What do i need to add to this water? This water is being used to anodize titanium.
2007-08-10 02:30:45 · 10 answers · asked by terminator2k4 1 in Science & Mathematics ➔ Chemistry
The amount is 1 gallon of distilled water.
2007-08-10 04:52:18 · update #1
Add Drain-O. Drain-O actually contains a very high concentration of Sodium Hydroxide (lye) solution. You could reasonably use this Drain-O, add it to water and get the pH close to 14.
2007-08-10 03:07:42 · answer #1 · answered by George H 2 · 0⤊ 0⤋
I would go with Red Devil Lye. Lye is just another name for NaOH. It will dissociate into Na+ and OH-, the OH- acts as your base deprotonating water and making more OH-.
We do need to know how much water you have and some other details before we can set up an equilibrium expression give you exact quantities though.
2007-08-10 03:05:49 · answer #2 · answered by ScienceNut 2 · 0⤊ 0⤋
Get a bunch of KOH or NaOH...also called lye. Put on some safety glasses and some gloves. Add it slowely cuz it will get pretty hot, add until no more goes into solution.
2007-08-10 06:00:41 · answer #3 · answered by tangelo316 1 · 0⤊ 0⤋
You need to tell me what base you are gonna use. From there, we can set up an equilibrium expression and determine how much base you need to add to the water. KOH or NaOH work well.
2007-08-10 02:36:03 · answer #4 · answered by Anonymous · 0⤊ 0⤋
You wont be able to buy analytic grade NaOH from anywhere, but you can purchase Red Devil Lye which is basically 99% pure NaOH.
2007-08-10 02:46:12 · answer #5 · answered by Dr Dave P 7 · 1⤊ 0⤋
Add a strong base.
2007-08-10 02:55:43 · answer #6 · answered by ag_iitkgp 7 · 0⤊ 0⤋
20 common laboratory apparatus chemistry functions
2016-05-18 21:59:48 · answer #7 · answered by antonette 3 · 0⤊ 0⤋
better add a strong acid like hidrocloric or sulphuric acid
2007-08-10 02:42:53 · answer #8 · answered by Anonymous · 0⤊ 1⤋
an alkali or something basic like sodium hydroxide.
2007-08-10 02:34:02 · answer #9 · answered by Dr. Eddie 6 · 1⤊ 0⤋
baking soda
2007-08-10 02:35:30 · answer #10 · answered by wolfcat 3 · 0⤊ 0⤋