How to make 50mL of 5% Potassium Dichromate solution?

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guruguhan

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Hello,

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
 

smieglitz

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Actually, I think it should be 2.5gm plus 47.5ml water to get a 5% solution since that should weigh 50gm total.

guruguhan said:
...I started with calculating the molar weights for water and potassium dichromate, (18.01528 and 294.1846)....in molar weights I can't think of it like that... but now (a second after writing that) I think I can...

Where's that avacado guy when you need him?

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.

Joe
 
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guruguhan

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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.)
 

Roger Hicks

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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.

Cheers,

Roger (www.rogerandfrances.com)
 
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smieglitz

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Googling turned this up:

http://dl.clackamas.edu/ch105-04/weight.htm

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.

Joe
 

Jordan

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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.
 

smieglitz

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Roger Hicks said:
Dear Joe,

That's a w/w solution, but as I said, most solutions are specified w/v.

Cheers,

Roger

Roger and Jordan,

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...

Joe
 

Gerald Koch

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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.
 

Jordan

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Joe,

I think Gerald answered your question well. W/v solutions are much easier to work with than w/w (although sometimes things like concentrated acids are specified in w/w). I always assume that photographic solutions are w/v unless specified otherwise.
 

Kirk Keyes

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smieglitz said:
Where's that avacado guy when you need him?

I think he's somewhere in California, maybe Mexico...
 

Kirk Keyes

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smieglitz said:
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 once suggested that people try to include the appropriate designation when discussing percent solutions, but I was accused of being too nit-picky. Perhaps, but it would remove some of the ambiguity.

There are actually 4 types of percent solutions:

w/w = weight per weight
w/v = weight per volume
v/w = volume per weight
v/v = volume per volume
 

sanking

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Kirk Keyes said:
I once suggested that people try to include the appropriate designation when discussing percent solutions, but I was accused of being too nit-picky. Perhaps, but it would remove some of the ambiguity.

There are actually 4 types of percent solutions:

w/w = weight per weight
w/v = weight per volume
v/w = volume per weight
v/v = volume per volume

Well, there are worse things to be called than nit-picky.

As others have noted, in photography percent solutions are nearly always made in w/v and unless there exists some contrary reason I see no point in designating otherwise.

Of course, the concept of percent solution w/v is not all that well understood, but Gerald Koch's explanation was concise and clear. Before Gerald's explanation we saw already how loose was the designation w/v.

Sandy
 

Kirk Keyes

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sanking said:
Well, there are worse things to be called than nit-picky.

Actually, I have been called worse than that here. I was trying to be nice about it.
 
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