[Decamethyl]

I got stuck somehow when I tried to understand the difference between the equivalence point and neutralization point through a specific example.
Just to get this clear, the theoretical definitions seem clear to me. EP: n (HA) = n (A-) which also means n (H3O+) = n (OH-)
Neutralization point: PH value = 7

But I somehow can't wrap my mind around it when trying to apply it to reaction equation, like this one

CO3HOO- + H3O+ + Na- + OH-  CO3HOO- + NA- + 2H20

What I don't understand here is that at the EP n (Na-) = n (CO3HOO-) which also means n (H30+) = n (OH-) which actually means that the balance is all to the right side of the equation.

But why is the PH value not = 7 here anyways ? Is it because CO3HOO- is still in there and could take an H from H20, so that it's PH value is at around 6 at the EP ?

[Borek]

Never heard abut the neutralization point.

What I know is end point (where we ended the titration) and equivalence point - where we added the stoichiometric amount of titrant.

EP: n (HA) = n (A-) which also means n (H3O+) = n (OH-)

Non sequitur. For a weak acid HA=A^- means pH=pKa. Unless you mean n_HA=n_BOH (number of moles of acid equal to number of moles of base, that's the stoichiometric equivalence). For weak acids it still doesn't mean [H₃O⁺]=[OH^-], as pH of a produced salt of a weak acid is never 7.

Neutralization point: PH value = 7

To be honest - useless idea, unless we are talking about titration of a strong acid/base with a strong base/acid (even then not exactly true).

[pcm81]

Never heard abut the neutralization point.

What I know is end point (where we ended the titration) and equivalence point - where we added the stoichiometric amount of titrant.

EP: n (HA) = n (A-) which also means n (H3O+) = n (OH-)

Non sequitur. For a weak acid HA=A^- means pH=pKa. Unless you mean n_HA=n_BOH (number of moles of acid equal to number of moles of base, that's the stoichiometric equivalence). For weak acids it still doesn't mean [H₃O⁺]=[OH^-], as pH of a produced salt of a weak acid is never 7.

Neutralization point: PH value = 7

To be honest - useless idea, unless we are talking about titration of a strong acid/base with a strong base/acid (even then not exactly true).

I think "Neutralization pont" is a term chem instructors came up with trying to sound cool and hipster like talking about spill cleanup etc...

[pcm81]

I got stuck somehow when I tried to understand the difference between the equivalence point and neutralization point through a specific example.
Just to get this clear, the theoretical definitions seem clear to me. EP: n (HA) = n (A-) which also means n (H3O+) = n (OH-)
Neutralization point: PH value = 7

But I somehow can't wrap my mind around it when trying to apply it to reaction equation, like this one

CO3HOO- + H3O+ + Na- + OH-  CO3HOO- + NA- + 2H20

What I don't understand here is that at the EP n (Na-) = n (CO3HOO-) which also means n (H30+) = n (OH-) which actually means that the balance is all to the right side of the equation.

But why is the PH value not = 7 here anyways ? Is it because CO3HOO- is still in there and could take an H from H20, so that it's PH value is at around 6 at the EP ?

CO3HOO- and NA- can take H+ from H2O hence pH is not 7.

[Borek]

Haven't noticed it earlier, no such thing as Na^- (at least not in water solutions).

[pcm81]

Haven't noticed it earlier, no such thing as Na^- (at least not in water solutions).

NA, not Na. N moles of Acid-, not Sodium.

[Borek]

That was my initial idea, but it is written and used so inconsistently (n(A-), Na-, NA-, n(Na-)) that I started to doubt that's what OP meant.