anonymoussse
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when solution is neutral, [OH-]=[H+], but the ions are still THERE. they're just there in equal amounts.

how about at equivalence point? is oh-=h+?????? still??
this time, HAVE THE OH- and H+ DISAPPEARED????????? (ALL REACTED, NO LONGER THERE)
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username1445490
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The term equivalence point means that the solutions have been mixed in exactly the right proportions according to the equation. This maybe the same as the neutral point when [OH-]=[H+]. This will be true when the salt formed does not react with water.
But if there is a weak acid or base involved then the equivalence point is not the same as the neutral point since the salt reacts with the water.
If a weak acid is used then at the equivalence point the solution will be slightly alkaline since the salt of the weak acid accepts H+ from water and some OH- is formed.
If a weak base is used then at the equivalence point the solution will be slightly acidic since the salt of the weak base donates H+ to water and some H3O+ is formed.

The H+ and OH- never all disappear. At the neutral point when [OH-]=[H+] and if the temperature is 298K then Ka is 1x10-14 so [H+] and [OH-] are 1x10-7 moldm-3 respectively and pH = 7.
Last edited by username1445490; 1 year ago
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anonymoussse
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(Original post by Madasahatter)
The term equivalence point means that the solutions have been mixed in exactly the right proportions according to the equation. This maybe the same as the neutral point when [OH-]=[H+]. This will be true when the salt formed does not react with water.
But if there is a weak acid or base involved then the equivalence point is not the same as the neutral point since the salt reacts with the water.
If a weak acid is used then at the equivalence point the solution will be slightly alkaline since the salt of the weak acid accepts H+ from water and some OH- is formed.
If a weak base is used then at the equivalence point the solution will be slightly acidic since the salt of the weak base donates H+ to water and some H3O+ is formed.

The H+ and OH- never all disappear. At the neutral point when [OH-]=[H+] and if the temperature is 298K then Ka is 1x10-14 so [H+] and [OH-] are 1x10-7 moldm-3 respectively and pH = 7.
I’m not sure I understand the formation of oh- and h+ from salt and water (I understand it, I don’t understand why) but I’ll remember that.

Also you know with weak acids, in buffers, the equilibrium gets pushed to the left (the conjugate salt of ch3cooh releases lots of ch3coo- so equilibrium shifts to the left (ch3cooh>ch3coo- + h+), therefore, doesn’t the ph of the weak acid get ever lower? (Less acidic)
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anonymoussse
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(Original post by Madasahatter)
The term equivalence point means that the solutions have been mixed in exactly the right proportions according to the equation. This maybe the same as the neutral point when [OH-]=[H+]. This will be true when the salt formed does not react with water.
But if there is a weak acid or base involved then the equivalence point is not the same as the neutral point since the salt reacts with the water.
If a weak acid is used then at the equivalence point the solution will be slightly alkaline since the salt of the weak acid accepts H+ from water and some OH- is formed.
If a weak base is used then at the equivalence point the solution will be slightly acidic since the salt of the weak base donates H+ to water and some H3O+ is formed.

The H+ and OH- never all disappear. At the neutral point when [OH-]=[H+] and if the temperature is 298K then Ka is 1x10-14 so [H+] and [OH-] are 1x10-7 moldm-3 respectively and pH = 7.
I’m not sure I understand the formation of oh- and h+ from salt and water (I understand it, I don’t understand why) but I’ll remember that.

Also you know with weak acids, in buffers, the equilibrium gets pushed to the left (the conjugate salt of ch3cooh releases lots of ch3coo- so equilibrium shifts to the left (ch3cooh>ch3coo- + h+), therefore, doesn’t the ph of the weak acid get ever lower? (Less acidic)

Question 2) the weak acid salt formed is soluble but if it was a strong acid it’d be insoluble right? Therefore it’s a precipitate?????
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username1445490
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(Original post by ggxsywes)
I’m not sure I understand the formation of oh- and h+ from salt and water (I understand it, I don’t understand why) but I’ll remember that.

Also you know with weak acids, in buffers, the equilibrium gets pushed to the left (the conjugate salt of ch3cooh releases lots of ch3coo- so equilibrium shifts to the left (ch3cooh>ch3coo- + h+), therefore, doesn’t the ph of the weak acid get ever lower? (Less acidic)

Question 2) the weak acid salt formed is soluble but if it was a strong acid it’d be insoluble right? Therefore it’s a precipitate?????
1. Formation of OH- and H+ ions by salts reacting with water:
The conjugate base of a weak acid will be a base eg CH3COO- from CH3COOH
CH3COO- is a strong enough base that it will be able to take H+ ions from H2O to form CH3COOH + OH- so a solution containing CH3COO- will be alkaline.
Whereas the conjugate base of a strong acid eg Cl- from HCl is such a weak base that it is unable to take H+ ions from H2O.so a solution containing Cl- ions will be neutral.
The conjugate acid of a weak base is an acid eg NH4+ from NH3. It can donate a proton to water to from H3O+ ions so a solution containing NH4+ ions will be acidic.

2. In a buffer you add the conjugate base (salt) to the weak acid. eg you add CH3COO- ions to CH3COOH. The effect of this is to push the equilibrium to the left and form more CH3COOH. This means the concentration of H+ ion decreases and the pH increases and the buffer will be less acidic than the acid on its own.

3. (Question 2) Solubility of the salt has nothing to do with whether it is the salt of a weak or strong acid.
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