[Needaask]

The pH of a weak acid strong base's salt such as CH3COONa is less than 7 because CH3COO- would undergo hydrolysis with water to form CH3COOH and OH-. Then also, the newly formed CH3COOH undergoes hydrolysis to produce CH3COO- and H3O+. But overall there is still more OH- making it more alkali than acidic causing the pH to be less than 7.

However, for a an acidic buffer with sodium acetate and ethanoic acid the pH is less than 7. Why would the production of H3O+ be favoured here instead of OH- like in the first case? And how should I determine if OH- is favoured or H+?

Thanks

[Borek]

The pH of a weak acid strong base's salt such as CH3COONa is less than 7

No, it is higher than 7.

[magician4]

in addition:

for a buffer,*⁾ the question of the pH resulting thereof is decided by the pKa of the acid involved: the pH of the respective buffer will be equal to the pKa of the acid.

as almost all carbonic acids will show a pKa smaller than 7 , for this class of buffers H⁺ always will win over OH^-.
if, on the other hand, we'd be talking even weaker acids - for example inorganic acids like HCN and thatlike , you'll find the opposite result by the end of the day: a HCN/CN^- buffer will show a pH of ~ 9.2


hence: take a look at the pKa, and you'll know who will win


regards

Ingo


*⁾
there are many uses of the word "buffer" in chemistry
in my answer, I refer to a 1:1 (mole/mole) solution acid:salt thereof , in relevant concentration (so the pK_W can be neglected). Furthermore the cation of the salt involved should be of next to no pH-relevance

[Needaask]

in addition:

for a buffer,*⁾ the question of the pH resulting thereof is decided by the pKa of the acid involved: the pH of the respective buffer will be equal to the pKa of the acid.

as almost all carbonic acids will show a pKa smaller than 7 , for this class of buffers H⁺ always will win over OH^-.
if, on the other hand, we'd be talking even weaker acids - for example inorganic acids like HCN and thatlike , you'll find the opposite result by the end of the day: a HCN/CN^- buffer will show a pH of ~ 9.2


hence: take a look at the pKa, and you'll know who will win


regards

Ingo


*⁾
there are many uses of the word "buffer" in chemistry
in my answer, I refer to a 1:1 (mole/mole) solution acid:salt thereof , in relevant concentration (so the pK_W can be neglected). Furthermore the cation of the salt involved should be of next to no pH-relevance

Actually, why would the pH be dependent on the pKa? I was thinking about this and I thought in the CH3COONa case, there would be total dissociation producing CH3COO- and the Kb of it is 5.6x10^-10 which is smaller than the Ka of its acid variant CH3COOH. So it would undergo hydrolysis to form the ethanoic acid and hydroxide ion. Then, the ethanoic acid would also hydrolyze to form H+ and the acetate ion. And even though the Ka is greater, the amount of H+ relative to the OH- is smaller?

While in the ethanoic acid and sodium acetate buffer, if we have equal concentrations of the acetate ion and ethanoic acid, Ka being greater than Kb would result in a more acidic solution as more H+ ions would form when compared to OH- ions? So if I had a lot more of the sodium acetate, the solution might actually become alkali instead of acidic?

I'm not too sure if I can explain it this way cos when I have the buffer, both the Ka and Kb reactions occur simultaneously so I'm unsure how to look at their effects simultaneously. Hope you can explain this. Thanks

[magician4]

Actually, why would the pH be dependent on the pKa?

it's a consequence of the law of mass action

if you wrote the law for any given (here: mono- )acid, you should write it like this:

[tex] K_a \ = \ \frac {[A^-] \ \cdot \ [H^+]}{[HA]} [/tex]

if you rearrange this slightly

[tex]K_a \ = \ [H^+] \ \cdot \ \frac {[A^-] }{[HA]}[/tex]

and write it in the logarithmic form:

[tex]log_{10} K_a \ = \ log_{10}[H^+] \ + \ log_{10} \frac {[A^-] }{[HA]}[/tex]

rearrange it again

[tex]- \ log_{10} [H^+] \ = \ - \ log_{10}K_a \ + \ log_{10} \frac {[A^-] }{[HA]}[/tex]

and then writhe "p" as shorthand for "-log₁₀" you'll end up with

[tex]pH \ = \ pK_a \ + \ log_{10} \frac {[A^-] }{[HA]}[/tex]

this is the Henderson-Hasselbalch equation.

Now, as for a buffer the ideal situation would be that [A^-]  = [HA] , the quotient [A-] / [HA] becomes 1 exactly. the logarithm of 1 being 0 on the other hand, for a given buffer pH = pKa +  0 = pKa will apply

that's why pKa rules the pH of a buffer


regards

Ingo

[Needaask]

Actually, why would the pH be dependent on the pKa?

it's a consequence of the law of mass action

if you wrote the law for any given (here: mono- )acid, you should write it like this:

[tex] K_a \ = \ \frac {[A^-] \ \cdot \ [H^+]}{[HA]} [/tex]

if you rearrange this slightly

[tex]K_a \ = \ [H^+] \ \cdot \ \frac {[A^-] }{[HA]}[/tex]

and write it in the logarithmic form:

[tex]log_{10} K_a \ = \ log_{10}[H^+] \ + \ log_{10} \frac {[A^-] }{[HA]}[/tex]

rearrange it again

[tex]- \ log_{10} [H^+] \ = \ - \ log_{10}K_a \ + \ log_{10} \frac {[A^-] }{[HA]}[/tex]

and then writhe "p" as shorthand for "-log₁₀" you'll end up with

[tex]pH \ = \ pK_a \ + \ log_{10} \frac {[A^-] }{[HA]}[/tex]

this is the Henderson-Hasselbalch equation.

Now, as for a buffer the ideal situation would be that [A^-]  = [HA] , the quotient [A-] / [HA] becomes 1 exactly. the logarithm of 1 being 0 on the other hand, for a given buffer pH = pKa +  0 = pKa will apply

that's why pKa rules the pH of a buffer


regards

Ingo

Hi thanks for the great post!

But I was actually wondering. In the case of the sodium acetate there would be both acetate ions and ethanoic acid molecules. However, the production of OH- reaction (Kb) is favoured. While in the buffer solution with both of them in the solution as well, why would the production of the H+ reaction (Ka) is favoured?

I understand the math in this case but I don't really get the theory on why more H+ is formed than OH- actually.

Thanks again !

[magician4]

But I was actually wondering.
(...) why would the production of the H+ reaction (Ka) is favoured?

a physicist once told me: a scientist should ask "how?" , i.e. try to explain interconnections, and not "why" , i.e. try to be philosophical about what we're finding. Ours is to describe nature and how it works, and not to give it a meaning instead.

what in last consequence you're doing is, to question nature why acetic acid does have the properties it actually has (i.e. the ratio of success with either OH^- or H⁺ production, if both the acid and it's salt were in the game)

now, you could step back some and try to explain the properties of acetic acid in aqueous solutions, maybe by applying some quantum mechanic calculations, too

but that's as far as we can reach in chemistry : to explain these properties by some underlying equations, which culminate in this specific K_a value for acetic acid  (with other acids having other K_a values belonging to): i.e. calculate the value for K_a from more basic features instead of just measuring it.

beyond that, maybe physicists do have some deeper insights how it is that matter behaves like it does: somehow the properties of acetic acid should emerge from that level, I take it


beyond that, I do have no answers

regards

Ingo