Dissociated Ionic Compounds

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Ruby Richter 2L
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Joined: Thu Jul 25, 2019 12:17 am

Dissociated Ionic Compounds

Postby Ruby Richter 2L » Sun Jan 12, 2020 12:10 pm

How do you use the net ionic equation to find the equilibrium constant for reactions? The textbook says it should be written using the activity for each type of ion but I'm not sure what that means.

CMaduno_1L
Posts: 102
Joined: Wed Sep 18, 2019 12:18 am

Re: Dissociated Ionic Compounds

Postby CMaduno_1L » Sun Jan 12, 2020 12:22 pm

In regards to the activity of each ion, I believe they are referring to the ion's respective concentration. I hope this helps!

Ruby Richter 2L
Posts: 103
Joined: Thu Jul 25, 2019 12:17 am

Re: Dissociated Ionic Compounds

Postby Ruby Richter 2L » Sun Jan 12, 2020 9:12 pm

CMaduno_1L wrote:In regards to the activity of each ion, I believe they are referring to the ion's respective concentration. I hope this helps!

That makes sense, in a problem would that value be given or is there a way to calculate it?

Sue Bin Park 2I
Posts: 52
Joined: Mon Jun 17, 2019 7:24 am

Re: Dissociated Ionic Compounds

Postby Sue Bin Park 2I » Sun Jan 12, 2020 9:19 pm

Ruby Richter 2L wrote:
CMaduno_1L wrote:In regards to the activity of each ion, I believe they are referring to the ion's respective concentration. I hope this helps!

That makes sense, in a problem would that value be given or is there a way to calculate it?


it would likely be given but i guess if they did want us to calculate the ions' concentrations they could provide us with a dissociation constant and the amount of ionic compound added to the soln. that would involve combining earlier chapters' skills so i'd think it's def feasible for an exam q but you probs wont see that in a chapter review or something.

Juliet Stephenson 4E
Posts: 100
Joined: Wed Sep 18, 2019 12:21 am

Re: Dissociated Ionic Compounds

Postby Juliet Stephenson 4E » Sun Jan 12, 2020 9:35 pm

Sue Bin Park 3D wrote:
Ruby Richter 2L wrote:
CMaduno_1L wrote:In regards to the activity of each ion, I believe they are referring to the ion's respective concentration. I hope this helps!

That makes sense, in a problem would that value be given or is there a way to calculate it?


it would likely be given but i guess if they did want us to calculate the ions' concentrations they could provide us with a dissociation constant and the amount of ionic compound added to the soln. that would involve combining earlier chapters' skills so i'd think it's def feasible for an exam q but you probs wont see that in a chapter review or something.


Could you go over how we could go from the dissociation constant and the amount of ionic compound add to the ion concentration? Would we just multiply them?

Sue Bin Park 2I
Posts: 52
Joined: Mon Jun 17, 2019 7:24 am

Re: Dissociated Ionic Compounds

Postby Sue Bin Park 2I » Wed Jan 22, 2020 6:41 pm

Juliet Stephenson 4E wrote:
Sue Bin Park 3D wrote:
Ruby Richter 2L wrote:That makes sense, in a problem would that value be given or is there a way to calculate it?


it would likely be given but i guess if they did want us to calculate the ions' concentrations they could provide us with a dissociation constant and the amount of ionic compound added to the soln. that would involve combining earlier chapters' skills so i'd think it's def feasible for an exam q but you probs wont see that in a chapter review or something.


Could you go over how we could go from the dissociation constant and the amount of ionic compound add to the ion concentration? Would we just multiply them?


It's pretty much like our acidity constants stuff. lets say A is a cation and B is an anion that dissolves to an extent. this is another dynamic equilibrium: the ionic compound dissociates and reassociates in aqueous solution. we are given Ksp.

AB (s) <-> A+(aq) + B-(aq)
Ksp = [A+][B-]
solve for x = [A+] = [B-]

if the coefficients are diff its just like in our other equilibrium problems we had.
CD(s) <-> 2C+(aq) + D-(aq)
Ksp = [C+]^2[D-] = (2x)^2(x)
solve for x = [D-] = .5[C+]

idk if he'd waste our time on this for a smaller test or midterm but yeah


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