Buffers question

[ilovemedi]

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So I'm using TBR and in the buffers section it talks about the equivalence point, which is when there are equal mole portions of acid and base. So in a buffer with weak conjugate pairs, why isnt the equivalence point always at a ph of 7. Is it because in weak buffers with conjugate pairs the acid/base ratio arent always equal or does it have to do with the fact that one conjugate pair may be relatively stronger?

In the book it said for a strong acid mixed with strong base, it would be at 7, so I'm wondering what it would be in a buffer with a weak acid and weak base...

 

[akim6890]

I'm not sure I fully understand your question but think about what a buffer does and why the equivalence point might change. A buffer prevents drastic changes in pH by keeping the proton concentration stable. So when you titrate a strong acid like HCl with a strong base like NaOH, the protons in solution from HCl bond with the OH- ions preventing changes in pH until there are no protons from HCl left at pH = 7 otherwise known as the equivalence point where the acid/base portions are equal. Any more NaOH added will cause the pH to go up by directly adding OH ions.

Now think about it with a strong acid and a weak base. Actually, what does it mean to be a weak base? It means in solution, there are fewer OH ions per mole of base added. If there are a bunch of protons in solution from my strong acid and I begin to add my weak base, it will have fewer OH- ions to soak up those protons from that strong acid and so it takes more weak base to neutralize less acid. Equivalence point = when weak base added is equal to strong acid present which will be? Less than 7. Why? Because there are still plenty of protons lying around from the poor job of the weak base creating an acidic solution. Opposite direction with a strong base and a weak acid and also, it doesn't matter which way you do it (adding base to acid or acid to base); it will still have the same result.

Now think about this... if I have 1L of 1M H2SO4 present in solution. How much NaOH must I add to reach pH = 7? Assume H2SO4 and NaOH are strong acid/base respectively.

 

[ilovemedi]

Let me put it to you like this, if you add a weak base to a weak acid, what would the equivalence point be? Would it be at a ph of 7 or does it depend on which is relatively stronger?

edit: to answer your question- 1L of 1 M NaOh because both are strong

 

[brlin]

I just hit this chapter today too. 🙂 From what I understand, the "equivalence point" of a weak acid and a weak base is technically just the buffer zone. True equivalence points occur only when you titrate a weak acid/base with a strong base/acid. Or strong/strong.

 

[brlin]

Edit: oops. Ignore this post. Sorry! But while I'm at it, I'll also point out that the weak acid/bases I were talking about would have to be a conjugate pair. A weak acid / base that are not a conjugate pair would not react at all

 

[akim6890]

I just hit this chapter today too. 🙂 From what I understand, the "equivalence point" of a weak acid and a weak base is technically just the buffer zone. True equivalence points occur only when you titrate a weak acid/base with a strong base/acid. Or strong/strong.

False. The equivalence point is when the acid equivalents equal the base equivalents. It has nothing to do with what combination of strong and weak. And sorry, that was a trick question. You need to keep in mind that H2SO4 if a diprotic acid and therefore has 2 equivalents of protons it can give up meaning it would take twice as much volume of NaOH to reach a pH of 7.

 

[akim6890]

Edit: oops. Ignore this post. Sorry! But while I'm at it, I'll also point out that the weak acid/bases I were talking about would have to be a conjugate pair. A weak acid / base that are not a conjugate pair would not react at all

This is also false. The conjugate pair is the buffer and it reacts with a much different strong acid or base to prevent change in pH. For example, consider the weak acid acetic acid and it's conjugate base sodium acetate. The dissociation in water would be
CH3COOH + H2O <--> H3O+ + CH3COO-

That is your buffer solution and when you add a strong base like NaOH, it takes the proton from H3O+ causing the equilibrium to shift to the right making more H3O+ and keeping the pH neutral. Likewise, if you added a strong acid like HCl, then the dissociated proton would react with acetate ion again keeping it neutral.

 

[ilovemedi]

they would not react as in they would not for a buffer, right? Also i have another question about that chapter: could you or anyone here clarify the equation 'ph range = pka +or - 1' means? I did not understand that paragraph at all. Is it just the ph range at which the buffer will tend to linger at? And does this equation work for all buffers or just buffers that are made from conjugate pairs (since they derived it using the henderson equation, which relies on conjugate pairs?)

