Physical Chemistry pH question

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Homework Statement


The dissociation constant Ka for the acid-base reaction HA-->(H+)+(A-) is given by Ka = products / reactants; but instead of listing Ka, people generally list pKa = -logKa (where log indicates the 10-base logarithm). Similarly, instead of using [H+], people generally use
pH = -log[H+] (equation 1)

b.) Let (theta) be the fraction of acid molecules that is protonated; so clearly, (theta) is a number between 0 and 1. Express (theta) as a function of [H+] and Ka; fully simplify your answer

2. Homework Equations
pH = -log[H+]

Ka = [H+][A-] / [HA]

pKa = -logKa

The Attempt at a Solution


[/B]
Pretty stuck on what I'm suppose to do. From looking at this and some titration graphs it seems that theta will be the 'Equivalents of H+ added'.

It says let theta be the fraction of acid molecules that is protonated, so does that just mean (Ka = [theta][A-]/[HA])?

And doing this as a function of [H+] = (theta) so then pH = -log[theta]?
 

Answers and Replies

  • #2
Borek
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Try to write formula defining θ (which I always called α BTW) using concentrations of the A- and HA.
 
  • #3
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Something like (theta) = (Ka)(HA) / (A-)?
 
  • #4
Borek
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That's not correct. How much acid in all? How much protonated acid? Ignore Ka and pH for now, just try to define θ.
 
  • #5
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[A-] is the concentration of deprotonated form of the acid,
and [HA] is the concentration of the protonated form of the acid

The theta is suppose to be the fraction of acid molecules that is protonated, so its none of these values right? since [HA] and [A-] are concentrations in mol/L, while the theta is suppose to be some number between 0 and 1 that is some type of equilibrium value where when it equals (0.5) pKa = pH where the amount of protonated and deprotonated are at equal amounts?

I'm sorry we haven't done anything with acids until this homework question...
 
  • #6
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is theta = [HA]/([HA]+[A-]) ? Or [HA]/([HA]+[A-]+[H+]) ?

(I'm stuck on the same question. Am I on the right track?)
 
  • #7
Borek
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[A-] is the concentration of deprotonated form of the acid,
and [HA] is the concentration of the protonated form of the acid

Right. So, what is the total concentration of acid (in all possible forms)?

So far it is not about acids or anything, it is just a basic fraction thing. If there are 20 mice and 30 rats in the lab, what fraction of the animals are rats? It is the same logic.

BTW: skelly got it right in one of their answers.
 
  • #8
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Is it the "Theta = [HA] / ([HA]+[A-])" that skelly posted? since that would give the fraction of protonated to unprotonated?
 
  • #9
Borek
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Exactly. Now you have to look for a way of expressing this fraction using Ka and [H+]. You will need to add another variable to your formulas - analytical concentration of the acid. You may call it CA. By definition

[tex]C_A = [HA] + [A^-][/tex]
 
  • #10
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So using that variable you could do "theta = HA /Ca" which you could turn into "(theta)(Ca) = HA" and then just substitute to get Ka = [(H+)(A-)] / [(ca)(theta)] ? I'm having a hard time thinking of another way to implement it.

 
  • #11
Borek
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What the stoichiometry of dissociation tells you about concentrations of H+ and A-?
 
  • #12
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They will always reach some sort of equilibrium, when they are multiplied they will always equal 1*10^-14?
 
  • #13
Borek
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I am asking not about H+ and OH-, but about H+ and A- (as in HA ↔ H+ + A-).
 
  • #14
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they will have equal amounts when dissociated?
 
  • #15
Borek
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Yes.

Now you should have everything. Express concentrations using θ and CA, plug them into Ka.
 
  • #16
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Should I not be setting θ=[HA]/Ca since Ca=[HA] + [A-], or Am I suppose to try substituting them into Ka separately?
 
  • #17
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You need to express theta in terms of [H+] and Ka. You already know theta = [HA] / ([HA] + [A-]) and what Ka is. Go ahead and use your algebra.
 
  • #18
Borek
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θ=[HA]/Ca

That's just a definition, you are asked to express the θ using Ka and pH.

Actually I just realized I was slightly off in my thinking about the problem, it is simpler than I thought, sorry about that. You don't need total concentration - that is, it is needed to define the θ, but there is no need for a separate variable, as it will cancel out in the final result.

Start solving Ka formula for the [itex]\frac{[HA]}{[A^-]}[/itex] - you will have this ratio on one side of the equation, and some combination of Ka and [H+] on the right. You also have an equation that you wrote much earlier:

[tex]\theta = \frac{[HA]}{[HA]+[A^-]}[/tex]

Solve these two equations for θ, eliminating [HA] and [A-] - it is a trivial algebra.
 
  • #19
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Ka / [H+] = [A-] / [HA]

1 / θ = ([HA] + [A-]) / [HA] ⇒ 1 / θ = 1+ ([A-]/[HA]) ⇒ (1-θ) / θ = [A-] / [HA]

Ka / [H+] = (1-θ) / θ ⇒ θ = 1 + Kaθ / [H+]
 
  • #20
Borek
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Ka / [H+] = [A-] / [HA]

1 / θ = ([HA] + [A-]) / [HA]

OK

θ = 1 + Kaθ / [H+]

Come on, you can do better. θ is still on both sides of the equation.
 
  • #22
Borek
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Don't guess, solve. It is a trivial algebra.
 
  • #23
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So trying to go from 1/θ = ([Ha]+[A-]) / [Ha] ⇒ 1/θ = [Ha]/[Ha] + [A-]/[Ha] is wrong because that is the part that I was skeptical about.
 
  • #24
Borek
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So trying to go from 1/θ = ([Ha]+[A-]) / [Ha] ⇒ 1/θ = [Ha]/[Ha] + [A-]/[Ha] is wrong

No, this is a perfectly valid operation.

Do you understand what it means to solve an equation for a variable? You were right up to

Ka / [H+] = (1-θ) / θ

It contains θ (which is your unknown) and two other variables. Just express θ using these two other variables and not θ itself.
 
  • #25
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Well I thought to solve for theta meant to get theta by itself on one side of the equation and have everything else on the other.

you're saying express theta using the other variables ( Ka / [H+] )? so does that mean just to simplify the part I was correct up to, to get...

[H+]/Ka = θ - 1
 

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