Morpholine in steam plant 1

[ChasBean1]

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

 

[25362]

You must use the ionization (dissociation) constant, which, BTW, is similar to that of hydrazine. The tabulated value pK_b for morpholine in water at 25 deg C is 5.67 (for hydrazine pK_b=5.77).

Thus, K_b=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.

 

[25362]

To convert 0.1M to ppm use morpholine's molecular weight, 87.12 g/mole:

(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.

 

[ChasBean1]

Thanks for the reply! I got lost after the "Assume as an exercise" paragraph, though... I just can't quite close the loop - not sure how x's represent three different terms(?)

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

 

[25362]

About the x's. In the reaction, as posted by you, the equilibrium constant Kb is equal to the product of formed morpholinium and hydroxyl (both equal molar concentrations as x) divided by the "remaining" unhydrolyzed morpholine (original-x). I took 0.1 M as the original molar concentration just as an example.

As for the Kb and pK_w 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

x²/(C-x)=Kb=2.1*10^-6​

Then C would be:

[(x²/2.1*10^-6]+x=C​

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.

 

[ChasBean1]

You were very helpful. Thank you so much for your input! -CB