Potassium Dichromate Bleach

[georgegrosu]

I sent an email yesterday to H2O and replied that he wants to make the discussion on forum. Let everyone know.
Well.
The whole discussion on forum is related not to if 35 or 37% concentration.
We had a bottle of battery acid too could not fully read the writing.
That's why I wrote the post 20 as - “ I Considered Concentration 35% by volume - solution of sulfuric acid. From Here We approximated I think is a part of sulfuric acid and water "
2 % difference in concentration is not big deviation as someone wrote here. ....That is just reckless....
The big difference comes from the fact that I took this 35 % as a percentage of volume concentration. Not percentage of mass concentration (I could not see the label).
H2O to come in simple thing. To be able to work around the world.
I. I considered mass percentage concentration = 37% (from H2O);
I consider the density of battery acid = 1.26 (from H2O).
I consider the volume percentage concentration = 37: 1.26% = 27.777 ml concentrated H2SO4.
27.777 ml concentrated H2SO4 ... ... ... ... ... ... ... 100 ml solution
12 ml concentrated H2SO4 ... ... ... ... ... ... ... ... ... X
X = (12 x 100): 27 777 = 43.20 ml acid battery
II. I consider mass percentage concentration = 35% (from George);
I consider the density of battery acid = 1.26 (from H2O).
I consider the volume percentage concentration = 35: 1.26% = 27.347 ml concentrated H2SO4.
27.347 ml concentrated H2SO4 ... ... ... ... ... ... ... 100 ml solution
12 ml concentrated H2SO4 ... ... ... ... ... ... ... ... ... X
X = (12 x 100): 27 347 = 43.88 ml acid battery
As seen from the above calculations is 1.5% difference.
When considering the concentration of 35% volume percentage concentration - error is 26%.
You can compare 1.5% with 26%? The error is 17 times higher.
The calculation can be done otherwise according a sketch that I sent yesterday's H2O (from Orwo).
First convert the mass percentage concentration in volume percentage concentration.
I consider the density of battery acid = 1.26 (from H2O).
Divide 37 : 1.26 = 27.777 ml concentrated H2SO4 per 100 ml solution.
These calculations can be done in several ways. It depends on what data you have available.
George

 

[georgegrosu]

I forgot to make a link to schema for calculating the quantities for the two solutions are mixed to form a solution of a given concentration. From Orwo
http://www.flickr.com/photos/21121448@N06/5396384582/
George

 

[Gerald C Koch]

The sulfuric acid can be replaced with sodiun bisufate which is easier and safer to use. Look in the APUG archives for the conversion factor. In the US it is available in hardware and pool suppliy stores for lowering swimming pool water pH.

 

[H2O]

...schema for calculating the quantities for the two solutions are mixed to form a solution of a given concentration. From Orwo
http://www.flickr.com/photos/21121448@N06/5396384582/....

My opinion about the cross rule was told in my past contribution.
It is based on exact parameters and equations which leads to the correct undoubted results.


Your considerations are based on manipulation by numbers without proof of some valid equation(s).
Are you able to prove your consideration by generaly equations?
Are you able to describe (to explain) your procedure step by step for grammar school students (for example)?

....
II. I consider mass percentage concentration = 35% (from George);
I consider the density of battery acid = 1.26 (from H2O).
I consider the volume percentage concentration = 35: 1.26% = 27.347 ml concentrated H2SO4.
27.347 ml concentrated H2SO4 ... ... ... ... ... ... ... 100 ml solution
12 ml concentrated H2SO4 ... ... ... ... ... ... ... ... ... X
X = (12 x 100): 27 347 = 43.88 ml acid battery
...

You divide 35 by 1.26. Why?
35 only means (in this case) that 100 g of solution contains 35 g of sulphuric acid (at "virtual" concentration 100 %) and 65 g of water.
1.26 is density of this solutions (g/mL). By means of this number can be calculated volume of this solution (79.4 mL).
Any your calculation is not supported by basic equations (the law of conservation of mass, definition of mass fraction and relation between density-volume-mass).

The almost correct results do not mean the correct! procedure.

Try to check your consideration from concentrated acid = 96 %.
The density is 1.84 g/mL.
Do you consider the volume percentage concentration = 96: 1.84 = 52.2 mL concentrated H2SO4????
How to continue?
52,2 mL concentrated H2SO4 ... ... ... ... ... ... ... 100 ml solution
12 ml concentrated H2SO4 ... ... ... ... ... ... ... ... ... X
???
I can't continue.



