Ok, so we have about 25 °F = 250 mg/l of hardness, expressed as CaCO3
(calcium carbonate). That will give us about 100 mg/l expressed as calcium,
or "fairly hard water". It is not so bad, it could be used if the pH was
lowered.

The quantity of citric acid required will be low, in the order of 50 - 100
mg of acid per liter of water. This is so little that it is difficult to
measure outside of a lab. I would suggest preparing a solution of citric
acid in water, using handy measures that you have at home, in the range of
10% by weight of acid in water. That will be about 2 tablespoon per cup.
Then try adding this a couple drops at a time to a liter (quart) of water.
Each drop will contains about 10 mg of acid. About 10 drops will be
required. Mix the water with the acid and let stand for a couple minutes.
Then check the pH. I can't predict the exact amount without having access to
your water, but you shouldn't too off.

Once you have the proper number of drops figured out, you can skip the pH
checking, it should be pretty much the same everytime.

By the way citric acid is great in the dishwashing machine if you have
problems with calcium deposits. A tablespoon once a week will do the job. It
is the same reaction, citric acide reacts with the calcium carbonate and
gives carbon dioxide (a gas) and calcium citrate, which is soluble in water.

I hope this helps.

Marc


"Reka" a écrit dans le message de
...
Okay, I'll see if I can ANSWER
--
Reka
http://www.rolbox.it/hukari/index.html
"A fanatic is one who can't change his mind and won't change the subject."
--Winston Churchill a post. My originals don't come through.
°F is equal to 5,6mg CaO/l, if that makes any sense to anyone. It
doesn't
to me. :-]
How much citric acid should I try???

Thanx,