hi all, i hope this is the right forum for this:
in an experiment i've been studying, there is the need for about 6.58 mL 37% HCl. i can only easily obtain 31.25% muriatic acid. i have calculated that there is .3784 grams less HCl and the same more water in the later at 6.58 mL. if i add another 1mL of 31.25% muriatic for a total of 7.58 mL i would then have an almost equal number of grams of HCL as in 6.58 mL 37% HCl (2.3687 g versus 2.4346 g), which would result in the same extraction results except for the extra water present in both the extra 5.75% difference between acids and with the addition of the extra 1 mL (.3784 g and .6875 g respectively for a total of an extra 1.0659 g of water, which is bad in this case). could this extra amount of water be removed by adding oven-dried epsom salts? these same dried salts can be used to make acetone anhydrous as called for in the experiement and i thought that i could also use it to remove the extra water in the HCl mentioned since it is intended to be mixed with the aforementioned dried acetone. i have read that 1 gram of dried epsom salts can absorb 1 gram of water, so after drying the acetone with the appropriate amount of dried epsom salts and filtering, could i add the HCL (with 1.0659 extra grams of water from the instructions due to the extra acid needed) and then add 1.0659 grams dried epsom salts to arrive at a HCl and acetone solution near equal to the one asked for with more concentrated HCl? i hope this makes sense and someone can help. thanks.
mycologos
in an experiment i've been studying, there is the need for about 6.58 mL 37% HCl. i can only easily obtain 31.25% muriatic acid. i have calculated that there is .3784 grams less HCl and the same more water in the later at 6.58 mL. if i add another 1mL of 31.25% muriatic for a total of 7.58 mL i would then have an almost equal number of grams of HCL as in 6.58 mL 37% HCl (2.3687 g versus 2.4346 g), which would result in the same extraction results except for the extra water present in both the extra 5.75% difference between acids and with the addition of the extra 1 mL (.3784 g and .6875 g respectively for a total of an extra 1.0659 g of water, which is bad in this case). could this extra amount of water be removed by adding oven-dried epsom salts? these same dried salts can be used to make acetone anhydrous as called for in the experiement and i thought that i could also use it to remove the extra water in the HCl mentioned since it is intended to be mixed with the aforementioned dried acetone. i have read that 1 gram of dried epsom salts can absorb 1 gram of water, so after drying the acetone with the appropriate amount of dried epsom salts and filtering, could i add the HCL (with 1.0659 extra grams of water from the instructions due to the extra acid needed) and then add 1.0659 grams dried epsom salts to arrive at a HCl and acetone solution near equal to the one asked for with more concentrated HCl? i hope this makes sense and someone can help. thanks.
mycologos
