i did it with a graphing software and by hand and got two different answers
#how to linearize this function y=-0.509e^(-0.02954x) +4.69
kind cloud
final sableBOT
i did it with a graphing software and by hand and got two different answers
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whole fern
i did it with a graphing software and by hand and got two different answers
I assume your model is y = ae^(bx) + c?
If so, then this can be linearized as ln(y - c) = ln(a) + bx, but this doesn't allow you to find parameters with linear optimization.
Nonlinear optimization still works, of course.
whole fern
any idea why i got two different answers?
Can you show your dataset and the equations you got?
kind cloud
ignore the blue one. the green one is ln(4.69-y)
kind cloud
i did it with a graphing software and by hand and got two different answers
this graph was made with the green y column and black x column
whole fern
Oh, it's a pH curve!
kind cloud
yes
whole fern
If so, can you provide some more context?
We can maybe try using a model with some more appropriate variables.
kind cloud
its the pH of water at different temperatures when I add CO2 to it
whole fern
Oh. So, basically, this is modeling of apparent Ka of H2CO3, kind of.
kind cloud
yeah a bit
whole fern
Alright. As per usual, we should have:
d(ln(Ka))/dT = ΔrH°/(RT^2)
Or, in other words:
d(ln(Ka))/d(1/T) = -ΔrH°/R
We can assume that ΔrH° doesn't depend on T, as our interval isn't very large. Then:
ln(Ka) = c - (ΔrH°/R)T^(-1)
Here c is some constant.
Then, let's work with Ka. Specifically, we want it in terms of pH.
kind cloud
oh i actually looked at this equation in my research but did not understand enough abt it...
whole fern
Well, it's just the van't Hoff equation.
Anyway, I won't bore you with what we already both know. We have:
Ka = [H3O+]^2/(C0 - [H3O+]) = 10^(-2pH)/(C0 - 10^(-pH))
ln(Ka) = -2ln(10)pH + ln(C0 - 10^(-pH))
Thus, our equation becomes:
-2ln(10)pH + ln(C0 - 10^(-pH)) = c - (ΔrH°/R)T^(-1)
So, -2ln(10)pH + ln(C0 - 10^(-pH)) = f(T^(-1)) is now a linear function.
So, first of all, how did you estimate the initial concentration of CO2?
Henry's law, perhaps?
whole fern
Oh. So, atmospheric pressure, I guess.
whole fern
Ok. Let me quickly look up Henry's constant for CO2.
Ok, found it. It doesn't change much from 273 K to 303 K, so I think we can somewhat assume that C0 doesn't depend on T, either.
Hold on, let me convert it to units we can use.
Ok, it should be 0.0314 M/atm. So, C0 = 0.0314 M.
Thus, our model should be Y = kX + b, where:
Y = -2ln(10)pH + ln(C0 - 10^(-pH))
C0 = 0.0314 M
X = T^(-1)
Try plotting that. If we didn't mess up anywhere, this should be a relatively straight line.
kind cloud
alright ill try this 👍
whole fern
Remember to convert T to absolute temperature!
kind cloud
ok
whole fern
I'll also try, I'm curious whether I've messed up somewhere or not.
kind cloud
i got this, but idk if its correct
whole fern
I have a similar picture. Doesn't look too linear, though... Hm.
whole fern
I also messed up the sign somewhere up there. Y should be -2ln(10)pH - ln(C0 - 10^(-pH)). That can also be simplified to Y = -ln(10)pH - ln(C0 10^pH - 1). Then that yields this. Still not that linear, though...
whole fern
Maybe I'm just misunderstanding the exercise, though.
I do kind of fear that C0 may depend on T, and quite a lot.
whole fern
Well, there is, at least, a way to check whether the slope is correct, at least.
We can find the enthalpy of dissociation of H2CO3 using van't Hoff.
Ok, so, according to 10.1039/A708075A, the enthalpy of dissociation of H2CO3 at 298.15 K is about 8-10 kJ/mol.
Let's see what we get in our case.
As established above, the slope of this line is k = -ΔrH°/R. So:
ΔrH° = -Rk = -0.008314 kJ/(mol*K)*1779 K = -14.8 kJ/mol
Well, that's not good. The sign is wrong.
Hm... I mean, this is somewhat odd, though.
As far as I know, dissociation of weak acids should be slightly endothermic.
So, as temperature increases, pH should decrease, as a larger fraction of acid dissociates.
But in your case pH increases with temperature.
kind cloud
but solubility of gases decreases with temperature
which means pH should be higher
with higher temp
whole fern
Ah, yeah, we're measuring the apparent constant, after all...
Not sure how to approach this, to be honest. I mean, this is a famously somewhat troublesome constant to determine.
kind cloud
it’s alright 👍