Ratio calculations and conversions mixing D-76 1:1 in recovery mode

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bascom49

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As I was mixing a batch of stock D-76 last night I overfilled my H2O by 200 ml making my stock solution 4l rather than 3.8l.
My math says that I now have a ration of 1 parts H2O to 19 parts D-76.

So to mix 500 ml of working solution at 1:1 I should mix 263.16 ml of "my" stock to 236.84 ml of H2O.

Right ?

Thanks Charles
 

runswithsizzers

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I believe that is correct.

D-76 package contents weigh 415g, so a stock solution contains 415g/3.8L = 109.21 g/L. And a 1:1 dilution to working strength contains 54.6 g/L of D-76

So your desired 500mL working solution needs to contain half that amount of D-76, or 27.3 g.

Your "stock" solution contains 415g/4 L, or 103.75 g/L

The volume of your "stock" solution needed to deliver the target amount of 27.3 g/500 mL is:
27.3 g [1000mL / 103.75 g] = 263.13 mL

You probably ought to get another opinion, though; it's early here, and I am under-caffeinated.
Edit: I previously wrote "236.13 mL" - now edited to what I intended to write" 263.13 mL
 
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Bill Burk

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I think the difference is not significant, and I would use it 1:1 anyway.

I mix a gallon stock but it’s always a generous gallon. I have five “quart” bottles, when I am on my last one I make a batch of four, filled to the top, with no air.

If I only made a gallon, the last bottle would be half air. So when I make more than a gallon the last bottle can still develop two tanks. I think my last bottle has a little air but I squeeze out what I can and use it first.

Oh. But I do have longer development times than anybody else. This could be one of the reasons.

(For example I use 13:30 for TMAX 400 for CI 0.62).
 

JPD

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You can mix it as usual, and if you want to be sure, develop the film for 10 seconds longer than usual if you want. The difference between 4000 ml and 3800 ml is just 1/20 so it's not a biggie.
 

MattKing

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I would round it off - 260 ml of "stock" plus 240 ml of diluting water.
 

runswithsizzers

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I would round it off - 260 ml of "stock" plus 240 ml of diluting water.
So true!

Depending on how carefully the 4 liters of water was measured, the calculated result may have one, two, or three significant figures (was it 4 L or 4.0 L or 4.00 L?). No way should I have included anything right of the decimal in my final answer, which, at best, cannot be any more precise than 263 mL.

But even if all three digits are significant, mathmatically, I'm pretty sure that last digit is not very significant from a practical point of view.
 
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