Chemistry Forums for Students > Analytical Chemistry Forum
Equivalence point and neutralization point
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- :lequil: 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.
--- Quote from: Decamethyl on May 19, 2019, 04:17:10 PM ---EP: n (HA) = n (A-) which also means n (H3O+) = n (OH-)
--- End quote ---
Non sequitur. For a weak acid HA=A- means pH=pKa. Unless you mean nHA=nBOH (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.
--- Quote ---Neutralization point: PH value = 7
--- End quote ---
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:
--- Quote from: Borek on May 19, 2019, 05:22:41 PM ---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.
--- Quote from: Decamethyl on May 19, 2019, 04:17:10 PM ---EP: n (HA) = n (A-) which also means n (H3O+) = n (OH-)
--- End quote ---
Non sequitur. For a weak acid HA=A- means pH=pKa. Unless you mean nHA=nBOH (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.
--- Quote ---Neutralization point: PH value = 7
--- End quote ---
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).
--- End quote ---
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:
--- Quote from: Decamethyl on May 19, 2019, 04:17:10 PM ---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- :lequil: 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 ?
--- End quote ---
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).
Navigation
[0] Message Index
[#] Next page
Go to full version