Physical Chemistry pH question

In summary, the dissociation constant for the acid-base reaction is Ka = products / reactants; pH = -log[H+]
  • #1
Jayjayjay
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0

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]?
 
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  • #2
Try to write formula defining θ (which I always called α BTW) using concentrations of the A- and HA.
 
  • #3
Something like (theta) = (Ka)(HA) / (A-)?
 
  • #4
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
[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
is theta = [HA]/([HA]+[A-]) ? Or [HA]/([HA]+[A-]+[H+]) ?

(I'm stuck on the same question. Am I on the right track?)
 
  • #7
Jayjayjay said:
[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
Is it the "Theta = [HA] / ([HA]+[A-])" that skelly posted? since that would give the fraction of protonated to unprotonated?
 
  • #9
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
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
What the stoichiometry of dissociation tells you about concentrations of H+ and A-?
 
  • #12
They will always reach some sort of equilibrium, when they are multiplied they will always equal 1*10^-14?
 
  • #13
I am asking not about H+ and OH-, but about H+ and A- (as in HA ↔ H+ + A-).
 
  • #14
they will have equal amounts when dissociated?
 
  • #15
Yes.

Now you should have everything. Express concentrations using θ and CA, plug them into Ka.
 
  • #16
Should I not be setting θ=[HA]/Ca since Ca=[HA] + [A-], or Am I suppose to try substituting them into Ka separately?
 
  • #17
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
Jayjayjay said:
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
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
Jayjayjay said:
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.
 
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  • #21
θ = ([H+] / Ka) + 1 ?
 
  • #22
Don't guess, solve. It is a trivial algebra.
 
  • #23
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
Jayjayjay said:
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
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
 
  • #26
Jayjayjay said:
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.

So do it.
 
  • #27
Didn't I do that with this

θ = ([H+] / Ka) + 1

⇒ θ = ([H+]+[Ka]) / [Ka]
 
  • #28
This is definitely wrong, check your math.

Sanity check: if θ=1+x/y and both x and y are positive, θ is larger than 1. By definition θ is lower than 1, so your answer must be wrong.
 
  • #29
is "Ka / [H+] = (1-θ) / θ" where I need to start, or is that where I initially messed up.
 
  • #30
Better to start again than search for mistake! You are going in circles or backwards in what is a two-line argument. By #8 or again in #18 you had the expression for θ - which is nothing but rewriting the definition of 1b in chemical/mathematical terms. Why go looking for 1/θ or θ/(1-θ) when you are asked for θ? On RHS of the said equation for θ you have one thing twice and another one once. In #1 you have a relation between those two things. Start again and use that relation to eliminate one of them!
 

1. What is pH and why is it important in physical chemistry?

pH is a measure of the acidity or basicity of a solution. In physical chemistry, it is important because it affects the properties and behavior of substances, such as their solubility and reactivity. pH is also a key factor in many chemical reactions and biological processes.

2. How is pH calculated?

pH is calculated using the negative logarithm of the concentration of hydrogen ions in a solution. The formula for pH is pH = -log[H+], where [H+] is the concentration of hydrogen ions in moles per liter.

3. What is the pH scale and what are its ranges?

The pH scale is a numerical scale that ranges from 0 to 14 and is used to measure the acidity or basicity of a solution. A pH of 7 is considered neutral, while values below 7 are acidic and values above 7 are basic. The scale is logarithmic, meaning that each unit change represents a tenfold difference in acidity or basicity.

4. How does temperature affect pH?

The pH of a solution can be affected by temperature due to changes in the equilibrium of ionization reactions. Generally, the pH of a solution decreases as temperature increases, meaning it becomes more acidic. However, this relationship can vary depending on the specific chemical reactions involved.

5. What are some common applications of pH in physical chemistry?

pH is used in a variety of applications in physical chemistry, such as in the production of pharmaceuticals, food and beverage industry, and environmental monitoring. It is also important in understanding the behavior of acids and bases in chemical reactions and in determining the optimal conditions for various experiments and processes.

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