RU_KLO
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Calculation for stardandard solutions for samples analitical chem.
hi,
Im trying to make standards for analytical chemistry (for calculating -more or less - by sight, concentrations)
They dont need to be "too" exact, but just answering the cuestion if X ion is in the sample and how much (by just seeing/comparing).
What I need is 10mg / 1ml solution of X ion or metal, but only 100ml solution. (I dont need to prepare a liter)
So I can apply this standard solution to the same procedure I use in the sample to see - more or less - how much of the of the corresponding ion/metal
is in the sample.
Or use the "standard solution" to check the procedure's sensibility.
Im using ACS "Control, Standard, and Stock Solutions
Part 3, Solutions and Mixtures Used in Tests
eISBN: 9780841230460" as guide, but the standard solutions are in the 0.1 or 0.01gr/ 1 ml. I need it to be 10mg/ml
this is simple because I dont have precise scales or ACS regents, and knowledge/experience. And higher concentration are difficult to miss. I dont
need perfect results but a good aproximation. Also you can always dilute....
Is this as easy as just multiply by 100/1000 the original reagent and then divide by 10 to get 100ml to from a liter? (because in the paper the
solutions are made by liter.)
Some are simple, because you dont do the second dilution, for example:
Aluminum (0.01 mg of Al in 1 mL)
Dissolve 0.100 g of metallic aluminum in 10 mL of dilute hydrochloric acid (1:1), and dilute with water to 100 mL. To 10 mL
of this solution, add 25 mL of dilute hydrochloric acid (1:1), and dilute with water to 1 L.
by dissolving 1g of metallic aluminum in 10 mL of dilute hydrochloric acid (1:1), and then dilute with water to 100 mL.
I have 10 mg/ml as needed.
but for example this one:
Antimony (0.1 mg of Sb in 1 mL)
Dissolve 0.2743 g of antimony potassium tartrate hemihydrate in 100 mL of water. Add 100 mL of hydrochloric acid, and
dilute with water to 1 L
How are the calculations done? I tried to use molar mas, but cannot get there.....
this is the info:
Antimony potassium tartrate, hemihydrate
631.84 g/mol
C8H6K2O13Sb2
Sb: 121.76 g/mol
how will you proceed to solve this to get 10mg/1ml of Sb?
Note: I have a Antimony metal, so maybe I can proceed like Aluminium by dissolving somehow 1g Antimony metal in "something" (maybe in tartaric acid +
KOH - I need a soluble compound that does not hydrolizes or precipitates from solution) and then adding water to 100 ml.
but how you proceed if you have Antimony potassium tartrate, hemihydrate.
to get from : 0.1 mg of Sb in 1 mL, 1 liter solution to 10mg of Sb in 1ml, 100ml solution.
Thanks
Attachment: Control, Standard, and Stock Solutions.pdf (82kB)
This file has been downloaded 9 times
Go SAFE, because stupidity and bad Luck exist.
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bnull
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Something was bothering me when I tried to calculate. Fortunately, I found what the problem was. There are two mistakes.
The first one is that you're using the wrong mass for the hemihydrate. Hemi- means "half"; monohydrate means one molecule of water per molecule of
salt, dihydrate means two, and so on, so hemihydrate means half molecule of water per molecule of salt. In reality, one molecule of water is shared by
two molecules of salt but, as we are usually more interested in the molar mass of the salt, we say there is half molecule of water per molecule of
salt. The molar mass of anhydrous antimony potassium tartrate is 613.83 g/mol. Add half molecule of water (18 g/2=9 g) and we have a molar mass of
622.83 g/mol of antimony potassium hemihydrate (let's call it tartar for convenience). So far so good.
To calculate the amount of tartar that contains exactly 1 g of antimony, we notice that there are two moles of antimony (2*121.76 g=243.52 g) in each
one mole of tartar (622.83 g), or Sbtartar=243.52 g622.83 g. Divide both values by the mass of antimony (243.52 g), so now we
have Sbtartar=1 g2.558 g, which means that there is 1 gram of antimony in each 2.558 grams of tartar. If you want 1 mg of Sb per
1 mL of solution, you dissolve 2.558 grams of tartar in water and make up to 1 L; if you want 0.1 mg of Sb per 1 mL of solution, you dissolve 0.2558 g
of tartar in water and make up to 1 L.
Did you notice something wrong? Yes, it is the number. The second mistake is in the paper. Multiply 0.2743 g by 10. We have 2.743 g. Multiply by the
mass of antimony in one molecule of antimony potassium tartrate, or 243.52*2.743 g. We now have 667.975 g (let's round it to 667.98 g). Subtract from
this number the molar mass of anhydrous antimony potassium tartrate: 667.98-613.83=54.145g. This, my dear fellow, is the mass of 3 moles of water of
hydration (the decimal part comes from rounding error). They used the molar mass of antimony potassium tartrate trihydrate, not
hemihydrate, in their calculations. You would never arrive at their number unless you used the molar mass of the trihydrate. It is a
trivial error but they didn't correct it.
So, anyway. To make a solution A with 10 mg Sb:1 mL, you dissolve 2.558 g of antimony potassium tartrate hemihydrate and make up to
100 mL. Multiply both sides of the ratio by 100:10 mg∗1001 mL∗100=1000 mg100 mL=1 g100 mL. To make a solution
B with 0.1 mg of Sb in 1 mL, you take 1 mL of A (which contains 10 mg of Sb) and dilute it to 100 mL. Multiply both
sides of the ratio by 100:0.1 mg∗1001 mL∗100=10 mg100 mL.
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RU_KLO
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thanks
the molar of the hemihydrate was taken from here:
https://pubchem.ncbi.nlm.nih.gov/compound/Antimony-potassium...
so it seems you cannot trust ACS papers nor pubchem...
Go SAFE, because stupidity and bad Luck exist.
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bnull
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Molecules with non-integer values are pretty tricky to represent with software. The figure given by PubChem has one molecule of water and one of
antimony potassium tartrate, and that was how their software calculated the mass. Perhaps they should include a least common multiple correction.
The ACS paper, on the other hand, it seems human error. Someone needed the molar mass, looked it up on Wikipedia, and used that of the trihydrate.
Nobody noticed, probably because the one who calculated the values wasn't the one who wrote the paper.
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