How to make 50mL of 5% Potassium Dichromate solution?
I'm having a bit of trouble figuring out how to make a 5% solution of potassium dichromate solution to use for contrast control in pt/pd. I started with calculating the molar weights for water and potassium dichromate, (18.01528 and 294.1846). And now my brain has shut down. Can I just add 2.5g of Potassium dichromate to 50mL of water? For some reason, I'm thinking because of the difference in molar weights I can't think of it like that... but now (a second after writing that) I think I can.
Anyways, confused...thanks in advance
Actually, I think it should be 2.5gm plus 47.5ml water to get a 5% solution since that should weigh 50gm total.
Where's that avacado guy when you need him?
Originally Posted by guruguhan
In practice, I doubt if there's much difference between adding 2.5 gm to 47.5 ml or to 50ml as long as you are consistent in the future when mixing the same concentration.
Thanks for the response. I would still like to know for sure how this is calculated if someone could post it (if when one reads x% solution they can just add x% of solute to solvent to arrive at such a solution.)
I'd be astonished if it were especially critical. Most things in photography aren't.
Besides, solutions are normally specified w/v (weight-for-volume) so a 5% solution would be 5 g in 100ml or 2.5g in 50 ml.
And as Joe says, consistency is all that matters anyway.
Last edited by Roger Hicks; 06-09-2006 at 04:41 AM. Click to view previous post history.
Googling turned this up:
Weight percent is calulated as:
(weight of solute/solution weight) X 100.
Therefore 5% K dichromate is 5gm K dichromate dissolved in 95 ml water or 2.5gm plus 47.5 ml water, etc., since water weighs 1gm per ml at STP.
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That's a w/w solution, but as I said, most solutions are specified w/v.
Last edited by Roger Hicks; 06-09-2006 at 09:13 AM. Click to view previous post history.
As the others said -- most photographic solutions are calculated as w/v percentages.
These concentrations could just as easily be expressed as g/L -- i.e., 5% (w/v) = 50 g/L.
No molar masses are needed.
Thanks a lot Roger & Jordan!
Roger and Jordan,
Originally Posted by Roger Hicks
Now you two have me confused. What is the clue as to whether a recipe is calling for a w/w or w/v measurement?
I would think if it was specified as "make up an X% solution of a certain chemical" that would be referencing a w/w measurement, which would yield a true percentage solution.
OTOH, if they specified "add Xgm solute to Yml solvent" then they're talking about w/v and since the units don't match , it isn't a true percentage although a simpler calculation for sure.
That may be the way most people are thinking about it but unless the directions specify the latter method, how would you know which is correct? FWIW, I've always taken "percent solution" to refer to the former.
Probably only matters if one is dropping $ 30/gm on a precious metal salt...
When you add 5 grams of salt to 95 grams of water you get a solution that weighs 100 grams but which occupies less than 100 cc. Therefore if you measured out 10 cc (10 ml) of this solution you would be getting more than 0.5 grams of salt. This is why w/w solutions are not commonly used. In order to get exactly 0.5 grams of salt you would have to weigh out the solution. The concentration of this solution is slightly greater than 5% w/v salt.
On the other hand, if you dissolve 5 grams of salt in enough water to make 100 cc of total solution then 10 cc would contain exactly 0.5 grams of salt. The concentration of this solution is exactly 5% w/v salt.
Notice in the above example that the salt is not just added to 100 cc of water because such a solution would have a volume greater than 100 cc. In this case the concentration is slightly less than 5% w/v salt.
There is also a third percentage solution and that is v/v which is used when a soluble liquid is mixed with water.
Hope that these 3 examples make things clearer.