Quote:
Originally posted by pHaestus
the pKa of ethylene glycol is 14.22, so it is basically completely present as C2H4(OH)2
(my weird notation to show that the OH on the alcohol groups are present and not O-)
Can just calculate pH using henderson-hasselbalch eqn:
pH = pKa + log (A-/HA )
oops too many unknowns; better just measure 
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It's not difficult to solve.
Assuming you use a 50% volumetric solution of ethylene glycol and water, and assuming volumetric additivity [edit-spelling] (probably off by 5-10% right there) you'd have a solution of 0.0180 molar ethylene glycol.
If the pKa is 14.22
Ka = 10^-14.22 = 6.0256 x 10^-15
Ka = [OH]*[C2H40H] / [C2H4(OH)2]
The concentrations of C2H4OH and OH- are equal and the concentration of C2H4(OH)2 doesn't change, so you can solve for them given the inital concentration of C2H4(OH)2.
6.0256 x 10^-15 ~ [OH-]^2 / 0.0180
Solving for the concentration of OH- gives 1.04 x 10^-8.
pOH = - log (1.04 x 10^-8) = 7.98
pH = 14-7.98 = 6.01
So if you have a 50% ethlyene glycol solution in water, you should have a pH of approximately 6.0.
If, after mixing the additives, you still have a pH below what you want, you might be able to add base to boost the pH, checking periodically with litmus paper.
If one of those additives is a buffer, though, you're doing your coolant a disservice.
Alchemy