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Activities and Common Ion Effect
Posted: Sun Feb 20, 2011 1:21 pm
by scholarman1988
Ksp = 5.6×10-23 for [Hg2]Br2 at 25 °C.
Calculate the molar concentration of bromide ions in a saturated mercury(I) bromide solution at 25 °C using the assumption that the solution is ideal -- i.e. the activity coefficients are 1.
So.....my guess is to just calculate the concentrations normally using the Ksp. But the activity coeffs are throwing me off. I'm not sure what it means.
Any suggestions?
Posted: Sun Feb 20, 2011 4:45 pm
by scholarman1988
this is what I did so far.
[Hg2]Br2 <=> 2(Hg^2+) + 2(Br^2+)
I 0 0
C 2x 2x
E
5.6e-23 = 16x^4
1.36e-6=x
therefore concentration of bromide is 2x
ie 2(1.36e-6) = 2.7355e-6
my friend did the same thing but his was a different compound and different x agebra, but his was right and mines still wrong :S
Posted: Sun Feb 20, 2011 8:18 pm
by expert
Your mistake in the dissociation process.
http://www.chemteam.info/Equilibrium/Ca ... Solub.html
Calculate according to
Hg
2Br
2 ⇔ Hg
2^2+ + 2Br-
Yes, the Hg ion is unusual. It's (Hg-Hg) with 2+ charge
Ksp = 5.6×10-23 = [Hg
2^2+]*[Br-]^2
Posted: Mon Feb 21, 2011 1:14 am
by scholarman1988
So if I understood this right, the equation should be
5.6e-23 = (x)(2x)^2
5.6e-23 = 4x^3
2.410e-8 = x
thus the concentration for bromide is 2.41e-8 M?
Cheers mate.
Posted: Mon Feb 21, 2011 12:34 pm
by expert
Yes, except 5.6e-23 = (x)(x)^2
Don't double Br- twice!
Posted: Mon Feb 21, 2011 12:38 pm
by scholarman1988
Hmmm that's interesting. I was looking at the link, then ho come problem 3 is different?
"Problem #3: Determine the Ksp of mercury(I) bromide (Hg2Br2), given that its molar solubility is 2.52 x 10¯8 moles per liter.
When Hg2Br2 dissolves, it dissociates like this:
Hg2Br2 (s) <===> Hg22+ (aq) + 2 Br¯ (aq)
Important note: it is NOT 2 Hg+. IT IS NOT!!! If you decide that you prefer 2 Hg+, then I cannot stop you. However, it will give the wrong Ksp expression.
The Ksp expression is:
Ksp = [Hg22+] [Br¯]2
There is a 1:1 ratio between Hg2Br2 and Hg22+, BUT there is a 1:2 ratio between Hg2Br2 and Br¯. This means that, when 2.52 x 10¯8 mole per liter of Hg2Br2 dissolves, it produces 2.52 x 10¯8 moles per liter of Hg22+, BUT 5.04 x 10¯8 moles per liter of Br¯ in solution.
Putting the values into the Ksp expression, we obtain:
Ksp = (2.52 e-8) (5.04 e-8)2 = 6.40 x 10¯23"
Posted: Mon Feb 21, 2011 5:57 pm
by expert
You are right, and I'm wrong. I didn't pay enough attention to my own reference. I'll try to be a better student from now on.
Posted: Mon Feb 21, 2011 8:16 pm
by scholarman1988
no worries mate, thanks for the help