Activities and Common Ion Effect

scholarman1988 · Post by **scholarman1988** » Sun Feb 20, 2011 1:21 pm

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?

scholarman1988 · Post by **scholarman1988** » Sun Feb 20, 2011 4:45 pm

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

expert · Post by **expert** » Sun Feb 20, 2011 8:18 pm

Your mistake in the dissociation process.
http://www.chemteam.info/Equilibrium/Ca ... Solub.html
Calculate according to
Hg₂Br₂ ⇔ Hg₂^2+ + 2Br-

Yes, the Hg ion is unusual. It's (Hg-Hg) with 2+ charge

Ksp = 5.6×10-23 = [Hg₂^2+]*[Br-]^2

scholarman1988 · Post by **scholarman1988** » Mon Feb 21, 2011 1:14 am

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.

expert · Post by **expert** » Mon Feb 21, 2011 12:34 pm

Yes, except 5.6e-23 = (x)(x)^2

Don't double Br- twice!

scholarman1988 · Post by **scholarman1988** » Mon Feb 21, 2011 12:38 pm

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"

expert · Post by **expert** » Mon Feb 21, 2011 5:57 pm

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.

scholarman1988 · Post by **scholarman1988** » Mon Feb 21, 2011 8:16 pm

no worries mate, thanks for the help