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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Shadow on April 27, 2012, 02:39:24 PM

Title: Degree of dissociation
Post by: Shadow on April 27, 2012, 02:39:24 PM

In a solution of zinc-chloride there are 1.806*10²² chloride ions and 11.56g of salt that didn't dissociate. Calculate the degree of dissociation in that solution. Is it 13%?

Title: Re: Degree of dissociation
Post by: Borek on April 27, 2012, 04:14:41 PM

You are on the right track, but you have ignored stoichiometry of the dissociation.

Title: Re: Degree of dissociation
Post by: Shadow on April 28, 2012, 06:03:52 AM

Calculated the n of the salt that didn't dissociate and added to it n(Cl^-)/2.
Then I divided n(Cl-)/2 with the number I got previously and I got 0.013=13%. What's wrong here?

Title: Re: Degree of dissociation
Post by: Shadow on April 29, 2012, 12:35:23 PM

How to get 15% here (the right answer)?
ZnCl₂ :rarrow: Zn²⁺+2Cl^-
I calculated the degree: n(ZnCl₂ that dissociated)/n(ZnCl₂ at the beginning)*100 and got 13%.
n(ZnCl₂ that dissociated) is equal to n(Cl^-)/2.
n(ZnCl₂ at the beginning) is equal to n(ZnCl₂ that dissociated) + n of the undissociated salt.
It can't be a mistake in calculation.

Title: Re: Degree of dissociation
Post by: Borek on April 29, 2012, 04:36:23 PM

Please post exact numbers you are working with. I am getting a different answer. Doesn't mean I am OK, but we need to compare details to be sure who is right.

Title: Re: Degree of dissociation
Post by: Shadow on April 30, 2012, 02:45:27 AM

n(ZnCl2 that dissociated) is equal to n(Cl-)/2, that is 1.806*10²²/6*10²³ all divided by 2, so it is 0.01505mol.
n(ZnCl2 at the beginning) is equal to n(ZnCl2 that dissociated) + n of the undissociated salt so it is 0.01505+11.56/(65+71)=0.10005mol
The degree is 0.01505/0.10005*100=15% not sure where I made a mistake last time, but I got the right one now. Thanks for help.