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...
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/