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Topic: Solubility product  (Read 4212 times)

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Offline Rutherford

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Solubility product
« on: June 17, 2012, 05:13:09 AM »
Calculate the concentration of Cl- ions that is required to start the precipitation of AgCl from Na3[Ag(S2O3)2] (c=1M). Ksp[Ag(S2O3)2]3-=4*10-14, Ksp(AgCl)=1*10-10.

I calculated the ionic strength by considering 3Na+ and 1[Ag(S2O3)2]3- ion:
I=0.5(3*12+1*32)=6
log f(Ag+)=-0.5z*sqrootI=5.96*10-2
log f(S2O32-)=-0.5z*sqrootI=1.26*10-5
Ksp[Ag(S2O3)2]3-=f(Ag+)c(Ag+)f(S2O32-)c(S2O32-)
c(Ag+)=c(S2O32-)=x
x=2.31*10-4mol/dm3 of Ag+ ions.
Ksp(AgCl)=f(Ag+)c(Ag+)f(Cl-)c(Cl-) where f(Cl-)=f(Ag+)
c(Cl-)=1.22*10-4mol/dm3 but the right answer is 4.6*10-6mol/dm3. How so?
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Offline Borek

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Re: Solubility product
« Reply #1 on: June 17, 2012, 05:28:50 AM »
Quote from: Raderford on June 17, 2012, 05:13:09 AM
Ksp[Ag(S2O3)2]3-=4*10-14

How come solution is 1M if Ksp is so low? Ar you sure it is is Ksp, are you sure about the value?

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I calculated the ionic strength

Did the question ask for the thermodynamic correction?

Quote
c(Ag+)=c(S2O32-)=x

Really?
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Offline Rutherford

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Re: Solubility product
« Reply #2 on: June 17, 2012, 05:58:36 AM »
1.It is the solubility product constant of [Ag(S2O3)2]3-, didn't understand you, what is its connection with the 1M Na3[Ag(S2O3)2] solution?
2.It didn't but I thought to calculate it because of the 3Na+ ions.
3.Oh there I've mistaken:
Ksp[Ag(S2O3)2]3-=f(Ag+)c(Ag+)f(S2O32-)2c(S2O32-)2
c(Ag+)=x
c(S2O32-)=2x but again didn't get the right result.
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Offline Borek

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Re: Solubility product
« Reply #3 on: June 17, 2012, 08:24:36 AM »
Quote from: Raderford on June 17, 2012, 05:58:36 AM
1.It is the solubility product constant of [Ag(S2O3)2]3-

It is not a solubility product, solubility product requires a solid - what is solid if you have an ion?

If anything, it is a reciprocal of a complex stability constant (formation constant).
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Offline Rutherford

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Re: Solubility product
« Reply #4 on: June 17, 2012, 09:44:27 AM »
Sorry then, my mistake. How can that help? Is the concentration of Ag(S2O3)23- included in the equilibrium?
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Offline Borek

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Re: Solubility product
« Reply #5 on: June 17, 2012, 11:52:19 AM »
Find out what the stability constant is.
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Offline AWK

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Re: Solubility product
« Reply #6 on: June 17, 2012, 12:59:31 PM »
log beta for Ag(S2O3)23– = 13,46 (thermodynamic value  for I=0) - this is close to your Ksp value if you change sign at power of 10.
Calculation of activity coefficient for I~6 using your formula is irrelevant (in fact there is no reliable method of activity coefficient calculation for such I value). See IUPAC Aq_solutions program (free).
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Offline Rutherford

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Re: Solubility product
« Reply #7 on: June 17, 2012, 01:18:09 PM »
Now you confused me. I used an aproximated version of: logf=(-0.51Z2*sqrI)/(1+sqrI). I will try to solve this tomorrow without the activities.
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Offline Borek

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Re: Solubility product
« Reply #8 on: June 17, 2012, 01:48:25 PM »
As AWK wrote, Debye-Huckel theory is good for I < 0.1.
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Offline Rutherford

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Re: Solubility product
« Reply #9 on: June 18, 2012, 08:04:46 AM »
Wasn't so hard to solve when I used concentrations only. Thanks for the help.
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