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# Using pKa values to predict the position of equilibrium

How to use pKa values to predict the position of equilibrium for organic acid-base reactions.

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• At , it was said that a specific proton in the molecule had a pKa value. Is there any meaning to refer a single proton as having a pKa value versus the entire molecule? In general chemistry, I only had teachers refer to an entire molecule as having a pKa value.
• Your Gen Chem teacher probably only used examples that had one acidic proton and could only be involved in one acid base reaction with the other reactants in the example. If you look at a molecule like Sulfuric acid, H2SO4, a molecule with more than one acidic hydrogen, it's going to have two pKa's, one for each proton. In these cases, you can avoid confusion if you talk about the pKa of the proton instead of the pKa of the molecule. It's not wrong to talk about the first and second pKa's of H2SO4 , but having the ability to talk about the pKa of a specific Hydrogen can make communication more clear. In other words, the chemistry doesn't change between the two sets of terminology, but referring to the pKa of a specific proton can be of value in clarifying communication.
• Where is the equation: pKeq = pKa(acid-left) - pKa(acid-right) derived? Is this in another video? Why is this true?
`Keq = [leftA-] [HrightA] / ( [HleftA] [rightA-] )`
rearrange and multiply top and bottom by `[H+]`
`Keq = ([leftA-][H+]/[HleftA]) * ([HrightA]/([rightA-][H+]))`
this is now the same as
`Keq = Ka(left) / Ka(right)`
take negative logs of both sides
`-log( Keq ) = -log( Ka(left) / Ka(right) )`
log properties
`-log( Keq ) = -log( Ka(left) ) - -log( Ka(right) )`
pK = -logK
`pKeq = pKa(left) - pKa(right)`
• In the previous video ('Using pKa table'), acetic acid was marked as a weak acid relative to hydrochloric acid - "relative" being the operative term.

For the above example, is acetic acid considered to be a weak/moderate/strong acid relative to the presence of sodium hydroxide? I suppose what I'm trying to ask is whether acid strength changes relative to the presence of the base and its strength. For example, does acetic acid's pKa value change (from 5) to something like 10 in the presence of a weak base because the base will not readily deprotonate acetic acid?
(1 vote)
• The pKa of the acid isn't dependent on the pKb or strength of the base. The equilibrium constant, Keq, which roughly measures how how readily the reactants will become the products, will.

Think of it like two children who want the same toy. Suzy has the toy. Billy wants to take it. How badly Billy wants the toy won't effect how much Suzy wants the toy, but it will effect who gets the toy in the end. (Example assumes that Suzy's desire of the toy doesn't change when she finds out Billy also wants it.)
• acetic acid basically is considered weak acid, then why is the k(eq) much greater than 1 ,even though acetic acid is a weak acid?
• K_a is much *less''than 1.
K_a = 0.000 0175, and this number is much less than 1.
• I'm terrible with math so this may be an incredibly dumb question but oh well:
in the first example to re-arrange the equation the - from -log was taken from the right to the left (makes sense) and resulted in 10^-.....
In the second equation the right side of the equation is positive - and yet we still end up with a 10^ -.....
can anybody explain why this is so? (probably a simple answer) :/
• Actually, in the second equation, the right side isn't positive. Since it's still a pKeq, it's still going to be -log(Keq). He just doesn't show that step. Rearrange the problem like you did with the first, and you get the same result as he did. I don't think that was a dumb question. :P
• This is the pattern I'm noticing so far(can someone confirm, or give me an example when this is not the case?):

When determining the direction of equilibrium, and say I don't know the pKa values between the acid and its conjugate base, I would compare the two and the one with more resonance structures will be the one that's a better acid. Since more resonance = more carbons for the electrons to localize to, making the oxygen that those electrons belonged to less negative, which lessens the oxygen's need for the proton to stick around. Whereas in the acid that doesn't have as many resonance structures (or any), the electrons will stay localized to the oxygen, making it less likely to give up that proton. , and thus, a worse acid.
• At , what does the crosshair-like symbol below the reversible double-headed arrow symbol indicate? It isn't present in the previous example. Thanks!