Dx/dy vs dy/dx

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  • #1
Borek
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I am reading an old paper and I have came to the passage shown in the picture:

gunnar.png


k, CA, CB and V0 are all just some constants.

I think I know what they did - they differentiated the first equation calculating dpH/dV and then took reciprocals of both sides of the equation. Is it a valid approach?

I am not able to get the same result solving the first equation for V and calculating dV/dpH then.

Bear in mind I am mathemathically challenged and English is my second language, I can be missing something obvious.
 
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  • #2
D H
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Yes, this is a valid approach. Let [itex]y=f(x)[/itex]. Obviously, [itex]dy/dx = df/dx[/itex]. Differentiating [itex]y=f(x)[/itex] with respect to y yields

[tex]1 = \frac{df}{dx}\,\frac{dx}{dy}[/tex]

Solving for [itex]dx/dy[/itex],

[tex]\frac{dx}{dy} = \frac 1 {df/dx} = \frac 1 {dy/dx}[/tex]
 
  • #3
Borek
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OK, thank you.

So going the other way (that is solving for V and differentiating), I should get the same result?
 
  • #4
epenguin
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OK, thank you.

So going the other way (that is solving for V and differentiating), I should get the same result?
On the way there I think you slightly simplify and easier see the calculation if you combine terms inside the bracket of first eq. into a single fraction and then express that as

pH = k - log(CAV0 - CBV) + log(V0 + V) :smile:
 
  • #5
Borek
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I have to solve for V, having two logs doesn't make it easier.

I tried both manually and with TI 89, but I can't get the same result as the one printed. Will try later again.
 
  • #6
epenguin
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Borek, I'll put this up here instead of PM as hopefully someone can check it in case of mistakes er typos. :biggrin:

For the differentiation

[tex]pH = k - log(\frac{C_AV_0 - C_BV}{V_0 + V})[/tex]

[tex] = k - log(C_AV_0 - C_BV) +log(V_0 + V) [/tex]

[tex] = k + \ln 10 [-\ln(C_AV_0 - C_BV) + \ln(V_0 + V)] [/tex]

Then differentiating

[tex]\frac{dpH}{dV} = [\frac{- C_B}{C_AV_0 - C_BV} + \frac{1}{V_0 + V}].\ln 10[/tex]

From this you can already see that the point where the LHS is infinite (the pH against V graph is vertical, or V against pH horizontal) is where
[tex]V = \frac{V_0C_A}{C_B}[/tex].


The RHS fractions can be combined giving

[tex] [\frac{\ln10.(C_A - C_B)V_0}{(C_AV_0 - C_BV)(V_0 + V)}] [/tex]

If you wanted to invert this into V = a function of [tex]\frac{dV}{dpH}[/tex], rearrange to

[tex](C_AV_0 - C_BV)(V_0 + V) = (C_A - C_B)V_0.\ln 10.\frac{dV}{dpH}[/tex]

call the RHS R, say, and then express V by solving the unlovely quadratic in V:

[tex] -V^2C_B + V V_0(C_A - C_B) + (C_A V_0^2 - R) = 0 [/tex]

You ought to get the same thing by inverting and differentiating but I think it is more tedious.

You said you want to invert the first equation anyway. I question whether you do :biggrin: - if it is for plotting it suffices to switch around axes when you plot rather than invert algebraically.

For inverting algebraically work with H rather than pH; and instead of k (which is probably a p something) I define k = - log M, (M = 10-k)

Then the first equation becomes

[tex]H = M + \frac{C_AV_0 - C_BV}{V_0 + V}[/tex]

which eventually transforms to

[tex] V = [\frac{C_A - C_B}{C_B + H - M} + 1]V_0[/tex]

or in terms of pH and k

[tex] V = [\frac{C_A - C_B}{C_B + 10^{-H} - 10^{-k}} + 1]V_0[/tex]
 
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  • #7
Borek
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From this you can already see that the point where the LHS is infinite (the pH against V graph is vertical, or V against pH horizontal) is where
[tex]V = \frac{V_0C_A}{C_B}[/tex].
Hardly surprising - that's the equivalence point :smile:

You said you want to invert the first equation anyway.
Whatever I was doing was just to make sure I understand what is going on. And today I had a time to try again - everything is OK.

call the RHS R, say, and then express V by solving the unlovely quadratic in V:

[tex] -V^2C_B + V V_0(C_A - C_B) + (C_A V_0^2 - R) = 0 [/tex]
Something is wrong, no need for quadratic here:

[tex] V = \frac {-(10^k - C_A 10^{pH}) V_0} {10^k + C_B 10^{pH}} [/tex]

For inverting algebraically work with H rather than pH
Nah, that's about dV/dpH - volume vs potentiometric answer.

This is from the Gran method original paper, BTW.

k (which is probably a p something)
No, that's log(activity coefficient).

Thank's for the help. Yesterday I was not sure about this reciprocal thing - it seemed logical, but I have learnt not to trust my instincts when it comes to anything more complicated than multiplication table.
 
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  • #8
epenguin
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I will work through it again with more calm and clearer head (also had computer trouble today). :blushing::biggrin:
We want to show the students how we make mistakes and are very 'umble. :approve:
 
  • #9
Redbelly98
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Borek, I'll put this up here instead of PM as hopefully someone can check it in case of mistakes er typos. :biggrin:

For the differentiation

[tex]pH = k - log(\frac{C_AV_0 - C_BV}{V_0 + V})[/tex]

[tex] = k - log(C_AV_0 - C_BV) +log(V_0 + V) [/tex]

[tex] = k + \ln 10 [-\ln(C_AV_0 - C_BV) + \ln(V_0 + V)] [/tex]

Then differentiating

[tex]\frac{dpH}{dV} = [\frac{- C_B}{C_AV_0 - C_BV} + \frac{1}{V_0 + V}].\ln 10[/tex]
Should that be +CB in the numerator (two minus-signs cancel each other)?
 

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