And densities of different solutions of H2SO4?
I think that it is not secret ;-)

[d is density where cm3 = mL]
Mass [%];d [g/cm3]
0.5;1.0016
1;1.0049
2;1.0116
3;1.0183
4;1.0250
5;1.0318
6;1.0385
7;1.0453
8;1.0522
9;1.0591
10;1.0661
12;1.0802
14;1.0947
16;1.1094
18;1.1245
20;1.1398
22;1.1554
24;1.1714
26;1.1872
28;1.2031
30;1.2191
32;1.2353
34;1.2518
36;1.2685
38;1.2855
40;1.3028
42;1.3205
44;1.3386
46;1.3570
48;1.3759
50;1.3952
52;1.4149
54;1.4351
56;1.4558
58;1.4770
60;1.4987
70;1.6105
80;1.7272
90;1.8144
92;1.8240
94;1.8312
96;1.8355
98;1.8361
100;1.8305

 

[georgegrosu]

The almost correct results do not mean the correct! procedure.

H2O I sent to you an email on 01-28-11 10:17 PM - preparation of battery acid.

260 ml 740 ml concentrated sulfuric acid and water.
Percentage Volumetric Concentration is 26%.

100 ml acid battery contains 26 ml of concentrated sulfuric acid (96%)
50 ml acid battery contains 13 ml of concentrated sulfuric acid (96%)
We need 12 ml of concentrated sulfuric acid (96%).
The difference between 13 ml and 12 ml of 1 ml, which represents about 10% of the 50 ml of battery acid.
If you subtract 5 ml (10% of 50 ml) from 50 ml = 45 ml get battery acid.
It is a calculation without a computer ... without tables. http://docs.google.com/viewer?a=v&q...5-kch6&sig=AHIEtbSaFro0Me5kivdeUobIZHwSfyGfwg
When working with volumetric concentration work more easily.
I conclude this discussion here.
Success.
George

 

[H2O]

.....
I consider the volume percentage concentration = 37: 1.26% = 27.777 ml concentrated H2SO4.
......
I consider the volume percentage concentration = 35: 1.26% = 27.347 ml concentrated H2SO4
.....

You didn't explain one your procedure (above) and you bring another.
It looks better.

You haven't read my posts carefully.
I spoke about the correctness of calculation!!
But the easy procedure doesn't mean the correct method.

Yes, the estimation belongs to the solution preparation and cross rule for volume percentage is possible to use.
But such procedure is not correct.
Only user decides what suits him or her. I offered the correct, applicable method supported by the exact equations.

The correct means true and the danger of every the estimation is that it can be far from true.
It is about approaches.
I think that it is important to know how to work correctly or truthfully and then the estimation can be used. Than I know how far I am from true. Only if I know the correct results I can decide if I can use the estimation or I can neglect some parameters in calculation and to make procedure easier.

So, you didn't understand several days ago how I reached 44,2 mL of battery acid.
Now you are able to estimate (not correct to calculate) dilution of sulfuric acid.
You wrote me that is problem for you to get table with densities ...
Now it seems the problem disappeared.

OK, I was succesful.

I recommend you to study:
http://www.chembuddy.com/?left=concentration&right=toc

and this is helpful too:
http://www.chembuddy.com/?left=CASC&right=concentration_and_solution_calculator

 

[bblhed]

Wow, thank you everyone that helped out in this discussion, I think that the lesson learned here is that you can't just edit chemical mixing formulas like you are changing a recipe for a wedding cake into a recipe to make a half dozen cupcakes.

What I gather from the discussion is that to make the B&W reversal bath I want to make that is stated as

1 L H2O
12 mL H2SO4
9.5g Potassium Dichromate

I need to do the following

Start with 800ml H2O
Add 44.2ml Battery acid
Add H2O to solution to make make 1L
Add 9.5g Potassium Dichromate

I will probably use 44ml of Battery acid because I do not have a pipette, but I believe that if I do have a problem using this mixture I can just adjust times until I get it right because I have a feeling that I will probably continue to use this mix for a long time unless R-9 or an equivalent comes along.

Thank you again everyone. Also, sorry for the delay in posting, I was busy shoveling my roof this weekend.

 

[H2O]

...you can't just edit chemical mixing formulas like you are changing a recipe for a wedding cake into a recipe to make a half dozen cupcakes....

))

It is important to keep correct formula for good cake too.
I think that good experimentator (cook ) must have the respect to any formula.

If you are able to measure 965 mL water (+/-) then you don't need to tune the total volume of solution.
By adding 45 mL battery acid (37 % mass; density 1.28 g/mL) and 9.5 g potassium dichromate you will gain the same volume of the solution (and almost the same acid strenght) as from the first formula.