 

[ilovemedi]

False. The equivalence point is when the acid equivalents equal the base equivalents. It has nothing to do with what combination of strong and weak. And sorry, that was a trick question. You need to keep in mind that H2SO4 if a diprotic acid and therefore has 2 equivalents of protons it can give up meaning it would take twice as much volume of NaOH to reach a pH of 7.

Interdasting... Two questions then...

1. what would the ph be of a solution containing equal molar equivalents of weak base and its weak conjugate acid? would it be 7? does it depend on their relative strengths?

2.As for your trick question... you got me! But are you sure it would be exaclty twice as much NaOH? After H2S04 gives up its first proton its no longer a strong acid, but rather a weak one. So wouldnt it be slightly less than twice as much??

 

[akim6890]

they would not react as in they would not for a buffer, right? Also i have another question about that chapter: could you or anyone here clarify the equation 'ph range = pka +or - 1' means? I did not understand that paragraph at all. Is it just the ph range at which the buffer will tend to linger at? And does this equation work for all buffers or just buffers that are made from conjugate pairs (since they derived it using the henderson equation, which relies on conjugate pairs?)

Ya so if you look at the equation pH = pKa + log(A-/HA) when A- and HA are equal meaning half the buffering capacity has been reached, pH = pKa. Therefore, the optimal buffering capability will be at that pH because there are equal parts of the conjugate acid/base pair to combat either a base or an acid added. Does that make sense? Think about it like a game of tug of war where people are being moved from one side to help the other. When there are an equal number of people (pH = pKa) or almost equal on either side, it's hard to move the middle of the rope (pH) but when you start moving people from one side to the other, it starts becoming much easier to move the middle of the rope one way or the other. Similar to that, the pH begins to move easier once the balance has shifted. As to why it's specifically +or- 1 from the pKa, I'm not totally sure. If I had to guess: in that range, the ratio of the pair is 10:1 or 1:10 so after that the buffering effect is minimal because there simply isn't any left to buffer.

 

[ilovemedi]

Thanks! Sorry for all the questions, my gen chem class barely touched on buffers/titrations so all of this is new to me.
If you dont mind i do have one last question. So in the chapter it talks about how a buffer can be made one of two ways: by mixing equal parts of a weak acid and its conjugate base or by titration of a weak reagent with a strong reagent. My question is, can the henderson-hasselbalch equation be used to determine the ph of a buffer reguardless of which of these ways the buffer was made?

I'm thinking yes but something i remember reading about only using using that equation for a conjugate system is throwing me off.

 

[DrDashInd]

The effect of both these methods is the same: they make a buffer (weak acid and it's conjugate base). During a titration when you add strong base, it deprotonates the weak acid - leaving the conjugate base, so you get left over weak acid and its' conj. base. Thus the effect of both methods is the same and pKa = pH-log(A-/HA) can be used in both instances.

 

[ChemTutor]

Interdasting... Two questions then...

1. what would the ph be of a solution containing equal molar equivalents of weak base and its weak conjugate acid? would it be 7? does it depend on their relative strengths?

...

depends on their relative strength.

i do not know if this is helpful, but this example is equivalent to dissolving the same molar amount of the corresponding salt in water.

 

[brlin]

False. The equivalence point is when the acid equivalents equal the base equivalents. It has nothing to do with what combination of strong and weak. And sorry, that was a trick question. You need to keep in mind that H2SO4 if a diprotic acid and therefore has 2 equivalents of protons it can give up meaning it would take twice as much volume of NaOH to reach a pH of 7.

Sorry, my wording might have been off. I meant that in the context of forming the buffer via neutralization of one reactant to form it's equivalent conjugate acid or base. As in you can only obtain an equivalence point during a titration when when you titrate a weak acid with a strong base or you titrate a weak base with a strong acid.

 

[brlin]

This is also false. The conjugate pair is the buffer and it reacts with a much different strong acid or base to prevent change in pH. For example, consider the weak acid acetic acid and it's conjugate base sodium acetate. The dissociation in water would be
CH3COOH + H2O <--> H3O+ + CH3COO-

That is your buffer solution and when you add a strong base like NaOH, it takes the proton from H3O+ causing the equilibrium to shift to the right making more H3O+ and keeping the pH neutral. Likewise, if you added a strong acid like HCl, then the dissociated proton would react with acetate ion again keeping it neutral.

I think we're saying the same thing. That a weak acid/base conjugate pair makes a buffer