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Need a chemist's advice
David Jenkins - 23/6/09 at 12:21 PM

No, not about a nasty skin disease, it's about making up a chemical solution!

I need to make a citric acid solution, which has been specified as "10% by weight". I have a small tub of citric acid in crystaline form, and a lot of water. Can someone advise how I can combine the two to get the solution I require?

I know three-fifths of f**k all about chemistry...

BTW: this solution is for passivating stainless steel, so it is something to do with Locosts...


BenB - 23/6/09 at 12:24 PM

Easy. Weigh out 10g of citric acid. Combine with 90ml of water. Job done


David Jenkins - 23/6/09 at 12:30 PM

Doh!

Even I can cope with that...

Cheers.


andyharding - 23/6/09 at 01:46 PM

Dunno if it's like other nasty acids I've used in the past but you should always add the acid to the water not the other way around or it might go bang in your face


David Jenkins - 23/6/09 at 01:48 PM

Not a problem with crystals...

Also, citric acid's quite mild in comparison with the other usual suspects like sulphuric or nitric.


iank - 23/6/09 at 02:58 PM

Still put the crystals into the liquid. It makes a weak acid that gets stronger as you add them, rather than a strong one that gets weaker if you get what I mean.
Less chance of splashing as well.


Mark Allanson - 23/6/09 at 04:42 PM

I don't think it is as simple as 10g crystal to 90ml of water. I seem to remember something about molecular weights being used which is the total atomic weight of the constituent atoms, eg a mole of oxygen is 16g (atomic weight 16) - but get someone else to confirm as it is a VERY long time since I learnt any chemistry, they called it alchemy in those days!


mads - 23/6/09 at 04:54 PM

Mark, what you are referring to is to do with concentration in terms of moles.

In the pharmaceutical world, 10% by weight would be 10g of substance A in 100g diluent so for David, as suggested it would be 10g Citric acid (as long as it is pure and classed as 100%) with 90mL water. (we would class water as 1mL = 1g)

We dont use Wikipedia at work as a reliable source of information but in this instance it is correct

HTH