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Well, certainly your nitric acid needs to be diluted. I like about
10 to 15 percent concentration. Instead, ask what concentration
solution, how much chemical to how much water, you should use. And
for that, I don't actually have an answer. But someone will...
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I’ll say it again. There is a very simple formula for arriving at a

desired concentration of any solution and I’ve posted it a few

times. PLEASE don’t be put off by the length of this post. It’s

simple mathematics which is very easy to understand and once it’s

understood that you can use it for any solution you need, you’ll want

to note it down somewhere - preferably in your workshop where it’s

easily accessible - laminated on the wall perhaps.

Final concentration X final volume = original concentration X

original volume (I may have quoted it the other way round before now,

but it doesn’t matter, one side of the equation equals the other so

it works both ways).

From what I can gather, individuals can purchase nitric acid up to a

concentration of 38%. Peter says that he likes 10-15%. So we can very

simply calculate how much water our nitric acid (38%) should be added

to.

We know our original concentration is 38% (in our example -

substitute whatever concentration you have purchased), and we know we

want to achieve a final concentration of 10%, so that’s not a

problem. However, with the volumes in our equation, we need to decide

one of the volumes so that we only have one unknown. It doesn’t

matter which one - it works for both. If we decide that we want to

add 20ml (0.2 litres) of 38% nitric acid, then we know the original

volume and it will be our FINAL volume which will become our unknown

(the answer to the equation).

I’ll abbreviate so it doesn’t go too far across the page. Final

concentration will be fin conc., final volume will be fin vol.,

original concentration will be orig conc., and original volume will

be orig vol. I’ll use the letter “Y” as our unknown (in this case to

represent the final volume). Our desired 10% solution will be

presented as 0.10 (ie 10 hundredths), our 38% as 0.38. It is simpler

to represent percentages thus, rather than having fractions such as

38/100 in our equation - obviously that works just the same but it

can get confusing. 20 ml will be presented as 0.20 (ie 0.20 litres)

and so on. Always be consistent with units, ie don’t mix ml with

litres within an equation.

fin conc. x fin vol. = orig conc. x orig. vol

0.10 x Y = 0.38 x 0.20

so, rearranging:

Y = (0.38 x 0.20) / 0.10

Y = 0.076 / 0.10

Y = 0.76 litres (76 ml)

IMPORTANT NOTE: 76 ml is the total final volume. This INCLUDES the

20 ml added acid, so you need to subtract that from 76 to find out

how much water is needed. 76 ml - 20 ml = 56 ml. Therefore, you would

need to add your 20 ml of 38% nitric acid to 56 ml water to achieve

your 10% solution.

If you want to use the equation the other way around and arrive at

say a final volume of 500 ml (0.5 litres), then Y (our unknown) will

now become our original volume (ie how much 38% nitric acid to add to

water).

fin conc. x fin vol. = orig conc. x orig. vol

0.10 x 0.50 = 0.38 x Y

so, rearranging:

Y = (0.10 x 0.5) / 0.38

Y = 0.05 / 0.38

Y = 0.13 litres (13 ml)

Again, IMPORTANT NOTE: Your 500 ml final volume is the total volume,

therefore you need to subtract 13 ml from 500 ml to determine how

much water you need to add the 13 ml of acid to. 500 ml - 13 ml = 487

ml water.

I hope this helps. It really is simple once you get your head around

what figures to substitute into the equation. Hubby says I should

post a blog about it rather than keep posting emails every time

someone asks the question. I may just do that. Then when the question

comes up again, I can post a link to my blog. But in the meantime,

this will do the job.

Any questions, drop me an email.

Helen

UK

http://www.hillsgems.co.uk

http://helensgems.ganoksin.com/blogs/