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# Buffers equation help!

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1. Hey guys,

a buffer solution is formed from the addition of propanoic acid to sodium hydroxide.

The reason why i need to form two equations is because the following question says :

When a further small amount of sodium hydroxide is added in the titration, the pH changes very little.

The answer talks about the OH- reacting with the H+ which is confusing.

Basically i need two equations which will help me understand this process. Cheers
2. Write the ionic equation first, that might make it easier.
You don't really need two.
3. hiya,
a base is a mixture of weak acid (propanoic acid) and its conjugate base and H+ (as its dissociated).

ok sodium hydroxide is a base, so upon the addition of a base (OH-) it would react with the H+ from the dissociation of acid (in the buffer) forming water. this would mean that more acid has to dissociate to make up for the deficit in H+ ions,

do that make sense?
Hope it helps
4. (Original post by James A)
Hey guys,

a buffer solution is formed from the addition of propanoic acid to sodium hydroxide.

The reason why i need to form two equations is because the following question says :

When a further small amount of sodium hydroxide is added in the titration, the pH changes very little.

The answer talks about the OH- reacting with the H+ which is confusing.

Basically i need two equations which will help me understand this process. Cheers
CH3CH2COOH + NaOH = CH3CH2COO-Na+ + H2O

THAT IS A WEAK ACID AND A SALT OF A WEAK ACID THUS A BUFFER SOLUTION

CH3CH2COO- + H+ = CH3CH2COOH

CH3CH2COOH + OH- = CH3CH2COO- + H2O

SO POTENTIAL pH CHANGE IS PREVENTED
5. You only need 1 equation. The second one is simply hydrogen ions reacting with OH ions to form water. Propanoic acid dissociating is your buffer solution. Don't consider NaOH. it's a reversible reaction since propanoic acid dissociates partially to give hydrogen ions and CH3CH2COO- ions. adding MORE hydroxide ions remove hydrogen ions from the solution, so the equilibrium shifts from left to right to produce more of the H+ ions and restoring the pH.
6. (Original post by hayjes02)
hiya,
a base is a mixture of weak acid (propanoic acid) and its conjugate base and H+ (as its dissociated).

ok sodium hydroxide is a base, so upon the addition of a base (OH-) it would react with the H+ from the dissociation of acid (in the buffer) forming water. this would mean that more acid has to dissociate to make up for the deficit in H+ ions,

do that make sense?
Hope it helps
Thanks, it does, but I'm struggling to figure out the equation of the dissociation of propanoic acid.

The correct way is:

CH3CH2COOH ⇌ CH3CH2COO- + H+

however i also stumble across books saying:

CH3CH2COOH + H20 ⇌ H30+ + CH3CH2COO-

why have they included the (h20) in this equation? I thought that when a weak acid dissociates, it needs H20 in the reaction to do so?
7. (Original post by James A)
Thanks, it does, but I'm struggling to figure out the equation of the dissociation of propanoic acid.

The correct way is:

CH3CH2COOH ⇌ CH3CH2COO- + H+

however i also stumble across books saying:

CH3CH2COOH + H20 ⇌ H30+ + CH3CH2COO-

why have they included the (h20) in this equation? I thought that when a weak acid dissociates, it needs H20 in the reaction to do so?
You know in reality H+ ions don't exist, it's H3O ions. Something dissociates because it reacts with water. Whether you consider the role of water or not should make no difference though
8. Oh and if that makes no sense, then think of HCl. In gaseous form it's just a gas. You test it with MOIST blue litmus paper. it has acidic properties ONLY when dissolved in water. That's because H+ ions are H3O+ ions really, but for simplicity we ignore the role of water
9. (Original post by James A)
Thanks, it does, but I'm struggling to figure out the equation of the dissociation of propanoic acid.

The correct way is:

CH3CH2COOH ⇌ CH3CH2COO- + H+

however i also stumble across books saying:

CH3CH2COOH + H20 ⇌ H30+ + CH3CH2COO-

why have they included the (h20) in this equation? I thought that when a weak acid dissociates, it needs H20 in the reaction to do so?

CH3CH2COOH + H20 ⇌ H30+ + CH3CH2COO-

Kc = [ CH3CH2COO-][ H3O+]/ [CH3CH2COOH][H2O]

Kc*[H2O] = Ka

therefore Ka = [ CH3CH2COO-][ H3O+]/ [CH3CH2COOH]

H+ exist as H3O+
but in the equation always use H+

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