Question
Posted on: March 25 2018Are the following bicarbonate & phosphate calculations correct
1 mol sodium citrate à 3 mol soda bicarbonate
Mr. Sodium citrate = 258.068 g/mol
Mr sodium bicarbonate = 84.006 g/mol
# mol NaHCO3 = 1g/84.006 g/mol = 0.0119039 mol
#mol NaCitr = 0.0119039 mol/3 = 0.00396797 mol
#g NaCitr = 0.00396797 mol X Mr = 1.02 g
In my opinion, this can also be just in capsules together with a mannitol or sorbitol.
Phosphate syrup: How much mmol phosphate is right?
Option 1
Na2HPO4 18.18 G
NaH2PO4 5g
ORANGE PEEL SYRUP 100ml
Aqua AD 300ml
Convert:
Mr Na2HPO4 = 141.957 g/mol
Mr NaH2PO4 = 119.976 g/mol
Mr PO43-= 94.97 g/mol
#mol Na2HPO4 = 18.18 g/141.957 g/mol = 0.12806695 mol
#mol NaH2PO4 = 5 g/119.976 g/mol = 0.04167500 mol
# mol PO43-= # mol Na2HPO4 + # mol NaH2PO4 = 0.16974195 mol
In 300 ml = 0.565 mmol/ml PO43-
Or
= 53.7 mg/ml PO43-
This is true: 0.56 mmol Elementary P or PO43-(phosphate) per mL, what oveencomes with 53.2 mg PO43 or 15.5 mg Elementary P (phosphorus)
Phosphate syrup, option 2: How much mmol phosphate is right? How much mmol K is right?
Na2HPO4 73g
Mono K phosphate 17.5 g
Raspberries Extract 7 g
After saccharin 0.3 g
Sorbitol 120 g
Aqua AD 300 ml
Convert:
Mr Na2HPO4 = 141.957 g/mol
Mr KH2PO4 = 136.086 g/mol
Mr PO43-= 94.97 g/mol
Mr K + = 39.09 g/mol
# mol Na2HPO4 = 7.3 g/141.957 g/mol = 0.051424023 mol à 73g/141.957 = 0514 mol
# mol KH2PO4 = 17.5 g/= 136.086 g/mol = 0.12859515 mol
# mol PO43-= # mol Na2HPO4 + # mol H2KPO4 = 0.18001918 mol à 0514 + 0128 mol = 0.64 mol
640 mmol at 300 mL thus gives 2.13 mmol P (Elemental phosphorus) or PO43-(phosphate) per mL, which corresponds to 66 mg P per mL and 202 mg PO43-
IN 300 ml = 0.6 mmol/ml PO43-
Or
= 56.99 mg/ml PO43-
# mol K + = # mol KH2PO4 = 0.12859515 mol
In 300 ml = 0.428651 mmol/ml K +
Or
= 16.755948 mg/ml K +
Answer
commented Prof Dr. I. Spriet (KU Leuven)
If I understand correctly, the question here is how much sodium citrate one should prescribe, if one wants to obtain a corresponding conversion to bicarbonate, according to a daily dose of 3 x1g bicarbonate. P >
Sodium citrate is converted to bicarbonate almost completely after ingestion. The solution used to serve sodium citrate in this indication is also known under ' SHOHL solution ' and consists of trisodium citrate. 2aqua 10g, citric acid. 1aqua 6.7 g and water up to 100 mL. The bicarbonate is made from sodium citrate As citric acid formed, each mL delivers 1mEq bicarbonate. A dose of sodium bicarbonate of 3x1g corresponds to 3 x 12 mEq bicarbonate. The content of citrate-ion (derived from both sodium citrate and citric acid) corresponds in the SHOHL solution with 2mEq/mL citrate. to make 3 x 12 mEq/Bicarbonate With 3 x 12 mL of the SHOHL solution. to give this in the form of capsules you have for 1 dose already 1200 mg of trisodium citrate. 2aqua and 804 mg citric acid required, which amounts to > 2g powder, and thus no longer to be used in 1 capsule. This is why this is always given in the form of a solution.
The other calculations are correct according to Prof. I. Spriet.