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Topic: Equilibrium Constant Equations  (Read 345 times)

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Offline Phoebe

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Equilibrium Constant Equations
« on: August 14, 2019, 04:01:35 AM »
Hello :)

This is my first post, nice to meet you all, I'm hoping you might share some chemistry wisdom with me.

I'm working through a couple of questions, and feel very uncertain about my progress, please tell me if I've gone rogue.

The first question asks me to write a balanced chemical equation for the dissociation of CH3COOH in water, including the phase state of each chemical species, my answer is:

                                              CH3COOH (aq) + H2O(l) :rarrow: CH3COO-(aq) + H3O-(aq)

The second question is:

Based on your chemical equation in the previous question, write an expression for corresponding equilibrium constant for acid dissociation Ka in terms of the concentrations of the species involved (hint: which one of the species is NOT included in the expression?)

This question is where I'm stuck - Have I written the original equation wrong?



The third question is:

The equilibrium constant for acid dissociation of acetic acid, Ka = 1.8x10-5. Explain what this tells us about the favoured direction of the equilibrium (i.e. does it favour reactants or products)?

I think: Acetic acid is a weak acid, meaning it will sparingly donate a proton (ionise) in water, The equilibrium lies to the left – it favours the reactants. Ka << 1

But again, my grip on this is (clearly) tenuous.

Grateful for any and all insight.

Offline AWK

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Re: Equilibrium Constant Equations
« Reply #1 on: August 14, 2019, 04:14:19 AM »
http://www.chembuddy.com/?left=pH-calculation&right=bronsted-lowry-theory
or any textbook that concerns acid-base theories.
AWK

Offline mjc123

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Re: Equilibrium Constant Equations
« Reply #2 on: August 14, 2019, 04:33:48 AM »
Your equation is correct, except it should be H3O+, not H3O-. Just a typo, I'm sure.

Offline Phoebe

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Re: Equilibrium Constant Equations
« Reply #3 on: August 14, 2019, 05:05:24 AM »
Thanks, Ok - I'll review acids and base theory and hope for clarity the second time around.

Thanks to you both for taking the time to look over my question.

Offline Babcock_Hall

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Re: Equilibrium Constant Equations
« Reply #4 on: August 14, 2019, 10:33:43 AM »
With respect to your third question, your discussion of where the equilibrium lies looks OK to me, although you could couch your answer in terms of the equilibrium concentrations of the products and reactants.

Offline sunkal

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Re: Equilibrium Constant Equations
« Reply #5 on: August 14, 2019, 02:43:46 PM »
Phoebe,

With respect to your second question, what you need to know is:
For the generic equation  A (aq) + C (aq) + D (l) = E (s) + F (aq), K = [F]eqm / ([A]eqm [C]eqm).  Notice that only the ones with (aq) appear in the equilibrium constant expression. The solid and the liquid do not appear. The reason is that their molarities are constants at the given temperature.  Note: I couldn't use B as a reactant, since when I put it in square brackets, it is understood as bolding, and everything after that appears in bold.

For the ionization of weak acid the K is symbolized as Ka. Otherwise, the above applies in that case also.

Offline Borek

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Re: Equilibrium Constant Equations
« Reply #6 on: August 14, 2019, 04:49:29 PM »
I couldn't use B as a reactant, since when I put it in square brackets, it is understood as bolding

https://www.chemicalforums.com/index.php?topic=59314.msg212349#msg212349
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline sunkal

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Re: Equilibrium Constant Equations
« Reply #7 on: August 14, 2019, 05:46:30 PM »
https://www.chemicalforums.com/index.php?topic=59314.msg212349#msg212349

Thank you very much, Borek. I understand there are two solutions for the issue I faced.

Offline Phoebe

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Re: Equilibrium Constant Equations
« Reply #8 on: August 18, 2019, 11:35:34 PM »
Hello everyone - Apologies for my slow reply, it's the end of semester for me and I had to drop chemistry for a few days to focus on more pressing deadlines.

I wanted to thank everybody who commented on my post. Chemistry is new to me and I'm really enjoying it, tho it does feel very unfamiliar, I've found that sometimes I encounter a question (like the chemical equilibrium one) phrased in such a way it seems impenetrable, when in fact the concept is quite simple.
 
It took me a while to digest but now I understand that because aA and bB don't fully react, their concentration remains greater than the products cC and dD, thus equilibrium favours the reactants. Also H2O and solids are not included in the expression, which would look something like:   

                 Ka = [CH3COO-] x [H3O+]
                         -------------------------------
                         [CH3COOH]


I really appreciate everyone's comments, they really helped crystallise the concepts for me. Thank you

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