Morpholine in steam plant
Morpholine in steam plant
(OP)
Greetings all:
Can anyone tell me how to derive ppm concentration of morpholine from condensate pH? I think they taught me this in the navy but it's been a while. I seem to remember:
C4H9N0 + H20 ---> C4H9NOH+ + OH-
morpholine water morpholinium hydroxyl
pH is -log[H30]
pOH is -log[OH-], right?
I would like to estimate how much morpholine it would take to obtain a pH of 8.0, 8.5, 9.0, and 9.5, assuming we start neutral.
Thanks for any advice on this! -CB
Can anyone tell me how to derive ppm concentration of morpholine from condensate pH? I think they taught me this in the navy but it's been a while. I seem to remember:
C4H9N0 + H20 ---> C4H9NOH+ + OH-
morpholine water morpholinium hydroxyl
pH is -log[H30]
pOH is -log[OH-], right?
I would like to estimate how much morpholine it would take to obtain a pH of 8.0, 8.5, 9.0, and 9.5, assuming we start neutral.
Thanks for any advice on this! -CB





RE: Morpholine in steam plant
Thus, Kb=2.1*10-6, meaning
[morpholinium][hydroxyl]/[morpholine]=2.1*10-6
Assume, as an exercise, a morpholine 0.1M concentration, thus
(x)(x)/(0.1-x)=2.1*10-6
since 0.1 >>x, we can assume with little error that
x=(0.1*2.1*10-6)0.5=0.000458.
pOH=-log 0.000458=3.34, or pH=14-3.34=10.7
The amount ionized is 0.000458*100/0.1=0.46%, i.e., the amount of the protonized morpholine (morpholinium) form.
I hope I didn't make a mistake.
Anyway, this is a quick way of estimating the pH of water solutions of weak bases. One can apply the reverse procedure to estimate concentrations from given pH values.
RE: Morpholine in steam plant
(0.1 mole/L) x (87.12 g/mole) = 8.7 g/L,
assuming 1 L= 1000 g, 8.7 g/L would be about 8700 ppm on a weight basis.
RE: Morpholine in steam plant
We have a known condensate pH of 9.38. I've found sources that agree with ~2.1 x 10-6 for Kb (found ~1.6 to 2.5). So what is the concentration of morpholine (not the morpholine/morpholinium ratio)?
Assuming neutral condensate pH prior to addition of morpholine. One other thing is that this is in 200-250°F (assume maybe ~120°C) water and/or steam. I would imagine this has an effect on the Kb.
I can't quite close the loop on how to get back to ppm (or g or mg/liter) morpholine. Any added clarification would be greatly appreciated. Thanks again, -CB
RE: Morpholine in steam plant
As for the Kb and pKw at higher temperatures, they certainly change. I don't have the values with me.
Molar concentrations are given in moles per liter of solution:
1 mol of morpholine is 87.12 g.
1 L of solution is 1,000 g. To get g/1,000,000 g, i.e., ppm, multiply by 1,000.
Thus 0.1 M = 0.1*87.12*1,000 = 8,712 ppm, rounded to 8,700 ppm.
To make the exercise backwards i/o to find C, the morpholine concentration:
At ambient temperature, since pH=9.38, pOH=14-9.38=4.62 then, [OH]=2.39*10-5=x
Then C would be:
C=[(2.39*10-5)2/2.1*10-6]+2.39*10-5=0.0002959 M
Expressed in ppm (mass): 0.0002959*87.12*1000=25.8 ppm, or rounded up to 26 ppm.
I wish I could be of more help. Good luck.
RE: Morpholine in steam plant