mL}$, while the chemical I have is $\ce{C8H4K2O12Sb2 * 3 H2O}$.

Should I add proportionately more of the trihydrate chemical to make up for the weight of water? I'm not sure what the exact active ingredient here is, and I realize that there are different chemical formulas besides the water content. But, I am hoping that if I account for the difference in weight of water, then I will get at least closer to the appropriate amount of necessary active reagent.

$\endgroup$
  • 3
    $\begingroup$ I changed the volume from $\pu{500 mL}$ to $\pu{400 mL}$ in your question according to the datasheet info I also provided a link for (I guess you used this one). $\pu{500 mL}$ is a suggested flask volume, not the volume of antimony potassium tartrate solution. $\endgroup$ – andselisk Jul 14 '17 at 15:05
  • $\begingroup$ The formatting was a nice change, but the 500mL final volume is correct (they're just suggesting using 400mL as a starting point before diluting to volume). Regardless, still unsure about what proper weight to use. Seems to be working well so far by just using the same 1.3715g as listed in SM, but using the chemical that we bought (trihydrate). I'm not entirely sure what this chemicals purpose is in the color-forming step, but it's not causing any major problems. I'm probably thinking too far into this. $\endgroup$ – H3vyM3741 Aug 2 '17 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.