Ph level questions

http://www.pulseinstruments.net/index.asp?PageAction=VIEWCATS&Category=1103&gclid=CK7dk7mw6ZkCFdxM5Qod_n4nRg

milo

Thanks for the link.

Here's a reply from one of those engineers unfortunately encumbered by a surfeit of training and knowledge, and therefore unsuitable for inventing innovative new technologies which contradict all that obsolete old knowledge.

A pound of potassium hydroxide dissolved in a gallon of water produces a roughly 2 molar solution:

atomic mass of K = 38 gm per mole

atomic mass of OH = 17 gm per mole

atomic mass of KOH = 38 gm/mole + 17 gm/mole = 55 gm/mole

1 lb KOH * 454 gm/1 lb = 454 gms of KOH

454 gm KOH * 1 mole KOH/55 gm KOH = 8.25 moles KOH

1 gallon = 3.785 liters

8.25 moles KOH/3.785 liters H2O = 2.2 molar

pOH is the negative logarithm of the OH ion concentration:

pOH = - log 2.2 = -0.342

At room temperature, the relationship of pH and pOH is given by:

pH + pOH = 14

This implies your pH is above 14, which is impossible, but it clearly shows you are, as you say, "off the chart." It may be that you cannot dissolve more than a certain amount of potassium hydroxide in water, that above the molarity that would yield a pH of 14, the KOH simply doesn't dissolve anymore.

But the bottom line is, your measurement is accurate, and it isn't due to any impurities; it's just the amount of KOH you poured into the water.

There are 2 different solutions. 1 has a pound of KoH and that is what I am using right now. The other had sodium hydroxide and iron oxide from the annode that it was destroying. The reason I want to test this fluid is the consistency of it. There is iron in the bottom of the jug that separated out but the water is still greenish almost like a 50/50 mix of antifreeze. It is more curiosity than anything with the old mix, but I would like to identify the saturation level of the new mix so I dont waste KoH.