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July 01, 2025, 01:57:45 pm

Author Topic: Addition of species on both sides of an equilibrium reaction  (Read 1663 times)  Share 

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kenhung123

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Addition of species on both sides of an equilibrium reaction
« on: October 19, 2010, 08:12:54 pm »
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When you add water and H+ to a weak acid, aren't you adding reactants and products? Then how can you predict the effects?
Question 1
If a small amount of 0.10 M hydrochloric acid, a strong acid, is added to a 0.10 M solution of ethanoic acid, a weak acid, and the temperature remains constant
A.   the equilibrium constant for the ionisation of CH3COOH (aq) increases
B.   the equilibrium concentration of CH3COO– (aq) decreases
C.   the equilibrium concentration of ethanoic acid decreases
D.   the pH increases

happyhappyland

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Re: Addition of species on both sides of an equilibrium reaction
« Reply #1 on: October 19, 2010, 10:22:54 pm »
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the conc of water does not change therefore it will have no affect on equilibrium thats why when you calculate Ka value you dont include it. Henceforth the answer is B.
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kenhung123

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Re: Addition of species on both sides of an equilibrium reaction
« Reply #2 on: October 19, 2010, 10:26:07 pm »
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Sorry but I don't really understand, when we dilute, doesn't the % ionisation increase? Water is present in the solution of HCl right?

happyhappyland

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Re: Addition of species on both sides of an equilibrium reaction
« Reply #3 on: October 19, 2010, 10:34:51 pm »
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No. % ionisation is (ch3coo-/ch3cooh)X100 Water has no role in it
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kenhung123

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Re: Addition of species on both sides of an equilibrium reaction
« Reply #4 on: October 19, 2010, 11:12:28 pm »
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CH3COOH + H2O <=> CH3COO- + H3O+

Adding water favours the forward reaction
Adding H3O+ favours the backward reaction

cama23

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Re: Addition of species on both sides of an equilibrium reaction
« Reply #5 on: October 20, 2010, 12:16:38 am »
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adding water doesnt favour the forward reaction as the concentration of water does not change!
i may be wrong but from memory adding water only produces a forward or back reaction when the amount of particles differs from reactants to products (favours more particles)

happyhappyland

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Re: Addition of species on both sides of an equilibrium reaction
« Reply #6 on: October 20, 2010, 08:52:31 am »
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CH3COOH + H2O <=> CH3COO- + H3O+

Adding water favours the forward reaction
Adding H3O+ favours the backward reaction

Yeh but h2o doesnt change... so you leave it out. Only increase h3o+ will favour a backwards reaction.
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stonecold

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Re: Addition of species on both sides of an equilibrium reaction
« Reply #7 on: October 20, 2010, 08:56:28 am »
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Is it because [H2O] is constant at ~55M, so adding or removing it won't really do anything?

And even though you are diluting everything, the number of particles on both sides of the reaction is the same no neither side will be favoured.
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fady_22

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Re: Addition of species on both sides of an equilibrium reaction
« Reply #8 on: October 20, 2010, 09:13:50 am »
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Is it because [H2O] is constant at ~55M, so adding or removing it won't really do anything?

And even though you are diluting everything, the number of particles on both sides of the reaction is the same no neither side will be favoured.
You aren't really diluting anything. The key word is "a small amount"-- the effect of dilution is very minimal.
The concentration of H+ increases when you add HCl.
Thus, the the equilibrium system will shift to the left, decreasing the concentration of CH3COO-. Hence, B.
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kyzoo

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Re: Addition of species on both sides of an equilibrium reaction
« Reply #9 on: October 20, 2010, 09:19:11 am »
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Is it because [H2O] is constant at ~55M, so adding or removing it won't really do anything?

And even though you are diluting everything, the number of particles on both sides of the reaction is the same no neither side will be favoured.

So then why does diluting a weak acid increase percentage ionisation?
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stonecold

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Re: Addition of species on both sides of an equilibrium reaction
« Reply #10 on: October 20, 2010, 09:19:52 am »
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Oh, lol, I didn't even read the question.  Yeah, you're just adding H+ so it favored the reverse.  H+ can't exist unless it is in water. :)

They'll never give us one of those questions where they do two things at once, or if the do, like in a trial exam I did once, the answer will be 'the equilibrium shift cannot be determined.'

Unless they both both favour the same reaction, then you can still determine the answer.
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stonecold

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Re: Addition of species on both sides of an equilibrium reaction
« Reply #11 on: October 20, 2010, 09:21:41 am »
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Is it because [H2O] is constant at ~55M, so adding or removing it won't really do anything?

And even though you are diluting everything, the number of particles on both sides of the reaction is the same no neither side will be favoured.

So then why does diluting a weak acid increase percentage ionisation?

I honestly don't know.  Can you please explain?

Is it something to do with the collision theory...
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fady_22

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Re: Addition of species on both sides of an equilibrium reaction
« Reply #12 on: October 20, 2010, 09:23:05 am »
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Is it because [H2O] is constant at ~55M, so adding or removing it won't really do anything?

And even though you are diluting everything, the number of particles on both sides of the reaction is the same no neither side will be favoured.

So then why does diluting a weak acid increase percentage ionisation?

Water is the solvent; it can't be regarded as a "particle".
Dilution still causes the forward reaction to be favoured, increasing percentage ionisation.

I like to omit water from ionisation reactions, just to stop this confusion.
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kyzoo

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Re: Addition of species on both sides of an equilibrium reaction
« Reply #13 on: October 20, 2010, 09:24:55 am »
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Is it because [H2O] is constant at ~55M, so adding or removing it won't really do anything?

And even though you are diluting everything, the number of particles on both sides of the reaction is the same no neither side will be favoured.

So then why does diluting a weak acid increase percentage ionisation?

Water is the solvent; it can't be regarded as a "particle".
Dilution still causes the forward reaction to be favoured, increasing percentage ionisation.

I like to omit water from ionisation reactions, just to stop this confusion.

O, so basically in CH3COOH <---> CH3COO- + H+
Dilution will cause shift the equilibrium to the side with more particles, hence right side?

Is that right?
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fady_22

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Re: Addition of species on both sides of an equilibrium reaction
« Reply #14 on: October 20, 2010, 09:25:50 am »
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Is it because [H2O] is constant at ~55M, so adding or removing it won't really do anything?

And even though you are diluting everything, the number of particles on both sides of the reaction is the same no neither side will be favoured.

So then why does diluting a weak acid increase percentage ionisation?

Water is the solvent; it can't be regarded as a "particle".
Dilution still causes the forward reaction to be favoured, increasing percentage ionisation.

I like to omit water from ionisation reactions, just to stop this confusion.

O, so basically in CH3COOH <---> CH3COO- + H+
Dilution will cause shift the equilibrium to the side with more particles, hence right side?

Is that right?

Yes. :)
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