Questions on how buffers are explained in TBR

[ravupadh]

I'm on the buffers section in the book and I have a few questions:

1. What is exactly meant by equivalents in this book? Do they mean equal molar concentrations of the weak acid and conjugate base? I understand that for a typical weak acid, having a weak acid with an equivalent of conjugate base added means that both compounds will be in roughly equal molar concentration units? For example, in HA + H2O -> A- + H3O+, if 2 moles of HA are used to make a solution, an equivalent of conjugate base would be 2 moles of A-? How does this come into play when you have diprotic or triprotic acids however? In that case would an equivalent have to be multiplied by its the stoichiometric coefficient?

2. Why do they state in the book that one way to make a buffer is to mix the weak acid and the weak base in roughly equal molar proportions, yet an acceptable pH range for the buffer calls for the conjugate pairs to be in a 10:1 or 1:10 ratio? Doesn't seem like 1:1 to me.

3. I understand that adding half an equivalent of strong base to weak acid will make a buffer, since half of the weak acid apparently reacts to form conjugate base and both conjugates will be in equal proportions. However, in the book they never show the equation for adding a half an equivalent of strong base to weak acid to make a buffer. In Example 5.7, they add 0.025 moles of KOH to 0.05 moles of HOAc. I understand this makes 0.025 moles of HA and 0.025 moles of A-, but what is the approporiate chemical equation for this? HA + H2O -> A- + H3O+ doesn't have -OH in it.

4. Also, referring back to the previous question and Example 5.7, shouldn't reacting OH with HA to make A- make H3O+ as well? And in that case, shouldn't H3O+ and A- be in equal molar concentrations since they have the same coefficient and both start out as 0? And so shouldn't Ka = (A-)(H3O+)/(HA) or (0.025)(0.025)/(0.025) = Ka? Yet that doesn't equal the Ka. So why is the H3O+ concentration different from A-'s (0.025) if both A- and H3O+ are on the same side of the equation and both start out as roughly 0 since it's a weak acid?

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[ilovemcat]

You have the right idea, except rather than concentration consider one equivalent to be moles of something you have.

 

[Mcat35]

the thing I didn't really like about TBR is how much stuff they teach you that you don't really need to know.. its more for someone thats gonna be a chemical engineer or something.

But anyways for MCAT acid base problems.

As long as you know that buffer = half equivalence point, flat portion of the graph, so yes you would only need 1/2 mole of base/acid since you only need 1/2 amount moles of OH/H+ to balance the H+/OH at half equaivalance.

Know what makes a buffer, and example NaHCO3 H2CO3 would make a buffer because its a weak acid + salt of the conjugate base.

Know henderson hasselbach... so you will know the PH at which there is buffering....


Anything else is pretty superfluous imo... you may disagree but the AAMC's I"ve taken its really basic on this topic.

 

[ravupadh]

I see. Thanks. I'm still confused however about the appropriate chemical equation that comes into play when you use a strong base with a weak acid to make a buffer. For example, let's say you add half an equivalent of NaOH to one equivalent of CH3COOH to make a buffer. I understand CH3COOH dissociates via this reaction: CH3COOH + H2O -> H3O+ + CH3COO-

However, where does the -OH from the NaOH come into play to make the CH3COO-? What's the appropriate chemical equation if you include -OH in the above reaction? I know NaOH dissociates via NaOH -> -OH + Na+ but that still doesn't show how it makes CH3COO- from CH3COOH. Any help would be appreciated. Thanks.

 

[Mcat35]

I see. Thanks. I'm still confused however about the appropriate chemical equation that comes into play when you use a strong base with a weak acid to make a buffer. For example, let's say you add half an equivalent of NaOH to one equivalent of CH3COOH to make a buffer. I understand CH3COOH dissociates via this reaction: CH3COOH + H2O -> H3O+ + CH3COO-

However, where does the -OH from the NaOH come into play to make the CH3COO-? What's the appropriate chemical equation if you include -OH in the above reaction? I know NaOH dissociates via NaOH -> -OH + Na+ but that still doesn't show how it makes CH3COO- from CH3COOH. Any help would be appreciated. Thanks.

ok, the -OH from NaOH would react with the Hydronium Ion from your equation: CH3COOH + H2O -> H3O+ + CH3COO- making the equation go backwards creating more water molecules and based on Le Chat's rules, you would also get more CH3COOH to not dissociate. If you only react 1/2 Mole OH- in this case based on stoichiometry, 1/2 mole of CH3COOH would not disscoiate, but the other half of it will react to form 1/2 mole of its conjugate base CH3COO- making conjugate base / weak acid to be exactly 1/1 so PH = PKA based on henderson hasselbach this is a buffer.

Hope that makes sense.

 

[ravupadh]

ok, the -OH from NaOH would react with the Hydronium Ion from your equation: CH3COOH + H2O -> H3O+ + CH3COO- making the equation go backwards creating more water molecules and based on Le Chat's rules, you would also get more CH3COOH to not dissociate. If you only react 1/2 Mole OH- in this case based on stoichiometry, 1/2 mole of CH3COOH would not disscoiate, but the other half of it will react to form 1/2 mole of its conjugate base CH3COO- making conjugate base / weak acid to be exactly 1/1 so PH = PKA based on henderson hasselbach this is a buffer.

Hope that makes sense.

Oh I see.. so the -OH doesn't come in and pull of an H from the CH3COOH?

But now that I think about it the way you put it wouldn't work for titrations. If you add one full equivalent of OH to CH3COOH, your solution will be entirely composed of CH3COO-, not CH3COOH as is in your methodology.

 

[Mcat35]

Oh I see.. so the -OH doesn't come in and pull of an H from the CH3COOH?



But now that I think about it the way you put it wouldn't work for titrations. If you add one full equivalent of OH to CH3COOH, your solution will be entirely composed of CH3COO-, not CH3COOH as is in your methodology.

well if there is no water you would just have dissociation of the Weak acid which is HA ---> H+ + A- This would be reversible since its a weak acid so its a double arrow. So this way your OH reacts with the H+ to form H2O. But generally it should be aqueous soltuion though, so therefore you'll have water as solvent.

that's why buffer only happens at half equivalence. if you add full equivalence you would not have a buffer.. you would be at the equivalence point of PH greater than 7 since it is a weak acid + a strong base. However at the half equivalence 1/2 mole OH- added you'll have conjugate base/weak acid in equal proportions.

At equivalence you should have nothing but CH3COONa left since all your weak acid has reacted.. Given off its H+ to form water molecules. So past the equivalence point, your solution would get more and more basic since you are just dumping more and more OH- in without anything to balance it out.